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Sequences and Series of Functions
Li Jiayi, Joanna
14 July, 2014

1

Contents
1

Review: Sequences and Series of Real Numbers

4

1
...


4

1
...
1

Sequences
...
1
...


4

1
...
3

Monotone Sequences
...
1
...


6

1
...
5

Cauchy’s Convergence Criterion
...


7

1
...
1

Series
...
2
...


8

1
...
3

Absolute and Conditