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Series & Sequences
Paul Dawkins
Calculus II
i
Table of Contents
Chapter 4 : Series & Sequences
...
179
Section 4-2 : More on Sequences
...
196
Section 4-4 : Convergence/Divergence of Series
...
211
Section 4-6 : Integral Test
...
229
Section 4-8 : Alternating Series Test
...
247
Section 4-10 : Ratio Test
...
258
Section 4-12 : Strategy for Series
...
265
Section 4-14 : Power Series
...
284
Section 4-16 : Taylor Series
...
302
Section 4-18 : Binomial Series
...
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Chapter 4 : Series & Sequences
In this chapter we’ll be taking a look at sequences and (infinite) series
...
However, we also need to understand some of the basics of sequences in
order to properly deal with series
...
Series is one of those topics that many students don’t find all that useful
...
However, series do play an important role in the
field of ordinary differential equations and without series large portions of the field of partial differential
equations would not be possible
...
Most of
the applications are beyond the scope of most Calculus courses and tend to occur in classes that many
students don’t take
...
Here is a list of topics in this chapter
...
We will focus on the basic terminology, limits of sequences and
convergence of sequences in this section
...
More on Sequences – In this section we will continue examining sequences
...
We will also determine a sequence is bounded below, bounded above and/or bounded
...
We will also give many of
the basic facts, properties and ways we can use to manipulate a series
...
Convergence/Divergence of Series – In this section we will discuss in greater detail the convergence and
divergence of infinite series
...
We will also give the Divergence Test for series in this section
...
We will examine Geometric Series, Telescoping Series, and
Harmonic Series
...
The Integral Test can be used on a infinite series provided the terms of the series
are positive and decreasing
...
© 2018 Paul Dawkins
http://tutorial
...
lamar
...
In order to use either
test the terms of the infinite series must be positive
...
Alternating Series Test – In this section we will discuss using the Alternating Series Test to determine if
an infinite series converges or diverges
...
A proof of the Alternating Series Test is also given
...
Ratio Test – In this section we will discuss using the Ratio Test to determine if an infinite series
converges absolutely or diverges
...
A proof of
the Ratio Test is also given
...
The Root Test can be used on any series, but unfortunately will not always yield
a conclusive answer as to whether a series will converge absolutely or diverge
...
Strategy for Series – In this section we give a general set of guidelines for determining which test to use
in determining if an infinite series will converge or diverge
...
A summary of all the various tests, as well as conditions that must be met to use them, we discussed in
this chapter are also given in this section
...
Power Series – In this section we will give the definition of the power series as well as the definition of
the radius of convergence and interval of convergence for a power series
...
Power Series and Functions – In this section we discuss how the formula for a convergent Geometric
Series can be used to represent some functions as power series
...
However, use of this
formula does quickly illustrate how functions can be represented as a power series
...
Taylor Series – In this section we will discuss how to find the Taylor/Maclaurin Series for a function
...
We also derive some well known formulas for Taylor series of
e x , cos ( x ) and sin ( x ) around x = 0
...
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Applications of Series – In this section we will take a quick look at a couple of applications of series
...
We will also see how we can use the first few terms of a power series to
approximate a function
...
In addition, when n is not an integer an
extension to the Binomial Theorem can be used to give a power series representation of the term
...
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Section 4-1 : Sequences
Let’s start off this section with a discussion of just what a sequence is
...
The list may or may not have an infinite number of terms in
them although we will be dealing exclusively with infinite sequences in this class
...
In the notation above we need to be very careful with the subscripts
...
There is a variety of ways of denoting a sequence
...
{a1 , a2 , , an , an+1 ,}
{an }
{an }n=1
∞
In the second and third notations above an is usually given by a formula
...
First, note the difference between the
second and third notations above
...
Next, we used a starting point of
n = 1 in the third notation only so we could write one down
...
A sequence will start where ever it needs to start
...
© 2018 Paul Dawkins
http://tutorial
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∞
n + 1
(a)
2
n n =1
∞
n +1
( −1)
(b)
n
2
n = 0
(c) {bn }n =1 , where
∞
bn = nth digit of π
Solution
∞
n + 1
(a)
2
n n =1
To get the first few sequence terms here all we need to do is plug in values of n into the formula given
and we’ll get the sequence terms
...
∞
( −1)n +1
(b)
n
2
n = 0
This one is similar to the first one
...
∞
n +1
1
( −1)
1 1 1
=
−1, , − , , − ,
n
2
n =0 2 4 8 16
Note that the terms in this sequence alternate in signs
...
(c) {bn }n =1 , where
∞
bn = nth digit of π
This sequence is different from the first two in the sense that it doesn’t have a specific formula for
each term
...
Each term should be the nth digit of π
...
14159265359
The sequence is then,
{3,1, 4,1,5,9, 2, 6,5,3,5,}
In the first two parts of the previous example note that we were really treating the formulas as functions
that can only have integers plugged into them
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n +1
f (n) =
g (n)
=
n2
( −1)
n +1
2n
This is an important idea in the study of sequences (and series)
...
Before delving further into this idea however we need to get a couple more ideas out of the way
...
To graph the sequence {an } we plot the points
( n, an ) as n ranges over all possible values on a graph
...
The first few points on the graph are,
2
n n =1
3 4 5 6
, 3, , 4, , 5, ,
4 9 16 25
(1, 2 ) , 2,
The graph, for the first 30 terms of the sequence, is then,
This graph leads us to an important idea about sequences
...
We then say that zero is the limit (or
sometimes the limiting value) of the sequence and write,
n +1
=
lim an lim
=
0
n →∞
n →∞ n 2
This notation should look familiar to you
...
In fact, if you recall, we said earlier that we could think of sequences as functions in
some way and so this notation shouldn’t be too surprising
...
Working Definition of Limit
© 2018 Paul Dawkins
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We say that
lim an = L
n →∞
if we can make an as close to L as we want for all sufficiently large n
...
2
...
Again, in other words, the
value of the
an ’s get larger and larger without bound as n approaches infinity
...
We say that
lim an = −∞
n →∞
if we can make an as large and negative as we want for all sufficiently large n
...
an ’s are negative and get larger and larger without bound as n
The working definitions of the various sequence limits are nice in that they help us to visualize what the
limit actually is
...
Let’s give those before proceeding
Precise Definition of Limit
1
...
We say that lim an = ∞ if for every number M > 0 there is an integer N such that
n →∞
an > M
whenever
n>N
3
...
Note that both definitions tell us that in order for a limit to exist and have a finite value all the sequence
terms must be getting closer and closer to that finite value as n increases
...
If lim an exists and is finite we say that the sequence is convergent
...
Note that sometimes we will say the sequence
© 2018 Paul Dawkins
http://tutorial
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lamar
...
Get used to the terms “convergent” and “divergent” as we’ll be seeing them quite a bit throughout this
chapter
...
Theorem 1
Given the sequence {an } if we have a function f ( x ) such that f ( n ) = an and lim f ( x ) = L then
x →∞
lim an = L
n →∞
This theorem is basically telling us that we take the limits of sequences much like we take the limit of
functions
...
We will more often just treat the limit as if it were a limit of a function and take the limit as we
always did back in Calculus I when we were taking the limits of functions
...
Properties
If {an } and {bn } are both convergent sequences then,
1
...
lim can = c lim an
3
...
lim n n →∞ , provided lim bn ≠ 0
n →∞ b
n →∞
lim bn
n
n →∞
5
...
© 2018 Paul Dawkins
http://tutorial
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Squeeze Theorem for Sequences
If
an ≤ cn ≤ bn
for all n > N for some N and lim
=
=
an lim
bn L then lim cn = L
...
As we’ll see not all sequences can be written as functions that we can actually take the limit of
...
While we can always write these sequence
terms as a function we simply don’t know how to take the limit of a function like that
...
Theorem 2
If lim an = 0 then lim an = 0
...
This theorem is easy enough to prove so let’s do that
...
Theorem 3
∞
The sequence {r n } converges if −1 < r ≤ 1 and diverges for all other values of r
...
© 2018 Paul Dawkins
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Case 1 : r > 1
We know from Calculus I that lim r x = ∞ if r > 1 and so by Theorem 1 above we also know that
x →∞
lim r = ∞ and so the sequence diverges if r > 1
...
Case 3 : 0 < r < 1
We know from Calculus I that lim r x = 0 if 0 < r < 1 and so by Theorem 1 above we also know that
x →∞
lim r = 0 and so the sequence converges if 0 < r < 1 and in this case its limit is zero
...
Case 5 : −1 < r < 0
First let’s note that if −1 < r < 0 then 0 < r < 1 then by Case 3 above we have,
n
lim
=
rn
lim
=
r
0
n →∞
n →∞
Theorem 2 above now tells us that we must also have, lim r n = 0 and so if −1 < r < 0 the sequence
n →∞
converges and has a limit of 0
...
Recall that in order of this limit to exist the
n→∞
terms must be approaching a single value as n increases
...
So, the sequence diverges for r = −1
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Case 7 : r < −1
In this case we’re not going to go through a complete proof
...
If we do that the sequence becomes,
{r }
n ∞
n =0
{
=( −2 )
}
n ∞
n =0
=
{1, −2, 4, −8,16, −32,}n=0
∞
So, if r = −2 we get a sequence of terms whose values alternate in sign and get larger and larger and
so lim ( −2 )n doesn’t exist
...
So, the sequence diverges for r = −2
...
Let’s take a look at a couple of examples of limits of sequences
...
If the sequence converges
determine its limit
...
See the Limits At Infinity, Part I section of the Calculus I notes for a review
of this if you need to
...
© 2018 Paul Dawkins
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∞
e2n
(b)
n n =1
We will need to be careful with this one
...
The
problem is that L’Hospital’s Rule only works on functions and not on sequences
...
Let’s define
e2 x
f ( x) =
x
and note that,
f (n) =
e2n
n
Theorem 1 says that all we need to do is take the limit of the function
...
More often than not we just do L’Hospital’s Rule on the sequence terms without first converting to x’s
since the work will be identical regardless of whether we use x or n
...
∞
n
−1)
(
(c)
n
n =1
We will also need to be careful with this sequence
...
However, technically we can’t take the limit of
sequences whose terms alternate in sign, because we don’t know how to do limits of functions that
exhibit that same behavior
...
As we will see in the next section, and in later sections, our intuition can lead us
astray in these problems if we aren’t careful
...
We will need to use Theorem 2 on this problem
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( −1)
lim =
n
n →∞
1
lim
= 0
n →∞ n
n
Therefore, since the limit of the sequence terms with absolute value bars on them goes to zero we
know by Theorem 2 that,
( −1)
lim
n
n →∞
n
=0
which also means that the sequence converges to a value of zero
...
So, by Theorem 3 this sequence diverges
...
Theorem 2 only works if the limit is zero
...
The
last part of the previous example is a good example of this (and in fact this warning is the whole reason
that part is there)
...
So, be careful using this Theorem 2
...
Before moving onto the next section we need to give one more theorem that we’ll need for a proof
down the road
...
n →∞
Proof of Theorem 4
Let ε > 0
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=
Now, let N max {2 N1 , 2 N 2 + 1} and let n > N
...
n →∞
© 2018 Paul Dawkins
http://tutorial
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...
In this section we want to take a quick look
at some ideas involving sequences
...
Given any sequence {an } we have the following
...
We call the sequence increasing if
an < an +1 for every n
...
We call the sequence decreasing if
an > an +1 for every n
...
If {an } is an increasing sequence or {an } is a decreasing sequence we call it monotonic
...
If there exists a number m such that m ≤ an for every n we say the sequence is bounded
below
...
5
...
The number M is sometimes called an upper bound for the sequence
...
If the sequence is both bounded below and bounded above we call the sequence bounded
...
In other words, a sequence that increases for three terms and then decreases for the rest of
the terms is NOT a decreasing sequence! Also note that a monotonic sequence must always increase or
it must always decrease
...
We’ll make the point about lower bounds, but we could just as easily make it about
upper bounds
...
Note
however that if we find one number m to use for a lower bound then any number smaller than m will
also be a lower bound
...
In other words, there are an infinite
number of lower bounds for a sequence that is bounded below, some will be better than others
...
I don’t necessarily need the best lower bound, just a
number that will be a lower bound for the sequence
...
© 2018 Paul Dawkins
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(a) {− n 2 }
n =0
∞
(b)
{( −1)
}
n +1 ∞
n =1
∞
2
(c)
2
n n =5
Solution
∞
(a) {− n 2 }
n =0
This sequence is a decreasing sequence (and hence monotonic) because,
− n 2 > − ( n + 1)
2
for every n
...
We
can use any positive number or zero as the bound, M, however, it’s standard to choose the smallest
possible bound if we can and it’s a nice number
...
Therefore, while the sequence is bounded above it is not bounded
...
(b)
{( −1)
}
n +1 ∞
n =1
The sequence terms in this sequence alternate between 1 and -1 and so the sequence is neither an
increasing sequence or a decreasing sequence
...
The sequence is bounded however since it is bounded above by 1 and bounded below by -1
...
∞
2
(c)
2
n n =5
This sequence is a decreasing sequence (and hence monotonic) since,
2
2
>
2
2
n
( n + 1)
© 2018 Paul Dawkins
http://tutorial
...
lamar
...
Also, since the
sequence is a decreasing sequence the first sequence term will be the largest and so we can see that
the sequence will also be bounded above by
2
25
...
We can also take a quick limit and note that this sequence converges and its limit is zero
...
As we noted in the previous section our intuition can often
lead us astray with some of the concepts we’ll be looking at in this chapter
...
∞
n
(a)
n + 1 n =1
∞
n3
(b) 4
n + 10000 n =0
Solution
∞
n
(a)
n + 1 n =1
We’ll start with the bounded part of this example first and then come back and deal with the
increasing/decreasing question since that is where students often make mistakes with this type of
sequence
...
The sequence is therefore bounded
below by zero
...
The sequence is then bounded above by one
...
Now let’s think about the monotonic question
...
This will not
always be the case and in this case we would be wrong
...
To determine the increasing/decreasing nature of this sequence we will need to resort to Calculus I
techniques
...
=
f ( x)
x
=
f ′( x)
x +1
1
( x + 1)
2
We can see that the first derivative is always positive and so from Calculus I we know that the
function must then be an increasing function
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Therefore because n < n + 1 and f ( x ) is increasing we can also say that,
a=
n
n
=
n +1
f ( n ) < f ( n + 1=
)
n +1
= an +1
n +1+1
an < an +1
⇒
In other words, the sequence must be increasing
...
Since the sequence is increasing the first term in the sequence must be the
1
smallest term and so since we are starting at n = 1 we could also use a lower bound of 2 for this
sequence
...
Some are better than others however
...
Before moving on to the next part there is a natural question that many students will have at this
point
...
First, notice that, as with the previous part, the sequence terms are all positive and will all be less
than one (since the numerator is guaranteed to be less than the denominator) and so the sequence is
bounded
...
As with the last problem, many students
will look at the exponents in the numerator and denominator and determine based on that that
sequence terms must decrease
...
Let’s take a look at the first few terms to see this
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1
a1 =
≈ 0
...
005678
10081
1
a5 =
≈ 0
...
02766
12401
729
a9 =
≈ 0
...
0007987
1252
4
a4 =
≈ 0
...
019122
1412
32
a8 =
≈ 0
...
05
a10 =
=
20
The first 10 terms of this sequence are all increasing and so clearly the sequence can’t be a decreasing
sequence
...
Now, we can’t make another common mistake and assume that because the first few terms increase
then whole sequence must also increase
...
This sequence is neither decreasing or increasing
...
In this case we’ll need the following function and its derivative
...
1607, x = − 4 30000 ≈ −13
...
By plugging in some test values of x we can quickly determine that the derivative
is positive for
0 < x < 4 30000 ≈ 13
...
Likewise, we
can see that the derivative is negative for
decreasing in this range
...
16 and so the function will be
So, our sequence will be increasing for 0 ≤ n ≤ 13 and decreasing for n ≥ 13
...
Finally, note that this sequence will also converge and has a limit of zero
...
Our
intuition will often not be sufficient to get the correct answer and we can NEVER make assumptions
© 2018 Paul Dawkins
http://tutorial
...
lamar
...
As the last part has shown there are
sequences which will increase or decrease for a few terms and then change direction after that
...
In other words, there is no
“magical” value of n for which all we have to do is check up to that point and then we’ll know what the
whole sequence will do
...
In this case
increasing n only changed (in fact increased) the denominator and so we were able to determine the
behavior of the sequence based on that
...
In cases like
this there is no way to determine which increase will “win out” and cause the sequence terms to
increase or decrease and so we need to resort to Calculus I techniques to answer the question
...
Theorem
If {an } is bounded and monotonic then {an } is convergent
...
It does not say that if a sequence is not bounded and/or not
monotonic that it is divergent
...
The sequence in that example was
not monotonic but it does converge
...
If {an } is bounded above and
increasing then it converges and likewise if {an } is bounded below and decreasing then it converges
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Section 4-3 : Series - The Basics
In this section we will introduce the topic that we will be discussing for the rest of this chapter
...
So just what is an infinite series? Well, let’s start with a sequence {an }n =1 (note
∞
the n = 1 is for convenience, it can be anything) and define the following,
s1 = a1
s=
a1 + a2
2
s3 = a1 + a2 + a3
s4 = a1 + a2 + a3 + a4
n
sn = a1 + a2 + a3 + a4 + + an = ∑ ai
i =1
The sn are called partial sums and notice that they will form a sequence, {sn }n =1
...
The most common names are :
series notation, summation notation, and sigma notation
...
If you need a quick refresher on summation notation see the review of summation
notation in the Calculus I notes
...
We want to take a look at the limit of the sequence of partial sums, {sn }n =1
...
Had our original sequence started at 2 then our infinite series
∞
would also have started at 2
...
It is important to note that
∞
∑a
i =1
i
is really nothing more than a convenient notation for lim
n →∞
n
∑a
i =1
i
so we
do not need to keep writing the limit down
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If the sequence of partial sums, {sn }n =1 , is convergent and its limit is finite then we also call the infinite
∞
series,
∞
∑ a convergent and if the sequence of partial sums is divergent then the infinite series is also
i =1
i
called divergent
...
This implies that an infinite series is just an infinite sum of
terms and as we’ll see in the next section this is not really true for many series
...
This section is going to be devoted mostly to notational issues as well as making sure we can do some
basic manipulations with infinite series so we are ready for them when we need to be able to deal with
them in later sections
...
If we
ever need to work with both infinite and finite series we’ll be more careful with terminology, but in most
sections we’ll be dealing exclusively with infinite series and so we’ll just call them series
...
So for example the following series are all the same
...
∞
3
∞
3
∞
= ∑
= ∑
∑
i +1
k +1
n
2
2
k 0=
n 0
=i 0=
3
+1
2
etc
...
So far we’ve used n = 0 and n = 1 but the index could have
an to represent an infinite series in which the starting
started anywhere
...
When we drop the initial value of the index we’ll also drop the
infinity from the top so don’t forget that it is still technically there
...
In these facts/theorems the starting point of the series will not affect the
result and so to simplify the notation and to avoid giving the impression that the starting point is
important we will drop the index from the notation
...
Note however, that if we do put an initial value of the index on a series in a fact/theorem it is there
because it really does need to be there
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Now that some of the notational issues are out of the way we need to start thinking about various ways
that we can manipulate series
...
We have the following properties
...
∑ ca
n
∑b
n
n
are both convergent series then,
, where c is any number, is also convergent and
∑ ca
n
2
...
The first property is simply telling us that we can always factor a multiplicative constant out of an
infinite series and again recall that if we don’t put in an initial value of the index that the series can start
at any value
...
The second property says that if we add/subtract series all we really need to do is add/subtract the
series terms
...
Before we move on to a different topic let’s discuss multiplication of series briefly
...
( 2 + x ) ( 3 − 5 x + x 2 ) =6 − 7 x − 3x 2 + x3
Yeah, it was just the multiplication of two polynomials
...
In
doing the multiplication we didn’t just multiply the constant terms, then the x terms, etc
...
Multiplying infinite series (even though we said we can’t think of an infinite series as an infinite sum)
needs to be done in the same manner
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To do this multiplication we would have to distribute the a0 through the second term, distribute the a1
through, etc then combine like terms
...
n
∞ ∞ ∞
where
=
a
b
c
c
=
ai bn −i
∑
n
∑ n ∑ n ∑ n
=
i =0
n 0=
n 0 =
n0
We also can’t say a lot about the convergence of the product
...
The reality is that multiplication of series is a
somewhat difficult process and in general is avoided if possible
...
Until then, don’t worry about multiplying series
...
To be honest this is not a
topic that we’ll see all that often in this course
...
Despite the fact that we won’t use it much in this course doesn’t mean
however that it isn’t used often in other classes where you might run across series
...
The basic idea behind index shifts is to start a series at a different value for whatever the reason (and
yes, there are legitimate reasons for doing that)
...
This means that we can’t just change the n = 2 to n = 0 as this would add in two
new terms to the series and thus change its value
...
We’ll start by defining a new index, say i, as
follows,
i= n − 2
Now, when n = 2 , we will get i = 0
...
Next, we can solve this for n to get,
n= i + 2
We can now completely rewrite the series in terms of the index i instead of the index n simply by
plugging in our equation for n in terms of i
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n+5 ∞ n+7
= ∑ n+2
∑
2n
n 2=
n 0 2
=
∞
To convince yourselves that these really are the same summation let’s write out the first couple of terms
for each of them,
n+5
=
n
n=2 2
∞
n+7
=
∑
n+2
n =0 2
∞
∑
7 8 9 10
+ + + +
2 2 23 2 4 25
7 8 9 10
+ + + +
2 2 23 2 4 25
So, sure enough the two series do have exactly the same terms
...
The method given above is the technically correct
way of doing an index shift
...
This will always work in this manner
...
Likewise, if we increase the initial
value of the index by a set amount, then all the n’s in the series term will decrease by the same amount
...
Example 1 Perform the following index shifts
...
n =1
(b) Write
n2
as a series that starts at n = 3
...
∞
=
∑ ar n−1
∞
ar ( + )−
=
∑
n 1=
n 0
=
n 1 1
∞
∑ ar
n
n 0
=
(b) For this problem we want to increase the initial value by 2 and so all the n’s in the series term
must decrease by 2
...
This final topic is really more about alternate ways
to write series when the situation requires it
...
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Let’s start with the following series and note that the n = 1 starting point is only for convenience since
we need to start the series somewhere
...
We could have stripped out more terms if we wanted to
...
∞
∞
∑ an = a1 + a2 + ∑ an
=
n 1=
n 3
∞
∞
∑ an = a1 + a2 + a3 + a4 + ∑ an
=
n 1=
n 5
Being able to strip out terms will, on occasion, simplify our work or allow us to reuse a prior result so it’s
an important idea to remember
...
In general, we can write a series as follows,
∞
an
∑=
n= 1
N
∑ an +
n= 1
∞
∑
n= N +1
an
We’ll leave this section with an important warning about terminology
...
So, once again, a sequence is a list of numbers while a series is a single number, provided it makes sense
to even compute the series
...
However, since they are different beasts this just won’t work
...
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...
Before worrying about convergence and divergence of a series we wanted
to make sure that we’ve started to get comfortable with the notation involved in series and some of the
various manipulations of series that we will, on occasion, need to be able to do
...
So, it is now time to start talking about the convergence and divergence of a series as this will be a topic
that we’ll be dealing with to one extent or another in almost all of the remaining sections of this
chapter
...
We’ll start with a sequence {an }n =1 and again note that we’re starting the sequence at n = 1 only for
∞
the sake of convenience and it can, in fact, be anything
...
∞
An infinite series, or just series here since almost every series that we’ll be looking at will be an infinite
series, is then the limit of the partial sums
...
We do, however, always need to remind ourselves
that we really do have a limit there!
If the sequence of partial sums is a convergent sequence (i
...
its limit exists and is finite) then the series
is also called convergent and in this case if lim sn = s then,
n →∞
∞
∑a
i =1
i
= s
...
e
...
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...
Example 1 Determine if the following series is convergent or divergent
...
∞
∑n
n =1
Solution
To determine if the series is convergent we first need to get our hands on a formula for the general
term in the sequence of partial sums
...
So, to determine if the series is convergent we will first need to see if the sequence of partial sums,
∞
n ( n + 1)
2 n =1
is convergent or divergent
...
The limit of the sequence terms is,
n ( n + 1)
= ∞
n →∞
2
lim
Therefore, the sequence of partial sums diverges to ∞ and so the series also diverges
...
In general finding a formula for the general term in the sequence of partial
sums is a very difficult process
...
This also means that
we’ll not be doing much work with the value of series since in order to get the value we’ll also need to
know the general formula for the partial sums
...
Also, the remaining examples we’ll be looking at in this section
will lead us to a very important fact about the convergence of series
...
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...
If it converges determine its
sum
...
However, in this section we are more interested in the general
idea of convergence and divergence and so we’ll put off discussing the process for finding the formula
until the next section
...
If it converges determine its
sum
...
Let’s just write down the first few partial sums
...
So, it looks like the sequence of partial sums is,
{sn }n=0 = {1, 0,1, 0,1, 0,1, 0,1,}
∞
and this sequence diverges since lim sn doesn’t exist
...
n →∞
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...
If it converges determine its
sum
...
sn
=
1
n
∑ 3=
i −1
i =1
3
1
1 − n
2 3
Again, do not worry about knowing this formula
...
In this case the limit of the sequence of partial sums is,
3
1 3
lim s=
lim 1 − n=
n
n →∞
n →∞ 2
3 2
The sequence of partial sums is convergent and so the series will also be convergent
...
There is only going to be one type of series where you will need to determine this formula
and the process in that case isn’t too bad
...
So, we’ve determined the convergence of four series now
...
Let’s go back and examine the series terms for each of these
...
lim n = ∞
this series diverged
1
=0
n →∞ n − 1
this series converged
lim ( −1) doesn't exist
this series diverged
1
=0
n →∞ 3n −1
this series converged
n →∞
lim
2
n
n →∞
lim
Notice that for the two series that converged the series term itself was zero in the limit
...
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...
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...
n →∞
Proof
First let’s suppose that the series starts at n = 1
...
Then the partial sums are,
n −1
n
sn −1 = ∑ ai = a1 + a2 + a3 + a4 + + an −1
sn = ∑ ai = a1 + a2 + a3 + a4 + + an −1 + an
i 1 =i 1
Next, we can use these two partial sums to write,
a=
sn − sn −1
n
Now because we know that
∑a
n
is convergent we also know that the sequence {sn }n =1 is also
∞
convergent and that lim sn = s for some finite value s
...
n →∞
We now have,
lim an = lim ( sn − sn −1 ) = lim sn − lim sn −1 = s − s = 0
n →∞
n →∞
n →∞
n →∞
Be careful to not misuse this theorem! This theorem gives us a requirement for convergence but not a
guarantee of convergence
...
If lim an = 0 the series may
n →∞
actually diverge! Consider the following two series
...
The first series diverges
...
Again, as noted above, all this theorem does is give us a requirement for a series to converge
...
If the series terms do not go to
zero in the limit then there is no way the series can converge since this would violate the theorem
...
Divergence Test
If lim an ≠ 0 then
n →∞
∑a
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n
will diverge
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Again, do NOT misuse this test
...
If the series terms do happen to go to zero the series may or may not
converge! Again, recall the following two series,
∞
1
∑n
diverges
n =1
∞
1
∑n
n =1
2
converges
One of the more common mistakes that students make when they first get into series is to assume that
an will converge
...
Example 5 Determine if the following series is convergent or divergent
...
If it’s clear that the terms don’t go to zero
use the Divergence Test and be done with the problem
...
4n 2 − n 3
1
=
− ≠0
3
n →∞ 10 + 2n
2
lim
The limit of the series terms isn’t zero and so by the Divergence Test the series diverges
...
You will need to keep track of all these tests, the conditions under which they can be
used and their conclusions all in one place so you can quickly refer back to them as you need to
...
As we saw in the
previous section if
∑ an and
∑ bn are both convergent series then so are
∑ can and
Furthermore, these series will have the following sums or values
...
∞
∑ ( an ± bn )= ∑ an ± ∑ bn
=
n k
=
n k=
n k
We’ll see an example of this in the next section after we get a few more examples under our belt
...
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...
In the first case if
an is divergent then
can will also be divergent (provided c isn’t zero of course) since multiplying
∑
∑
a series that is infinite in value or doesn’t have a value by a finite value (i
...
c) won’t change the fact that
an and bn be
the series has an infinite or no value
...
Now, since the main topic of this section is the convergence of a series we should mention a stronger
an is said to converge absolutely if
an also converges
...
A series
∑
∑
convergence is stronger than convergence in the sense that a series that is absolutely convergent will
also be convergent, but a series that is convergent may or may not be absolutely convergent
...
At this point we don’t really have the tools at hand to properly investigate this topic in detail nor do we
have the tools in hand to determine if a series is absolutely convergent or not
...
When we finally have the tools in hand to discuss this topic in more
detail we will revisit it
...
The idea is mentioned here only because we
were already discussing convergence in this section and it ties into the last topic that we want to discuss
in this section
...
It’s now time to briefly discuss this
...
A rearrangement of a series is exactly what it
might sound like, it is the same series with the terms rearranged into a different order
...
∞
∑a
n
n =1
= a1 + a2 + a3 + a4 + a5 + a6 + a7 +
A rearrangement of this series is,
∞
∑a
n =1
n
= a2 + a1 + a3 + a14 + a5 + a9 + a4 +
The issue we need to discuss here is that for some series each of these arrangements of terms can have
different values despite the fact that they are using exactly the same terms
...
It can be shown that,
∞
∑
n =1
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( −1)
n
n +1
1 1 1 1 1 1 1
=1 − + − + − + − + =ln 2
2 3 4 5 6 7 8
(1)
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...
So, let’s multiply this by 12 to get,
1 1 1 1 1 1 1 1
1
− + − + − + − + =ln 2
2 4 6 8 10 12 14 16
2
(2)
Now, let’s add in a zero between each term as follows
...
Repeating terms in a series will not
affect its limit however and so both (2) and (3) will be the same
...
The values however are definitely different despite the fact that the
terms are the same
...
This is a very real result and we’ve not made any logic
mistakes/errors
...
Facts
Given the series
1
...
∑a
of ∑ a
2
...
Again, we do not have the tools in hand yet to determine if a series is absolutely convergent and so
don’t worry about this at this point
...
There are times when we can (i
...
the
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...
e
...
As a final note, the fact above tells us that the series,
∞
∑
n =1
( −1)
n +1
n
must be conditionally convergent since two rearrangements gave two separate values of this series
...
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...
Actually, special may not be the
correct term
...
The third type is
divergent and so won’t have a value to worry about
...
So, let’s get started
...
If we start with the first form it can be shown that the partial sums are,
a (1 − r n )
a
ar n
=
sn
=
−
1− r
1− r 1− r
The series will converge provided the partial sums form a convergent sequence, so let’s take the limit of
the partial sums
...
However, note that we can’t let r = 1 since this will give division by zero
...
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Therefore, a geometric series will converge if −1 < r < 1 , which is usually written r < 1 , its value is,
∞
∞
ar
=
ar
∑=
∑
n −1
n
=
n 1=
n 0
a
1− r
Note that in using this formula we’ll need to make sure that we are in the correct form
...
Likewise, if the series starts at n = 1
then the exponent on the r must be n − 1
...
If they converge give the value of
the series
...
However, notice that both parts of the series
term are numbers raised to a power
...
We will just need to decide which form is the correct form
...
It will be fairly easy to get this into the correct form
...
One of the n’s
in the exponent has a negative in front of it and that can’t be there in the geometric form
...
∞
=
∑ 9− n+2 4n+1
∞
4n +1
∑
n−2
n 1 9
=
9−( n − ) 4n +1
∑=
2
n 1=
n 1
=
∞
Now let’s get the correct exponent on each of the numbers
...
∞
4n +1 ∞ 4n −142
=
∑
∑
9n − 2 n 1 9n −19−1
=
n 1=
n + 2 n +1
4
∑ 9−=
=
n 1
∞
Now, rewrite the term a little
...
Therefore, since r < 1 we know the series
will converge and its value will be,
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...
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...
In this case the
series starts at n = 0 so we’ll need the exponents to be n on the terms
...
Also note that r ≥ 1
and so this series diverges
...
We can now do some examples
...
(a)
∞
∑9
− n+2
4n +1
n =0
(b)
∞
∑9
− n+2
4n +1
n =3
Solution
(a)
∞
∑9
− n+2
4n +1
n =0
In this case we could just acknowledge that this is a geometric series that starts at n = 0 and so we
could put it into the correct form and be done with it
...
Let’s notice that if we strip out the first term from this series we arrive at,
∞
∞
∞
92 41 + ∑ 9− n + 2 4n +1 =
324 + ∑ 9− n + 2 4n +1
∑ 9− n+2 4n+1 =
=
n 0 =
n 1=
n 1
From the previous example we know the value of the new series that arises here and so the value of
the series in this example is,
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...
However, we can start with the series used in the previous example and strip terms out of it
to get the series in this example
...
We will strip out the first two terms from the
series we looked at in the previous example
...
∞
∑9
− n+2
∞
∑9
4n +1 =
− n+2
1296
256
− 208 =
5
5
4n +1 − 208 =
=
n 3=
n 1
Notice that we didn’t discuss the convergence of either of the series in the above example
...
Consider the following series written in two separate ways (i
...
we stripped out a couple of terms from
it)
...
This means that it’s got a finite value and
adding three finite terms onto this will not change that fact
...
Likewise, suppose that
∞
∞
∑a
n =0
n
is also finite and so is
is convergent
...
n
∞
∑a
n =3
n
...
The
difference of a few terms one way or the other will not change the convergence of a series
...
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...
In this portion we are going to look
at a series that is called a telescoping series
...
Example 3 Determine if the following series converges or diverges
...
∞
1
∑
2
n = 0 n + 3n + 2
Solution
We first need the partial sums for this series
...
By now you should be fairly adept at this since
we spent a fair amount of time doing partial fractions back in the Integration Techniques chapter
...
So, what does this do for us? Well, let’s start writing out the terms of the general partial sum for this
series using the partial fraction form
...
This is the origin of the name
telescoping series
...
1
lim sn =lim 1 −
=1
n →∞
n →∞
n+2
The sequence of partial sums is convergent and so the series is convergent and has a value of
∞
∑n
n =0
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2
1
=1
+ 3n + 2
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...
Consider the following example
...
If it converges find its value
...
The partial sums are,
1 n 1
1
−
∑
2 i 1 i +1 i + 3
=i 1 =
sn=
=
=
n
1
2
1
2
∑ i + 1 − i + 3 =
1
1 1 1 1 1 1 1
1 1
1
−
− + − + − + + −
+
2 2 4 3 5 4 6
n n + 2 n + 1 n + 3
1 1 1
1
1
+ −
−
2 2 3 n + 2 n + 3
In this case instead of successive terms canceling a term will cancel with a term that is farther down
the list
...
Notice as well that in order
to help with the work a little we factored the 12 out of the series
...
There is no test that will tell us that we’ve got a telescoping
series right off the bat
...
The following series, for example, is not a
telescoping series despite the fact that we can partial fraction the series terms
...
Next, we need to go back and address an issue that was first raised in the previous section
...
Now that we have a
few more series in hand let’s work a quick example showing that
...
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Example 5 Determine the value of the following series
...
∞
∑ n
=
n 1
∞
∞
4
4
− n + 2 n +1
−
9
4
=
−
9− n + 2 4n +1
∑
∑
2
2
+ 4n + 3 =
n + 4n + 3 n 1
n 1=
∞
1
−
9− n + 2 4n +1
∑
2
n + 4n + 3 n 1
n 1=
=
∞
= 4∑
5 1296
= 4 −
5
12
3863
= −
15
We didn’t discuss the convergence of this series because it was the sum of two convergent series and
that guaranteed that the original series would also be convergent
...
Here is the harmonic series
...
The harmonic series is divergent and we’ll need to wait until the next section to show that
...
We’re also going to use the harmonic series to illustrate a couple of ideas
about divergent series that we’ve already discussed for convergent series
...
Example 6 Show that each of the following series are divergent
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∞
5
∞
1
∑ n = 5∑ n
n 1=
n 1
=
Now,
∞
1
∑n
is divergent and so five times this will still not be a finite number and so the series has to
n =1
be divergent
...
(b)
∞
1
∑n
n=4
In this case we’ll start with the harmonic series and strip out the first three terms
...
This subtraction will not
change the divergence of the series
...
Therefore, this series is divergent
...
So, some general rules about the convergence/divergence of a series are now in order
...
These are nice ideas
to keep in mind
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Section 4-6 : Integral Test
The last topic that we discussed in the previous section was the harmonic series
...
It is now time to prove that statement
...
So, we will be trying to prove that the harmonic series,
∞
1
∑n
n =1
diverges
...
Let’s start off by asking what the area
under f ( x ) =
1
on the interval [1, ∞ )
...
So, just how does that help us to prove that the harmonic series diverges? Well, recall that we can
always estimate the area by breaking up the interval into segments and then sketching in rectangles and
using the sum of the area all of the rectangles as an estimate of the actual area
...
We will break up the interval into subintervals of width 1 and we’ll take the function value at the left
endpoint as the height of the rectangle
...
So, the area under the curve is approximately,
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...
First, each of the rectangles overestimates the
actual area and secondly the formula for the area is exactly the harmonic series!
Putting these two facts together gives the following,
1 ⌠∞1
>
∞
dx =
⌡1 x
n =1 n
∞
A≈∑
Notice that this tells us that we must have,
∞
1
1
>
∞
⇒
=∞
∑
∑
n
=
n 1=
n 1 n
∞
Since we can’t really be larger than infinity the harmonic series must also be infinite in value
...
So, we’ve managed to relate a series to an improper integral that we could compute and it turns out
that the improper integral and the series have exactly the same convergence
...
When discussing the Divergence Test we
made the claim that
∞
1
∑n
n =1
2
converges
...
We will try to relate this to the area under f ( x ) =
Improper Integral section we know that,
1
is on the interval [1, ∞ )
...
We will once again try to estimate the area under this curve
...
Here is a sketch of this case,
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...
Notice however that the only difference is that we’re
missing the first term
...
With this
series however, this isn’t quite enough
...
So, let’s do a little more work
...
In
other words,
1 n +1 1
< ∑ = sn +1
i2 i 1 i2
=i 1 =
n
sn = ∑
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...
Therefore, the partial sums
form an increasing (and hence monotonic) sequence
...
In the second section on Sequences we gave a theorem that stated that a bounded and monotonic
sequence was guaranteed to be convergent
...
So, who cares right? Well recall that this means that the series must then also be
convergent!
So, once again we were able to relate a series to an improper integral (that we could compute) and the
series and the integral had the same convergence
...
Luckily for us we don’t need to do all this work every time
...
Integral Test
Suppose that f ( x ) is a continuous, positive and decreasing function on the interval [ k , ∞ ) and that
f ( n ) = an then,
1
...
If
∫
∫
∞
k
∞
k
f ( x ) dx is convergent so is
f ( x ) dx is divergent so is
∞
∑a
n
n=k
∞
∑a
n=k
n
...
A formal proof of this test can be found at the end of this section
...
First, the lower limit on the improper
integral must be the same value that starts the series
...
All that’s really required is that eventually the function is decreasing and positive
...
To see why this is true let’s suppose
that the series terms increase and or are negative in the range k ≤ n ≤ N and then decrease and are
positive for n ≥ N + 1
...
So, the original series will be convergent/divergent only if the second infinite series on the
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...
A similar argument can be made using the improper integral as well
...
In practice however, we only need to make sure that the function/series is
eventually a decreasing and positive function/series
...
There is one more very important point that must be made about this test
...
It will only give the convergence/divergence of the series
...
No value
...
All that the test gave us was that,
∞
1
∑n
n =1
2
<2
So, we got an upper bound on the value of the series, but not an actual value for the series
...
In a later section we look at estimating values of series, but even in that section
still won’t actually be getting values of series
...
1 π2
=
= 1
...
< 2
∑
2
6
n =1 n
∞
Let’s work a couple of examples
...
∞
1
∑
n = 2 n ln n
Solution
In this case the function we’ll use is,
f ( x) =
1
x ln x
This function is clearly positive and if we make x larger the denominator will get larger and so the
function is also decreasing
...
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...
Example 2 Determine if the following series is convergent or divergent
...
Now we need to check that the
function is decreasing
...
We can use our Calculus I knowledge to help us however
...
Picking a couple of test
points we can see that the function is increasing on the interval 0,
1
2
)
1
2
1
2
...
Therefore, eventually the function will be decreasing and that’s all that’s required for us to
use the Integral Test
...
We can use the Integral Test to get the following fact/test for some series
...
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Fact (The p –series Test)
If k > 0 then
∞
1
∑n
n=k
p
converges if p > 1 and diverges if p ≤ 1
...
This fact follows directly from the Integral Test and a similar fact we saw in the Improper Integral
section
...
Using the p-series test makes it very easy to determine the convergence of some series
...
∞
1
(a) ∑ 7
n=4 n
∞
1
(b) ∑
n
n =1
Solution
(a) In this case p= 7 > 1 and so by this fact the series is convergent
...
The last thing that we’ll do in this section is give a quick proof of the Integral Test
...
Proof of Integral Test
First, for the sake of the proof we’ll be working with the series
∞
∑a
n =1
n
...
Another way of dealing with the n = k is we could do an index shift and start the series at n = 1 and
then do the Integral Test
...
Also note that while we allowed for the first few terms of the series to increase and/or be negative in
working problems this proof does require that all the terms be decreasing and positive
...
This gives the
following figure
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Now, note that,
=
f ( 2 ) a=
f ( 3) a3
2
=
f ( n ) an
The approximate area is then,
A ≈ (1) f ( 2 ) + (1) f ( 3) + + (1) f ( n ) = a2 + a3 + an
and we know that this underestimates the actual area so,
n
∑a
i =2
Now, let’s suppose that
∫
∞
1
i
n
= a2 + a3 + an < ∫ f ( x ) dx
1
f ( x ) dx is convergent and so
∫
∞
1
f ( x ) dx must have a finite value
...
However, that’s easy enough to deal with
...
n
∑a
i =1
i
are
Next, because the terms are positive we also know that,
© 2018 Paul Dawkins
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...
lamar
...
So, we now know that the sequence of
∞
partial sums {sn }n =1 converges and hence our series
∞
∞
∑a
n =1
is convergent
...
The second part is somewhat easier
...
In this case the area is approximately,
A ≈ (1) f (1) + (1) f ( 2 ) + + (1) f ( n − 1) = a1 + a2 + an −1
Since we know this overestimates the area we also then know that,
n −1
sn −1 = ∑ ai = a1 + a2 + an −1 > ∫
i =1
Now, suppose that
∫
∞
1
n −1
1
f ( x ) dx
f ( x ) dx is divergent
...
However, because n − 1 → ∞ as n → ∞ we also know that
∫
n −1
1
f ( x ) dx → ∞
...
This in turn
tells us that sn → ∞ as n → ∞
...
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So, we now know that the sequence of partial sums, {sn }n =1 , is a divergent sequence and so
∞
a divergent series
...
If they are not then the test doesn’t work
...
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...
While the integral test is a nice test, it does force us to do improper integrals
which aren’t always easy and, in some cases, may be impossible to determine the convergence of
...
∞
∑3
n
n =0
1
+n
In order to use the Integral Test we would have to integrate
∞
1
⌠
dx
x
⌡0 3 + x
and we’re not even sure if it’s possible to do this integral
...
First, let’s note that the series terms are positive
...
Next let’s note that we must have x > 0 since we are integrating on the interval 0 ≤ x < ∞
...
So, if we drop the x from the
denominator the denominator will get smaller and hence the whole fraction will get larger
...
n +1
1
1
<
= sn +1
∑
i
i
3 +i i 0 3 +i
0=
n
sn = ∑
=i
Then since,
© 2018 Paul Dawkins
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...
lamar
...
Then from the second section on
sequences we know that a monotonic and bounded sequence is also convergent
...
This means that the series
itself,
∞
∑3
n
n =0
is also convergent
...
Then since the original series terms were positive (very
important) this meant that the original series was also convergent
...
In this case the
original series would have to take a value larger than the new series
...
This means that the original series must also be infinite and hence
divergent
...
Comparison Test
Suppose that we have two series
Then,
1
...
n
∑b
If ∑ a
∑a
n
and
∑b
n
with an , bn ≥ 0 for all n and an ≤ bn for all n
...
is divergent then so is ∑ b
...
Then if the larger series is convergent the smaller series must also
be convergent
...
Note as well that in order to apply this test we need both series to start at the same place
...
Do not misuse this test
...
The larger series may still diverge
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Recall that we had a similar test for improper integrals back when we were looking at integration
techniques
...
Note as well that the requirement that an , bn ≥ 0 and an ≤ bn really only need to be true eventually
...
As long as we eventually reach a point where an , bn ≥ 0 and an ≤ bn for all sufficiently
large n the test will work
...
If we then look
an (the same thing could be done for bn ) we get,
at
∑
∞
an
∑=
n= k
∑
∑a
N
n= k
n +
∞
∑
n= N +1
an
The first series is nothing more than a finite sum (no matter how large N is) of finite terms and so will be
finite
...
Let’s take a look at some examples
...
∞
n
∑
2
2
n =1 n − cos ( n )
Solution
Since the cosine term in the denominator doesn’t get too large we can assume that the series terms
will behave like,
n 1
=
n2 n
which, as a series, will diverge
...
Recall that from the comparison test with improper integrals that we determined that we can make a
fraction smaller by either making the numerator smaller or the denominator larger
...
So, if we drop the cosine term we will in fact be
making the denominator larger since we will no longer be subtracting off a positive quantity
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1
∞
∑n
n =1
diverges (it’s harmonic or the p-series test) by the Comparison Test our original series must also
diverge
...
∞
e− n
∑
2
n =1 n + cos ( n )
Solution
This example looks somewhat similar to the first one but we are going to have to be careful with it as
there are some significant differences
...
Therefore, the temptation at this point
is to focus in on the n in the denominator and think that because it is just an n the series will diverge
...
In this example, however,
we also have an exponential in the numerator that is going to zero very fast
...
So, let’s guess that this series will converge and we’ll need to find a larger series that will also
converge
...
This will, in turn, make the denominator smaller and so the term will get
larger or,
e− n
e− n
≤
n + cos 2 ( n ) n
Next, we know that n ≥ 1 and so if we replace the n in the denominator with its smallest possible
value (i
...
1) the term will again get larger
...
−
n =1
−x
We can notice that f ( x ) = e is always positive and it is also decreasing (you can verify that
correct?) and so we can use the Integral Test on this series
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∫
∞
1
e − x dx = lim ∫ e − x dx = lim ( −e − x ) = lim ( −e −t + e −1 )= e −1
t
t
t →∞ 1
t →∞
1
t →∞
Okay, we now know that the integral is convergent and so the series
∞
∑e n
−
must also be convergent
...
∞
∑e n
−
is larger than the original series we know that the original series must also
n =1
With each of the previous examples we saw that we can’t always just focus in on the denominator when
making a guess about the convergence of a series
...
We also saw in the previous example that, unlike most of the examples of the comparison test that
we’ve done (or will do) both in this section and in the Comparison Test for Improper Integrals, that it
won’t always be the denominator that is driving the convergence or divergence
...
One of the more common mistakes is to just focus in on the denominator and make a guess based just
on that
...
Let’s work another example of the comparison test before we move on to a different topic
...
∞
n2 + 2
∑
4
n =1 n + 5
Solution
In this case the “+2” and the “+5” don’t really add anything to the series and so the series terms
should behave pretty much like
n2 1
=
n4 n2
which will converge as a series
...
This means that we’ll either have to make the numerator larger or the denominator smaller
...
Doing this gives,
n2 + 2 n2 + 2
<
n4 + 5
n4
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...
lamar
...
However, this is actually the furthest that we need to go
...
n2 + 2 ∞ n2 ∞ 2
=
∑
∑ n4 + ∑
4
n 4=n 1 =
n 1
n 1 n
=
∞
∞
1
2
= ∑ 2 +∑ 4
n
n 1=
n 1 n
=
∞
As shown, we can write the series as a sum of two series and both of these series are convergent by
the p-series test
...
Recall that the sum of two convergent series will also be convergent
...
The comparison test is a nice test that allows us to do problems that either we couldn’t have done with
the integral test or at the best would have been very difficult to do with the integral test
...
Consider the following series
...
The original series converged because
the 3n gets very large very fast and will be significantly larger than the n
...
The fact that we are now subtracting the n off instead
of adding the n on really shouldn’t change the convergence
...
So, we would expect this series to converge
...
To use the comparison test on this series we would need to find a larger series that we could easily
determine the convergence of
...
If we drop
the n we will make the denominator larger (since the n was subtracted off) and so the fraction will get
smaller and just like when we looked at the comparison test for improper integrals knowing that the
smaller of two series converges does not mean that the larger of the two will also converge
...
The following
variant of the comparison test will allow us to determine the convergence of this series
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Limit Comparison Test
Suppose that we have two series
∑a
n
and
∑b
with an ≥ 0, bn > 0 for all n
...
e
...
e
...
The proof of this test is at the end of this section
...
To see why this is, consider the following two definitions
...
an
1
1
1
=
c lim
=
lim= =
n →∞ b
n →∞ b
b
n
n
lim n c
n →∞ a
an
n
In other words, if c is positive and finite then so is c and if c is positive and finite then so is c
...
Both definitions will give the same results from the test
so don’t worry about which series terms should be in the numerator and which should be in the
denominator
...
Also, this really is a comparison test in some ways
...
If c = 0 or c = ∞ we can’t say this and so the test fails to give any
information
...
Let’s see how this test works
...
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Example 4 Determine if the following series converges or diverges
...
On top
of that we will need to choose the new series in such a way as to give us an easy limit to compute for
c
...
We know that this series converges and there is a chance that since both series
have the 3n in it the limit won’t be too bad
...
1 3n − n
n →∞ 3n
1
n
= lim1 − n
n →∞
3
c = lim
Now, we’ll need to use L’Hospital’s Rule on the second term in order to actually evaluate this limit
...
n
Example 5 Determine if the following series converges or diverges
...
So, the terms in this series should behave as,
n2
n2
1
= =
7
1
3 7
n
n 3 n3
and as a series this will diverge by the p-series test
...
© 2018 Paul Dawkins
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...
lamar
...
1
Finally, to see why we need c to be positive and finite (i
...
c ≠ 0 and c ≠ ∞ ) consider the following two
series
...
Now compute each of the following limits
...
Clearly, both series do not have the same convergence
...
To see this consider the series,
∞
1
∑n
∞
1
∑n
3
=
n 1=
n 1
2
Both of these series converge and here are the two possible limits that the limit comparison test uses
...
© 2018 Paul Dawkins
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...
lamar
...
We’ll close out this section with proofs of the two tests
...
We only need to require that they
start at the same place so to help with the proof we’ll assume that the series start at n = 1
...
We’ll start off with the partial sums of each series
...
First, because an , bn ≥ 0 we know
that,
n
∑a + a
sn ≤ sn + an +1 =
n +1
∑a
=
i
n +1
=i 1 =i 1
tn ≤ tn + bn +1 =
n
∑ bi + bn+1 =
i
n +1
∑b =
i
=i 1 =i 1
= sn +1
⇒
sn ≤ sn +1
tn +1
⇒
tn ≤ tn +1
So, both partial sums form increasing sequences
...
With these preliminary facts out of the way we can proceed with the proof of the test itself
...
Since bn ≥ 0 we know that,
=
tn
n
∞
∑ bi ≤ ∑ bi
=i 1 =i 1
However, we also have established that sn ≤ tn for all n and so for all n we also have,
∞
sn ≤ ∑ bi
i =1
Finally, since
∞
∑b
n =1
n
is a convergent series it must have a finite value and so the partial sums, sn are
bounded above
...
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bounded sequence is also convergent and so {sn }n =1 is a convergent sequence and so
∞
convergent
...
Because an ≥ 0 we then know that we must have
sn → ∞ as n → ∞
...
So, {tn }n =1 is a divergent sequence and so
∞
∞
∑b
n =1
n
is divergent
...
an
a
we know that for large enough n the quotient n must be close to c and
bn
bn
so there must be a positive integer N such that if n > N we also have,
a
m< n
Multiplying through by bn gives,
Now, because c = lim
n →∞
mbn < an < Mbn
provided n > N
...
converges then so does
the Comparison Test
n
∑a
n
∑ Mb
n
and since an < Mbn for all sufficiently large n by
also converges
...
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Section 4-8 : Alternating Series Test
The last two tests that we looked at for series convergence have required that all the terms in the series
be positive
...
The test that we are going to look into in this section will be a test for alternating series
...
an =
bn ≥ 0
( −1) bn
n
an =
bn ≥ 0
( −1) bn
n +1
There are many other ways to deal with the alternating sign, but they can all be written as one of the
two forms above
...
If you should happen to run into a different form than the first two, don’t worry
about converting it to one of those forms, just be aware that it can be and so the test from this section
can be used
...
Then if,
∑a
n
and either an =
1
...
{bn } is a decreasing sequence
( −1)
n
bn or an =
( −1)
n +1
bn where
bn ≥ 0 for
n →∞
the series
∑a
n
is convergent
...
There are a couple of things to note about this test
...
Secondly, in the second condition all that we need to require is that the series terms, bn will be
eventually decreasing
...
All that is required is that eventually we will have
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bn ≥ bn +1 for all n after some point
...
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To see why this is consider the following series,
∞
∑ ( −1)
n
n =1
Let’s suppose that for 1 ≤ n ≤ N
bn
{bn } is not decreasing and that for n ≥ N + 1 {bn } is decreasing
...
The convergence of the series will depend solely on the convergence of the second
(infinite) series
...
On the other hand, if the second series is divergent
either because its value is infinite or it doesn’t have a value then adding a finite number onto this will
not change that fact and so the original series will be divergent
...
We
only need to require that the series terms will eventually be decreasing since we can always strip out the
first few terms that aren’t actually decreasing and look only at the terms that are actually decreasing
...
All we do is check
that eventually the series terms are decreasing and then apply the test
...
Example 1 Determine if the following series is convergent or divergent
...
( −1)
n +1
∞
n +1 1
=
( −1)
∑
∑
n
n
n 1=
n 1
=
∞
1
bn =
n
Now, all that we need to do is run through the two conditions in the test
...
© 2018 Paul Dawkins
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...
lamar
...
Also, the
( −1)
n+1
could be ( −1) or any other form of alternating sign and we’d still call it an Alternating
n
Harmonic Series
...
In general however, we will
need to resort to Calculus I techniques to prove the series terms decrease
...
Example 2 Determine if the following series is convergent or divergent
...
( −1)
n
∞
n2
n2
n
=
−
1
(
)
∑
∑
n2 + 5
n2 + 5
=
n 1=
n 1
∞
⇒
n2
bn =
n2 + 5
Let’s check the conditions
...
Since this condition
isn’t met we’ll need to use another test to check convergence
...
So, the divergence test requires us to compute the following limit
...
For a second let’s consider the following,
n2
( −1)=
lim
n
n →∞
n2 + 5
(
)
n2
n
lim ( −1) lim 2
n →∞
n→∞ n + 5
Splitting this limit like this can’t be done because this operation requires that both limits exist and
while the second one does the first clearly does not
...
So, let’s start with,
n2
( −1)=
lim
n
n →∞
© 2018 Paul Dawkins
n2 + 5
n2
n
lim ( −1) 2
n →∞
n + 5
http://tutorial
...
lamar
...
So, as n → ∞ the terms are alternating between positive and negative values that
are getting closer and closer to 1 and -1 respectively
...
Example 3 Determine if the following series is convergent or divergent
...
That won’t change how the test
works however so we won’t worry about that
...
n
n+4
The first is easy enough to check
...
It is not immediately clear that these terms will
decrease
...
Increasing the
numerator says the term should also increase while increasing the denominator says that the term
should decrease
...
Let’s start with the following function and its derivative
...
Note that x = −4 is not a
critical point because the function is not defined at x = −4
...
Using the test points,
f ′ (1) =
3
50
f ′ ( 5) = −
5
810
and so we can see that the function in increasing on 0 ≤ x ≤ 4 and decreasing on x ≥ 4
...
© 2018 Paul Dawkins
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...
lamar
...
Both conditions are met and so by the Alternating Series Test the series must be converging
...
Do not just make the assumption that the terms will be decreasing and let it go at
that
...
Example 4 Determine if the following series is convergent or divergent
...
To see
this we need to acknowledge that,
cos ( nπ ) =
( −1)
n
If you aren’t sure of this you can easily convince yourself that this is correct by plugging in a few
values of n and checking
...
It should be pointed out that the rewrite we did in previous example only works because n is an integer
and because of the presence of the π
...
Let’s close this section out with a proof of the Alternating Series Test
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Proof of Alternating Series Test
Without loss of generality we can assume that the series starts at n = 1
...
Also note that the assumption here is that we have an =
an =
( −1)
n
( −1)
n +1
bn
...
Finally, in the examples all we really needed was for the bn to be positive and decreasing eventually
but for this proof to work we really do need them to be positive and decreasing for all n
...
s2 = b1 − b2 ≥ 0
s4 = b1 − b2 + b3 − b4 = s2 + b3 − b4 ≥ s2
because b3 − b4 ≥ 0
s6 = s4 + b5 − b6 ≥ s4
because b5 − b6 ≥ 0
because b2 n −1 − b2 n ≥ 0
s2 n = s2 n − 2 + b2 n −1 − b2 n ≥ s2 n − 2
So, {s2n } is an increasing sequence
...
s2 n ≤ b1 for all n
...
So, let’s assume that its limit is s or,
lim s2 n = s
n →∞
Next, we can quickly determine the limit of the sequence of odd partial sums, {s2 n +1} , as follows,
lim s2 n +1 = lim ( s2 n + b2 n +1 ) = lim s2 n + lim b2 n +1 = s + 0 = s
n →∞
© 2018 Paul Dawkins
n →∞
n →∞
n →∞
http://tutorial
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lamar
...
This in turn tells
us that
∑a
n
is convergent
...
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Section 4-9 : Absolute Convergence/Divergence
When we first talked about series convergence we briefly mentioned a stronger type of convergence but
didn’t do anything with it because we didn’t have any tools at our disposal that we could use to work
problems involving it
...
First, let’s go back over the definition of absolute convergence
...
If
∑a
n
is convergent and
is divergent we call the series conditionally convergent
...
Fact
If
∑a
n
is absolutely convergent then it is also convergent
...
This means that we can then
say,
0 ≤ an + an ≤ 2 an
Now, since we are assuming that
∑a
n
is convergent then
∑2 a
n
is also convergent since we
can just factor the 2 out of the series and 2 times a finite value will still be finite
...
+ an ) − ∑ an
is the difference of two convergent series and so is also convergent
...
Series
that are absolutely convergent are guaranteed to be convergent
...
Let’s take a quick look at a couple of examples of absolute convergence
...
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Example 1 Determine if each of the following series are absolute convergent, conditionally
convergent or divergent
...
So, let’s see if it is an absolutely convergent series
...
( −1)
n
1
n
=
n 1=
n 1 n
∞
∑
∞
=∑
This is the harmonic series and we know from the integral test section that it is divergent
...
It is however conditionally convergent since the
series itself does converge
...
∞
∑
( −1)
n+2
∞
=∑
2
=
n 1=
n 1
n
1
n2
This series is convergent by the p-series test and so the series is absolute convergent
...
sin n
3
n =1 n
∞
(c)
∑
© 2018 Paul Dawkins
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...
lamar
...
First, this is NOT an alternating series and so we can’t use
any tools from that section
...
This means that we need to check the convergence of the following series
...
sin n
n3
Therefore, the original series is absolutely convergent (and hence convergent)
...
When we first discussed the convergence
of series in detail we noted that we can’t think of series as an infinite sum because some series can have
different sums if we rearrange their terms
...
We closed that section off with the following fact,
Facts
Given the series
1
...
∑a
of ∑ a
2
...
Now that we’ve got the tools under our belt to determine absolute and conditional convergence we can
make a few more comments about this
...
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First, as we showed above in Example 1a an Alternating Harmonic is conditionally convergent and so no
matter what value we chose there is some rearrangement of terms that will give that value
...
Next, we showed in Example 1b that,
∞
∑
( −1)
n+2
n2
is absolutely convergent and so no matter how we rearrange the terms of this series we’ll always get the
same value
...
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Section 4-10 : Ratio Test
In this section we are going to take a look at a test that we can use to see if a series is absolutely
convergent or not
...
Before proceeding with the test let’s do a quick reminder of factorials
...
If n is an integer such that n ≥ 0 then n factorial is defined as,
n ! = n ( n − 1)( n − 2 ) ( 3)( 2 )(1)
if n ≥ 1
0! = 1
by definition
Let’s compute a couple real quick
...
For instance,
5!= 5 ( 4 )( 3)( 2 )(1)= 5 ⋅ 4!
4!
=
=
5! 5 ( 4 )( 3)( 2 )(
1) 5 ( 4 ) ⋅ 3!
3!
In general, we can always “strip out” terms from a factorial as follows
...
Also, when dealing with factorials we need to be very careful with parenthesis
...
© 2018 Paul Dawkins
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...
lamar
...
This is often one of the more
common mistakes that students make when they first run across factorials
...
Ratio Test
Suppose we have the series
∑a
n
...
if L < 1 the series is absolutely convergent (and hence convergent)
...
if L > 1 the series is divergent
...
if L = 1 the series may be divergent, conditionally convergent, or absolutely convergent
...
Notice that in the case of L = 1 the ratio test is pretty much worthless and we would need to resort to a
different test to determine the convergence of the series
...
If they are not there it will be
impossible for us to get the correct answer
...
Example 1 Determine if the following series is convergent or divergent
...
Here are the series terms an
...
( −10 )
=
2( n +1) +1
4
( ( n + 1) + 1)
n +1
=
an +1
© 2018 Paul Dawkins
( −10 )
2 n +3
4 ( n + 2)
n +1
http://tutorial
...
lamar
...
So,
42 n +1 ( n + 1)
( −10 )
42 n +3 ( n + 2 ) ( −10 )n
−10 ( n + 1)
42 ( n + 2 )
n +1
L = lim
n →∞
= lim
n →∞
n +1
10
lim
16 n→∞ n + 2
10
=
<1
16
=
So, L < 1 and so by the Ratio Test the series converges absolutely and hence will converge
...
Make sure
that you do this canceling
...
Example 2 Determine if the following series is convergent or divergent
...
Here is
the limit
...
We simply can’t do the limit with the
factorials in it
...
If we do that with the numerator (in this case
because it’s the larger of the two) we get,
L = lim
( n + 1) n !
5 n!
n →∞
at which point we can cancel the n! for the numerator an denominator to get,
L = lim
n →∞
( n + 1) = ∞ > 1
5
So, by the Ratio Test this series diverges
...
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Example 3 Determine if the following series is convergent or divergent
...
( n + 1) ( 2n − 1)!
L = lim
n →∞ 2 ( n + 1) − 1 !
(
) n2
2
( n + 1) ( 2n − 1)!
= lim
n →∞ ( 2n + 1) !
n2
2
( n + 1)
( 2n − 1)!
= lim
n →∞ ( 2n + 1)( 2n )( 2n − 1) !
n2
2
n + 1)
(
= lim
n →∞ 2n + 1 2n
(
)( ) ( n2 )
2
= 0 <1
So, by the Ratio Test this series converges absolutely and so converges
...
∞
9n
∑
n =1
( −2 )
n +1
n
Solution
Do not mistake this for a geometric series
...
So, let’s compute the limit
...
© 2018 Paul Dawkins
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...
lamar
...
If we hadn’t
used them we would have gotten L =− 92 < 1 which would have implied a convergent series!
Now, let’s take a look at a couple of examples to see what happens when we get L = 1
...
In both of these examples we
will first verify that we get L = 1 and then use other tests to determine the convergence
...
∞
∑n
n =0
Solution
Let’s first get L
...
We will need to resort to another test for this series
...
1
=
=
lim bn lim
0
2
n →∞
n →∞ n + 1
1
1
bn =
bn +1
>
=
2
n + 1 ( n + 1)2 + 1
The two conditions are met and so by the Alternating Series Test this series is convergent
...
Example 6 Determine if the following series is convergent or divergent
...
( n + 3)( 2n + 7 )
2n + 7
n+3
=
= 1
L lim = lim
n →∞ 2 ( n + 1) + 7 n + 2
n →∞ ( 2n + 9 )( n + 2 )
Again, the ratio test tells us nothing here
...
In fact that probably should have been our first choice on this one anyway
...
© 2018 Paul Dawkins
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...
lamar
...
There is one more thing that we should note about the ratio test before we move onto the next section
...
This will always happen with rational expression involving only polynomials or polynomials under
radicals
...
Also, in the second to last example we saw an example of an alternating series in which the positive
term was a rational expression involving polynomials and again we will always get L = 1 in these cases
...
Proof of Ratio Test
First note that we can assume without loss of generality that the series will start at n = 1 as we’ve
done for all our series test proofs
...
To do this let’s first note that because L < 1 there is some number r such that L < r < 1
...
Just why is this important? Well we can now look at
the following series
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This is a geometric series and because 0 < r < 1 we in fact know that it is a convergent series
...
However since,
N
∞
an
∑=
∑ an +
n= 1
we know that
∞
∑a
n =1
n
k=
1
n= 1
∞
∑
n= N +1
an
is also convergent since the first term on the right is a finite sum of finite terms
∞
∑a
and hence finite
...
Next, we need to assume that L > 1 and we’ll need to show that
L = lim
n →∞
∑a
n
is divergent
...
This in turn means that,
lim an ≠ 0
n →∞
Therefore, by the Divergence Test
∑a
n
is divergent
...
To do this we just need a series for each case
...
∞
1
∑n
n =1
∞
∑
( −1)
n
n =1
∞
1
∑n
absolutely convergent
2
n
conditionally convergent
divergent
n =1
© 2018 Paul Dawkins
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lamar
...
As with the Ratio Test this
test will also tell whether a series is absolutely convergent or not rather than simple convergence
...
Define,
n a
=
L lim
=
lim an
n
n →∞
n →∞
1
n
Then,
1
...
2
...
3
...
A proof of this test is at the end of the section
...
Also note that, generally for the series we’ll be dealing with
in this class, if L = 1 in the Ratio Test then the Root Test will also give L = 1
...
Fact
1
lim n n = 1
n →∞
Let’s take a look at a couple of examples
...
∞
nn
∑
1+ 2 n
n =1 3
Solution
There really isn’t much to these problems other than computing the limit and then using the root test
...
L = lim
n →∞
n
3
n
1
n
1+ 2 n
n
= lim
n →∞
3
1
+2
n
=
∞
= ∞ >1
32
So, by the Root Test this series is divergent
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Example 2 Determine if the following series is convergent or divergent
...
1
n n
5n − 3n
5n − 3n3
=
=
L= lim 3
lim
n →∞
n →∞ 7 n 3 + 2
7n + 2
3
−3 3
=
<1
7
7
Therefore, by the Root Test this series converges absolutely and hence converges
...
Example 3 Determine if the following series is convergent or divergent
...
L
= lim
n →∞
( −12 )
n
n
n
1
n n
12 12
= lim =
= 12 > 1
n →∞ 1
1
n
n
After using the fact from above we can see that the Root Test tells us that this series is divergent
...
Also note that this proof is very similar to the proof of the Ratio
Test
...
To do this let’s first note that because L < 1 there is some number r such that L < r < 1
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∞
∑r
n
n =0
is a geometric series and because 0 < r < 1 we in fact know that it is a convergent series
...
However since,
N −1
∞
an
∑=
∞
∑ an + ∑ an
=
n 1 =
n 1=
n N
we know that
∞
∑a
n =1
n
is also convergent since the first term on the right is a finite sum of finite terms
∞
∑a
and hence finite
...
Next, we need to assume that L > 1 and we’ll need to show that
n a
=
L lim
=
lim an
n
n →∞
n →∞
∑a
n
is divergent
...
Finally, we need to assume that L = 1 and show that we could get a series that has any of the three
possibilities
...
We’ll leave the details of checking to you
but all three of the following series have L = 1 and each one exhibits one of the possibilities
...
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...
Note that these are a general set of guidelines and because some series can have more than one test
applied to them we will get a different result depending on the path that we take through this set of
guidelines
...
Once this has been done
you can identify the test that you feel will be the easiest for you to use
...
1
...
e
...
Note that you should only do the
n →∞
Divergence Test if a quick glance suggests that the series terms may not converge to zero in
the limit
...
Is the series a p-series (
∞
1
)
or
a
geometric
series
(
ar n or
∑ np
∑
n =0
∞
∑ ar
n −1
)? If so use the fact
n =1
that p-series will only converge if p > 1 and a geometric series will only converge if r < 1
...
3
...
4
...
e
...
Remember however, that in order to use
the Comparison Test and the Limit Comparison Test the series terms all need to be positive
...
Does the series contain factorials or constants raised to powers involving n? If so, then the
Ratio Test may work
...
6
...
7
...
n
8
...
© 2018 Paul Dawkins
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lamar
...
If more than one test can be used try to use the test
that will be the easiest for you to use and remember that what is easy for someone else may not be easy
for you!
Also, just so we can put all the tests into one place here is a quick listing of all the tests that we’ve got
...
If
2
...
If
n
2
...
∑a
...
Then,
n
n=k
∑a
...
is convergent then so is
n
n=k
n
n
n
Limit Comparison Test
Suppose that we have two series
∑a
n
and
∑b
n
with an , bn ≥ 0 for all n
...
e
...
e
...
Alternating Series Test
Suppose that we have a series
n
...
lim bn = 0 and,
2
...
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Calculus II
Ratio Test
Suppose we have the series
264
∑a
n
...
if L < 1 the series is absolutely convergent (and hence convergent)
...
if L > 1 the series is divergent
...
if L = 1 the series may be divergent, conditionally convergent, or absolutely convergent
...
Define,
n a
=
=
L lim
lim an
n
n →∞
n →∞
1
n
Then,
1
...
2
...
3
...
© 2018 Paul Dawkins
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...
lamar
...
This is usually a very difficult thing to do and we still aren’t going to talk about how to find the value of a
series
...
Often that is all that you
need to know
...
It doesn’t make any sense to talk about the value of a series that doesn’t converge and so we
will be assuming that the series we’re working with converges
...
So, let’s start with the series
∞
∑a
n =1
n
(the starting point is not important, but we need a starting point to
do the work) and let’s suppose that the series converges to s
...
In other words,
lim sn = s
n →∞
Now, just what does this mean for us? Well, since this limit converges it means that we can make the
partial sums, sn , as close to s as we want simply by taking n large enough
...
We can just take a partial sum and use that as an
estimation of the value of the series
...
First, how good is the estimation? If we don’t have an idea of how good the estimation is then it really
doesn’t do all that much for us as an estimation
...
We won’t always be able to do this, but if we can that will be nice
...
Let’s first
start with the full series and strip out the first n terms
...
Recall that we can use any letter for the index and it won’t change the value
...
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Now, notice that the first series (the n terms that we’ve stripped out) is nothing more than the partial
sum sn
...
Finally let’s acknowledge that we also know the value of the series since we are
assuming it’s convergent
...
Of course, we can’t get our hands on the actual value of the remainder because we don’t have the
actual value of the series
...
Once
we’ve got an estimate on the value of the remainder we’ll also have an idea on just how good a job the
partial sum does of estimating the actual value of the series
...
We’ll go through each one
separately
...
e
...
) and when using the tests we noted that all we really needed was for them to
eventually meet the preconditions in order for the test to work
...
If there are a few terms at the start where the preconditions aren’t met we’ll need to strip those terms
out, do the estimate on the series that is left and then add in the terms we stripped out to get a final
estimate of the series value
...
We derived the integral test by using the fact that the series could be thought of as an
estimation of the area under the curve of f ( x ) where f ( n ) = an
...
As we’ll soon see if we can get an upper and lower bound on the value of the remainder we can use
these bounds to help us get upper and lower bounds on the value of the series
...
So, let’s first recall that the remainder is,
Rn =
© 2018 Paul Dawkins
∞
∑a
i= n +1
i
= an +1 + an + 2 + an +3 + an + 4 +
http://tutorial
...
lamar
...
We can see that the remainder, Rn , is the area estimation and it will overestimate the exact area
...
Rn ≥ ∫
∞
n +1
f ( x ) dx
(2)
Next, we could also estimate the area by starting at x = n , taking rectangles of width 1 again and then
using the right endpoint as the height of the rectangle
...
This is shown in the following sketch
...
This leads to the following inequality,
© 2018 Paul Dawkins
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...
lamar
...
This will in turn give us an upper bound and a lower bound on just how good the partial sum, sn , is as an
estimation of the actual value of the series
...
First, we’ll start with the fact that
=
s sn + Rn
Now, if we use (2) we get,
s = sn + Rn ≥ sn + ∫
∞
n +1
f ( x ) dx
Likewise if we use (3) we get,
∞
s = sn + Rn ≤ sn + ∫ f ( x ) dx
n
Putting these two together gives us,
sn + ∫
∞
n +1
∞
f ( x ) dx ≤ s ≤ sn + ∫ f ( x ) dx
(4)
n
This gives an upper and a lower bound on the actual value of the series
...
Let’s work an example with this
...
Solution
First, for comparison purposes, we’ll note that the actual value of this series is known to be,
1 π2
=
= 1
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Using n = 15 let’s first get the partial sum
...
580440283
Note that this is “close” to the actual value in some sense but isn’t really all that close either
...
These are fairly simple integrals, so we’ll leave it to you to verify the
values
...
580440283 +
16
15
1
...
647106950
1
...
s ≈ 1
...
So, that is how we can use the Integral Test to estimate the value of a series
...
Comparison Test
In this case, unlike with the integral test, we may or may not be able to get an idea of how good a
particular partial sum will be as an estimate of the exact value of the series
...
First, let’s remind ourselves on how the comparison test actually works
...
Therefore, we found a
bn that converged and an ≤ bn for all n
...
What we want to do is determine how good of a job the partial sum,
n
sn = ∑ ai
i =1
will do in estimating the actual value of the series
∑a
n
...
Let’s actually write down the remainder for both series
...
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Now, since an ≤ bn we also know that
Rn ≤ Tn
When using the comparison test it is often the case that the bn are fairly nice terms and that we might
actually be able to get an idea on the size of Tn
...
If we are unable to get an idea of the size of Tn then using the comparison test to help with estimates
won’t do us much good
...
2n
...
So,
2n
2n 1
≤
=
4n + 1 4n 2
and
n
n
1
∑
n =0 2
is a geometric series and converges because r = 12 < 1
...
=
s15
2n
1
...
Since our second
series is a geometric series we can compute this directly as follows
...
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...
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The series on the left is in the standard form and so we can compute that directly
...
Doing the work gives,
n
n
n
15
∞
1
1
1
=
∑
∑ − ∑
2 n 0=
2 n 0 2
=
n 16 =
1
=
− 1
...
000030518
So, according to this if we use
s ≈ 1
...
000030518
and that’s not too bad
...
383093004
∑
n
n =0 4 + 1
∞
and so we can see that the actual error in our estimation is,
Error =
Actual − Estimate =
1
...
383062486 =
0
...
This will not always happen and so we shouldn’t expect that to happen in all cases
...
Before moving on to the final part of this section let’s again note that we will only be able to determine
how good the estimate is using the comparison test if we can easily get our hands on the remainder of
the second term
...
Alternating Series Test
Both of the methods that we’ve looked at so far have required the series to contain only positive terms
...
However, with that
said there is one case where it isn’t too bad
...
Once again we will start off with a convergent series
a ∑ ( −1)
∑=
n
n
bn which in this case happens to
be an alternating series that satisfies the conditions of the alternating series test, so we know that
bn ≥ 0 and is decreasing for all n
...
We want to know how good of an estimation of the actual series value will
the partial sum, sn , be
...
© 2018 Paul Dawkins
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...
lamar
...
Let’s take a look at an example
...
Solution
This is an alternating series and it does converge
...
8224670336
12
( −1)
n2
n
= −0
...
00390625
16
So, our estimation will have an error of no more than 0
...
In this case the exact value is
known and so the actual error is,
R15 =s − s15 =0
...
It will often be the case that
the actual error will be less than the estimated error
...
Ratio Test
This will be the final case that we’re going to look at for estimating series values and we are going to
have to put a couple of fairly stringent restrictions on the series terms in order to do the work
...
We’ll also be adding on another restriction in a bit
...
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273
In this case we’ve used the ratio test to show that
L = lim
n →∞
and found that L < 1
...
To do this we computed
an +1
an
As with the previous cases we are going to use the remainder, Rn , to determine how good of an
estimation of the actual value the partial sum, sn , is
...
{ }
1
...
If rn is an increasing sequence then,
Rn ≤
an +1
1− L
Proof
Both parts will need the following work so we’ll do it first
...
Rn =
∞
∑a
i= n +1
i
= an +1 + an + 2 + an +3 + an + 4 +
a
a
a
= an +1 1 + n + 2 + n +3 + n + 4 +
an +1 an +1 an +1
Next, we need to do a little work on a couple of these terms
...
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...
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Calculus II
274
Rn= an +1 (1 + rn +1 + rn +1rn + 2 + rn +1rn + 2 rn +3 +)
Okay now let’s do the proof
...
So,
Rn ≤
For the second part we are assuming that
an +1
1 − rn +1
{r } is increasing and we know that,
n
an +1
L
=
lim rn lim
=
n →∞
n →∞ a
n
and so we know that rn < L for all n
...
So,
Rn ≤
an +1
1− L
{ }
Note that there are some restrictions on the sequence rn and at least one of its terms in order to use
these formulas
...
Let’s take a look at an example of this
...
Solution
First, let’s use the ratio test to verify that this is a convergent series
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L= lim
n →∞
n + 1 3n
n +1 1
= lim
=
<1
n +1
n n→∞ 3n
3
3
So, it is convergent
...
s15
=
15
n
∑=
3
n =0
n
0
...
=
rn
n +1 3
n +1 1 1
=
= 1 +
n +1
n
3
3n
3 n
n
The last rewrite was just to simplify some of the computations a little
...
Also note that
r16 =
1
3
(1 + 161 ) < 1
...
0000005755187
1 − r16 1 − 13 (1 + 161 )
So, it looks like our estimate is probably quite good
...
∞
and so we can compute the actual error
...
000000575
This is less than the upper bound, but unlike in the previous example this actual error is quite close to
the upper bound
...
There is usually no way
of knowing ahead of time which it will be and without the exact value in hand there will never be a way
of determining which it will be
...
Often series that we used ratio test on are also
alternating series and so if that is the case we can always resort to the previous material to get an upper
bound on the error in the estimation, even if we didn’t use the alternating series test to show
convergence
...
© 2018 Paul Dawkins
http://tutorial
...
lamar
...
It’s now time to
start looking at some specific kinds of series and we’ll eventually reach the point where we can talk
about a couple of applications of series
...
A power series about a, or just power
series, is any series that can be written in the form,
∞
∑ c ( x − a)
n =0
n
n
where a and cn are numbers
...
The first thing to
notice about a power series is that it is a function of x
...
In all the prior sections we’ve only allowed numbers in the series and
now we are allowing variables to be in the series as well
...
Everything that we know about series still holds
...
The
difference is that the convergence of the series will now depend upon the values of x that we put into
the series
...
Before we get too far into power series there is some terminology that we need to get out of the way
...
This number is called the radius of
R
...
Note that the series may or may not converge if x − a =
at these points will not change the radius of convergence
...
These two concepts are fairly closely tied together
...
a−R< x
power series converges
x < a − R and x > a + R
power series diverges
The interval of convergence must then contain the interval a − R < x < a + R since we know that the
power series will converge for these values
...
Therefore, to completely identify the interval of convergence all that we have to do is
determine if the power series will converge for x= a − R or x= a + R
...
© 2018 Paul Dawkins
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...
lamar
...
In this case the power series becomes,
∞
∑ cn ( a − a ) =
n
∞
∞
∞
∑ cn ( 0 ) = c0 ( 0 ) + ∑ cn ( 0 ) = c0 + ∑ 0 = c0 + 0 = c0
=
n 0=
n 0
0
n
n
=
n 1 =
n 1
and so the power series converges
...
It is important to note that no matter what else is happening in the power series we are guaranteed to
get convergence for x = a
...
Let’s work some examples
...
Example 1 Determine the radius of convergence and interval of convergence for the following
power series
...
To
determine the remainder of the x’s for which we’ll get convergence we can use any of the tests that
we’ve discussed to this point
...
From this we can get the
radius of convergence and most of the interval of convergence (with the possible exception of the
endpoints)
...
Most of the
power series that we’ll be looking at are set up for one or the other
...
( −1) ( n + 1)( x + 3)
L = lim
n +1
n →∞
= lim
n →∞
4n +1
n +1
4n
( −1) ( n )( x + 3)
n
n
− ( n + 1)( x + 3)
4n
Before going any farther with the limit let’s notice that since x is not dependent on the limit it can be
factored out of the limit
...
The limit is then,
© 2018 Paul Dawkins
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...
lamar
...
So, we have,
1
x +3 <1
4
1
x +3 >1
4
⇒
x+3 < 4
series converges
⇒
x+3 > 4
series diverges
We’ll deal with the L = 1 case in a bit
...
These are exactly the conditions required for the radius of convergence
...
Now, let’s get the interval of convergence
...
−4 < x + 3 < 4
−7 < x < 1
So, most of the interval of validity is given by −7 < x < 1
...
Note that these values of x will
correspond to the value of x that will give L = 1
...
x = −7 :
In this case the series is,
∞
∑
( −1)
4
n
n
n
=
( −4 )
∞
∑
( −1)
4
n
n 1=
n 1
=
∞
n
n
n
( −1)
n
4n
=
∑ ( −1) ( −1) n
n
( −1) ( −1)
n
n
n =1
n
1
=
( −1) =
2n
∞
= ∑n
n =1
This series is divergent by the Divergence Test since lim n = ∞ ≠ 0
...
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279
This series is also divergent by the Divergence Test since lim ( −1) n doesn’t exist
...
The interval of convergence is
then,
−7 < x < 1
In the previous example the power series didn’t converge for either endpoint of the interval
...
The power series could converge
at either both of the endpoints or only one of the endpoints
...
2n
n
( 4 x − 8)
∑
n =1 n
∞
Solution
Let’s jump right into the ratio test
...
2 4x − 8 < 1
series converges
2 4x − 8 > 1
series diverges
We need to be careful here in determining the interval of convergence
...
In other words, we need to factor a 4 out of the absolute value
bars in order to get the correct radius of convergence
...
Now, let’s find the interval of convergence
...
© 2018 Paul Dawkins
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...
lamar
...
x=
15
:
8
The series here is,
n
∞
2n 15
2n 1
−
=
8
∑
∑ −
n 2
2
=
n 1=
n 1 n
∞
2n ( −1)
=∑
2n
n =1 n
∞
∞
=∑
( −1)
n
n
n
n
n =1
This is the alternating harmonic series and we know that it converges
...
So, the power series converges for one of the endpoints, but not the other
...
The interval of convergence for this power series is then,
15
17
≤x<
8
8
We now need to take a look at a couple of special cases with radius and intervals of convergence
...
∞
∑ n !( 2 x + 1)
n
n =0
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lamar
...
( n + 1)!( 2 x + 1)
n
n →∞
n !( 2 x + 1)
( n + 1) n ! ( 2 x + 1)
= lim
n +1
L = lim
n!
n →∞
=+
2 x 1 lim ( n + 1)
n →∞
At this point we need to be careful
...
We’ll have L = ∞ > 1 provided x ≠ − 12
...
If you think about it we actually already knew that
however
...
Remember that we get a from ( x − a ) , and notice the coefficient of the x must
n
be a one!
In this case we say the radius of convergence is R = 0 and the interval of convergence is x = − 12 ,
and yes we really did mean interval of convergence even though it’s only a point
...
∞
∑
( x − 6)
n =1
n
nn
Solution
In this example the root test seems more appropriate
...
In these cases, we say that the radius of convergence is R = ∞ and interval of convergence is
−∞ < x < ∞
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So, let’s summarize the last two examples
...
Likewise, if the power series
converges for every x the radius of convergence is R = ∞ and interval of convergence is −∞ < x < ∞
...
Example 5 Determine the radius of convergence and interval of convergence for the following
power series
...
That’s not really important to the problem, but it’s worth
pointing out so people don’t get excited about it
...
In this case it is 2n rather than the
standard n
...
This one seems set up for the root test again so let’s use that
...
The radius of convergence requires an exponent of 1 on
the x
...
Notice that we didn’t bother to put down the inequality for divergence this time
...
We will usually skip that part
...
First from the inequality we get,
− 3
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...
x= − 3:
Here the power series is,
∞
∑
(− 3)
(( − 3 ) )
2 n
2n
∞
=∑
( −3)
( −3)
n
∞
3)
(
=∑
n
n
n =1 ( −1) ( 3 )
n
n 1=
n 1
=
=
n
∞
∑ ( −1)
n
n =1
This series is divergent by the Divergence Test since lim ( −1) doesn’t exist
...
( 3 )=
2n
∞
∑
( −3)
∞
∑ ( −1)
n
n
=
n 1=
n 1
which is divergent
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Section 4-15 : Power Series and Functions
We opened the last section by saying that we were going to start thinking about applications of series
and then promptly spent the section talking about convergence again
...
With this section we will start talking about how to represent functions with power series
...
Let’s start off with one that we already know how to do, although when we first ran across this series we
didn’t think of it as a power series nor did we acknowledge that it represented a function
...
Now, if we take a = 1 and r = x this becomes,
∞
=
∑ xn
n =0
1
1− x
(1)
provided x < 1
Turning this around we can see that we can represent the function
f ( x) =
with the power series
∞
∑x
n
1
1− x
(2)
(3)
provided x < 1
n =0
This provision is important
...
This means the equality in (1) will only
hold if x < 1
...
Note as well that we can also use this to
acknowledge that the radius of convergence of this power series is R = 1 and the interval of
convergence is x < 1
...
We will be representing many functions as power series and
it will be important to recognize that the representations will often only be valid for a range of x’s and
that there may be values of x that we can plug into the function that we can’t plug into the power series
representation
...
© 2018 Paul Dawkins
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...
lamar
...
So, let’s jump into a couple of examples
...
g ( x) =
1
1 + x3
Solution
What we need to do here is to relate this function back to (2)
...
Recall that the x in (2) is simply a variable and can represent anything
...
Therefore, all we need to do is replace
the x in (3) and we’ve got a power series representation for g ( x )
...
All we need to do now is a little simplification
...
This usually won’t
happen
...
Example 2 Find a power series representation for the following function and determine its interval
of convergence
...
The difference is the numerator and at first glance
that looks to be an important difference
...
However, now that we’ve worked the first example this one is actually very simple since we can use
the result of the answer from that example
...
© 2018 Paul Dawkins
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lamar
...
The last step is to bring the coefficient into the series and we’ll be done
...
We typically only want a single x in a power series
...
This is an
important idea to remember as it can often greatly simplify our work
...
f ( x) =
x
5− x
Solution
So, again, we’ve got an x in the numerator
...
f ( x) = x
1
5− x
If we had a power series representation for
1
5− x
we could get a power series representation for f ( x )
...
We’ll first notice that in order to use (2) we’ll need the number in the denominator
to be a one
...
g ( x) =
1 1
5 1− x
5
Now all we need to do to get a power series representation is to replace the x in (3) with x5
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Now let’s do a little simplification on the series
...
All we need to do for this is to multiply the power series
representation for g ( x ) by x and we’ll have it
...
So, hopefully we now have an idea on how to find the power series representation for some functions
...
We now need to look at some further manipulation of power series that we will need to do on occasion
...
Let’s start with differentiation of the power series,
∞
f ( x ) = ∑ cn ( x − a ) = c0 + c1 ( x − a ) + c2 ( x − a ) + c3 ( x − a ) +
n
2
3
n =0
Now, we know that if we differentiate a finite sum of terms all we need to do is differentiate each of the
terms and then add them back up
...
Nicely enough for us however, it is known that if the power series representation of f ( x ) has a radius
of convergence of R > 0 then the term by term differentiation of the power series will also have a
radius of convergence of R and (more importantly) will in fact be the power series representation of
f ′ ( x ) provided we stay within the radius of convergence
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Again, we should make the point that if we aren’t dealing with a power series then we may or may not
be able to differentiate each term of the series to get the derivative of the series
...
It has been changed from n = 0 to n = 1
...
Notice however, that since the n=0 term of the above series is also zero, we could start the
series at n = 0 if it was required for a particular problem
...
We can now find formulas for higher order derivatives as well now
...
Once again, notice that the initial value of n changes with each differentiation in order to acknowledge
that a term from the original series differentiated to zero
...
Just as with the differentiation, when we’ve got an infinite
series we need to be careful about just integration term by term
...
In other words,
n
⌠ ∞
=
f
x
dx
(
)
∑ cn ( x − a ) dx
∫
⌡ n =0
=
∞
∑ ∫ c ( x − a)
n =0
n
∞
= C + ∑ cn
n =0
n
dx
( x − a)
n +1
n +1
Notice that we pick up a constant of integration, C, that is outside the series here
...
© 2018 Paul Dawkins
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...
lamar
...
Now, let’s see how we can use these facts to generate some more power series representations of
functions
...
g ( x) =
Solution
To do this problem let’s notice that
1
(1 − x )
(1 − x )
=
2
1
2
d 1
dx 1 − x
Then since we’ve got a power series representation for
1
1− x
all that we’ll need to do is differentiate that power series to get a power series representation for
g ( x)
...
© 2018 Paul Dawkins
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...
lamar
...
h=
( x ) ln ( 5 − x )
Solution
In this case we need to notice that
⌠ 1 dx =
− ln ( 5 − x )
⌡ 5− x
and then recall that we have a power series representation for
1
5− x
Remember we found a representation for this in Example 3
...
A good choice is x = 0 since
that will make the series easy to evaluate
...
In fact, there is no way
to bring it into the series so don’t get excited about it
...
© 2018 Paul Dawkins
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...
lamar
...
The problem with the approach in that section is that everything came down to needing to be able to
relate the function in some way to
1
1− x
and while there are many functions out there that can be related to this function there are many more
that simply can’t be related to this
...
So, for the time being, let’s make two assumptions
...
Now that we’ve assumed that a power series representation exists we need to determine what the
coefficients, cn , are
...
Let’s first just evaluate everything
at x = a
...
Unfortunately, there isn’t any
other value of x that we can plug into the function that will allow us to quickly find any of the other
coefficients
...
Let’s continue with this idea and find the second derivative
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c2 =
Using the third derivative gives,
f ′′ ( a )
2
f ′′′ (=
x ) 3 ( 2 ) c3 + 4 ( 3)( 2 ) c4 ( x − a ) +
=
f ′′′ ( a ) 3 ( 2 ) c3
=
⇒
c3
f ′′′ ( a )
3( 2)
Using the fourth derivative gives,
f(
4)
4 ( 3)( 2 ) c4 + 5 ( 4 )( 3)( 2 ) c5 ( x − a )
( x) =
f ( ) (a)
4 ( 3)( 2 )
4
f ( ) ( a ) 4 ( 3)( 2 ) c4
=
4
c4
=
⇒
Hopefully by this time you’ve seen the pattern here
...
f (n) ( a )
cn =
n!
0
This even works for n = 0 if you recall that 0! = 1 and define f ( ) ( x ) = f ( x )
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Before working any examples of Taylor Series we first need to address the assumption that a Taylor
Series will in fact exist for a given function
...
To determine a condition that must be true in order for a Taylor series to exist for a function let’s first
define the nth degree Taylor polynomial of f ( x ) as,
Tn ( x )
=
n
∑
i =0
f (i ) ( a )
i
( x − a)
i!
Note that this really is a polynomial of degree at most n
...
We’ll see a nice application of
Taylor polynomials in the next section
...
th
Next, the remainder is defined to be,
Rn=
( x ) f ( x ) − Tn ( x )
So, the remainder is really just the error between the function f ( x ) and the nth degree Taylor
polynomial for a given n
...
Theorem
Suppose that f=
( x ) Tn ( x ) + Rn ( x )
...
In general, showing that lim Rn ( x ) = 0 is a somewhat difficult process and so we will be assuming that
n →∞
this can be done for some R in all of the examples that we’ll be looking at
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Now let’s look at some examples
...
Solution
This is actually one of the easier Taylor Series that we’ll be asked to compute
...
This is one of the few
functions where this is easy to do right from the start
...
Solution
There are two ways to do this problem
...
We’ll work both solutions since the longer one has some nice ideas that we’ll
see in other examples
...
However, unlike the first one
we’ve got a little more work to do
...
−x
=
f ( ) ( x ) e=
f ( ) ( 0) 1
0
0
−e − x
−1
f ( ) ( x) =
f ( ) ( 0) =
1
1
−x
=
f ( 2) ( x ) e=
f ( 2) ( 0 ) 1
f(
3)
3
−e − x
f ( ) ( 0) =
−1
( x) =
f
(n)
© 2018 Paul Dawkins
( x) =
( −1) e− x
n
f
(n)
( 0) =
( −1)
n
n=
0,1, 2,3
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...
lamar
...
n
We often won’t be able to get a general formula for f ( ) ( x ) so don’t get too excited about getting
n
that formula
...
So, in this case we’ve got general formulas so all we need to do is plug these into the Taylor Series
formula and be done with the problem
...
However, in this
case there is a much shorter solution method
...
Let’s do the same thing with this one
...
So, all we need to do is replace the x in the Taylor Series that we found in the first example with “-x”
...
2
Example 3 Find the Taylor Series for f ( x ) = x 4e −3 x about x = 0
...
In this example, unlike the previous example, doing this directly would be significantly longer
and more difficult
...
Also, we’ll pick on the exponential function one
© 2018 Paul Dawkins
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lamar
...
This will be the final Taylor Series for exponentials in
this section
...
Solution
n
Finding a general formula for f ( ) ( −4 ) is fairly simple
...
Example 5 Find the Taylor Series for f ( x ) = cos ( x ) about x = 0
...
f ( ) ( x ) cos
x
f ( ) ( 0) 1
=
=
0
0
0
f ( ) ( x) =
f ( ) ( 0) =
− sin x
1
1
2
f ( ) ( 0) =
− cos x
−1
( x) =
3
3
f ( ) ( x ) sin
x
f ( ) ( 0) 0
=
=
4
4
f ( ) ( x ) cos
x
f ( ) ( 0) 1
=
=
5
5
0
f ( ) ( x) =
f ( ) ( 0) =
− sin x
6
6
f ( ) ( x) =
f ( ) ( 0) =
− cos x
−1
f(
2)
In this example, unlike the previous ones, there is not an easy formula for either the general
derivative or the evaluation of the derivative
...
So, let’s plug what we’ve got into the Taylor series and see what we get,
© 2018 Paul Dawkins
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...
lamar
...
This doesn’t really help us to get a general
formula for the Taylor Series
...
cos x =
1 −
1 2 1 4 1 6
x + x − x +
2!
4!
6!
n =0
=
n 1=
n 2=
n 3
By renumbering the terms as we did we can actually come up with a general formula for the Taylor
Series and here it is,
−1) x 2 n
(
cos x = ∑
( 2n ) !
n =0
∞
n
This idea of renumbering the series terms as we did in the previous example isn’t used all that often, but
occasionally is very useful
...
Example 6 Find the Taylor Series for f ( x ) = sin ( x ) about x = 0
...
0
0
f ( ) ( x ) sin
x
f ( ) ( 0) 0
=
=
f ( ) ( x ) cos
x
f ( ) ( 0) 1
=
=
1
1
2
0
f ( ) ( 0) =
− sin x
( x) =
3
3
f ( ) ( x) =
f ( ) ( 0) =
− cos x
−1
4
4
f ( ) ( x ) sin
x
f ( ) ( 0) 0
=
=
5
5
f ( ) ( x ) cos
x
f ( ) ( 0) 1
=
=
6
6
0
f ( ) ( x) =
f ( ) (0) =
− sin x
f(
2)
So, we get a similar pattern for this one
...
© 2018 Paul Dawkins
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...
lamar
...
So renumbering the terms as we did in the
previous example we get the following Taylor Series
...
Example 7 Find the Taylor Series for f ( x ) = ln ( x ) about x = 2
...
=
f ( 0) ( x ) ln=
f ( 0) ( 2 ) ln 2
( x)
1
1
=
f (1) ( 2 )
x
2
1
1
− 2
− 2
f ( 2) ( x ) =
f ( 2) ( 2 ) =
x
2
2
2
=
f ( 3) ( x ) =
f ( 3) ( 2 )
3
x
23
2 ( 3)
2 ( 3)
− 4
− 4
f ( 4) ( x ) =
f ( 4) ( 2 ) =
x
2
2 ( 3)( 4 )
2 ( 3)( 4 )
=
f ( 5) ( x ) =
f ( 5) ( 2 )
5
x
25
=
f (1) ( x )
( −1) ( n − 1)!
( −1) =
( n − 1)!
=
f ( x)
f (n) ( 2 ) =
n
n
n
(n)
n +1
n +1
x
2
1, 2,3,
Note that while we got a general formula here it doesn’t work for n = 0
...
In order to plug this into the Taylor Series formula we’ll need to strip out the n = 0 term first
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∞
∑
ln ( x )
=
n =0
f (n) ( 2 )
n
( x − 2)
n!
∞
f ( 2) + ∑
=
f(
n)
( 2)
n!
n =1
( x − 2)
n
( −1) ( n − 1)! x − 2 n
ln ( 2 ) + ∑
=
(
)
n
n +1
∞
n! 2
n =1
∞
ln ( 2 ) + ∑
=
( −1)
n2
n =1
n +1
n
( x − 2)
n
Notice that we simplified the factorials in this case
...
Also, do not get excited about the term sitting in front of the series
...
Example 8 Find the Taylor Series for f ( x ) =
Solution
Again, here are the derivatives and evaluations
...
x2
3
( −1) ( n + 1)!
n
x n+2
f
(n)
( −1) ( n + 1)! = n + 1 !
( −1) =
( )
n+2
( −1)
n
Notice that all the negative signs will cancel out in the evaluation
...
Here is the Taylor Series for this function
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1
=
x2
∞
∑
n =0
∞
∑
=
f ( n ) ( −1)
n
( x + 1)
n!
( n + 1)!
n!
n =0
=
( x + 1)
∞
∑ ( n + 1)( x + 1)
n
n
n =0
Now, let’s work one of the easier examples in this section
...
x3 − 10 x 2 + 6 about x = 3
...
f ( 0) ( 3) =
−57
f ( 0) ( x ) =
x 3 − 10 x 2 + 6
f ( ) ( x) =
f ( ) ( 3) =
3 x 2 − 20 x
−33
1
1
2
f ( ) ( 3) =
6 x − 20
−2
( x) =
3
3
f ( ) ( x ) 6=
f ( ) ( 3) 6
=
( 4)
(n)
f=
f=
( 3) 0 n ≥ 4
( x) 0
f(
2)
This Taylor series will terminate after n = 3
...
Here is the Taylor Series for this one
...
We leave it like it is
...
The problem is that they are beyond the scope of this course and so aren’t
covered here
...
So, we’ve seen quite a few examples of Taylor Series to this point and in all of them we were able to find
general formulas for the series
...
To see an example of one that doesn’t
have a general formula check out the last example in the next section
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Before leaving this section there are three important Taylor Series that we’ve derived in this section that
we should summarize up in one place
...
∞
ex = ∑
n =0
xn
n!
−1) x 2 n
(
cos x = ∑
( 2n ) !
n =0
n
∞
−1) x 2 n +1
(
sin x = ∑
n = 0 ( 2n + 1) !
∞
© 2018 Paul Dawkins
n
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...
lamar
...
There are in fact many applications of series, unfortunately most of them are beyond the scope of this
course
...
If you are
interested in seeing how that works you can check out that chapter of my Differential Equations notes
...
One of the more
commonly used methods in that subject makes use of Fourier Series
...
In these applications it is very difficult, if not impossible,
to find the function itself
...
While the differential equations applications are beyond the scope of this course there are some
applications from a Calculus setting that we can look at
...
⌠ sin x dx
⌡ x
Solution
To do this we will first need to find a Taylor Series about x = 0 for the integrand
...
We already have a Taylor Series for sine about x = 0 so we’ll just use that as
follows,
2 n +1
∞
−1) x 2 n
(
sin x 1 ∞ ( −1) x
= =
∑ ( 2n + 1)! ∑
x
x n 0=
=
n 0 ( 2n + 1) !
n
n
We can now do the problem
...
This idea of deriving a series representation for a function instead of trying to find the function itself is
used quite often in several fields
...
© 2018 Paul Dawkins
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...
lamar
...
It’s more an application of
partial sums
...
In the Estimating
the Value of a Series we used a partial sum to estimate the value of a series
...
The main difference is that we will now be
using the partial sum to approximate a function instead of a single value
...
Recall that the nth degree Taylor Polynomial of f ( x ) is given by,
Tn ( x )
=
n
∑
i =0
f (i ) ( a )
i
( x − a)
i!
Let’s take a look at example of this
...
Solution
Here is the general formula for the Taylor polynomials for cosine
...
© 2018 Paul Dawkins
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...
lamar
...
In fact, by the time we get to T8 ( x ) the only difference is
right at the ends
...
Also, the larger the interval the higher degree Taylor polynomial we need to get a good
approximation for the whole interval
...
Notice
that because the Taylor series for cosine doesn’t contain any terms with odd powers on x we get the
following Taylor polynomials
...
Sometimes this will happen although that was not
really the point of this
...
It will never be more than n, but it can be less than n
...
However, the previous section was getting too long so the example is in this section
...
So, in that sense it does belong in
this section
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Example 3 Find the first three non-zero terms in the Taylor Series for f ( x ) = e x cos x about x = 0
...
However, as
we noted prior to this example we want to use this example to illustrate how we multiply series
...
n
∞ x n ∞ ( −1) x 2 n
e cos x = ∑ ∑
n ! n 0 ( 2n ) !
=
n 0=
x
We’re not going to completely multiply out these series
...
The problem statement says that we want the first three non-zero
terms
...
If none of
the terms are zero this would mean that the first three non-zero terms would be the constant term, x
term, and x 2 term
...
If we’ve assumed wrong it will be very easy to
fix so don’t worry about that
...
x 2 x 4
x 2 x3 x 4
e x cos x =1 + x + + + + 1 − + +
2 6 24
2 24
Note that we do need to acknowledge that these series don’t stop
...
Just for a second however, let’s suppose that each of these did stop and ask
ourselves how we would multiply each out
...
In other words, we would first multiply every term in
the second series by 1, then every term in the second series by x, then by x 2 etc
...
Do you see why?
Each of the terms that we neglected to write down have an exponent of at least 5 and so multiplying
by 1 or any power of x will result in a term with an exponent that is at a minimum 5
...
So, let’s start the multiplication process
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x2 x4
x 2 x3 x 4
e x cos x =1 + x + + + + 1 − + +
2 6 24
2 24
2
4
3
5
2
4
x
x
x
x
x
x
x6
=1 − + + + x − + + + − + +
24
2
24
2 4 48
2
Second Series × 1
+
Second Series × x
2
Second Series × x 2
x3 x5 x 7
x4 x6
x8
− +
+ + − +
+ +
6
12
144
24
48
576
3
Second Series × x 6
4
Second Series × x 24
Now, collect like terms ignoring everything with an exponent of 5 or more since we won’t have all
those terms and don’t want them either
...
It looks like we over guessed and ended up with four non-zero terms, but that’s okay
...
In other words, there is no reason to completely redo all the work
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Calculus II
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Section 4-18 : Binomial Series
In this final section of this chapter we are going to look at another series representation for a function
...
Binomial Theorem
If n is any positive integer then,
(a + b)
n
n
n n −i i
=
∑
a b
i =0 i
= a n + na n −1b +
where,
n ( n − 1) n − 2 2
a b + + nab n −1 + b n
2!
n n ( n − 1)( n − 2 ) ( n − i + 1)
=
i 1, 2,3, n
=
i!
i
n
=1
0
This is useful for expanding ( a + b ) for large n when straight forward multiplication wouldn’t be easy
n
to do
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Example 1 Use the Binomial Theorem to expand ( 2 x − 3)
4
Solution
There really isn’t much to do other than plugging into the theorem
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There is an extension to this however
that allows for any number at all
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Calculus II
308
If k is any number and x < 1 then,
(1 + x )
where,
k
∞
k n
=
∑
x
n =0 n
k ( k − 1) 2 k ( k − 1)( k − 2 ) 3
=1 + kx +
x +
x +
2!
3!
k k ( k − 1)( k − 2 ) ( k − n + 1)
n 1, 2,3,
=
=
n!
n
k
=1
0
So, similar to the binomial theorem except that it’s an infinite series and we must have x < 1 in order
to get convergence
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Example 2 Write down the first four terms in the binomial series for
Solution
So, in this case k =
1
2
9− x
and we’ll need to rewrite the term a little to put it into the form required
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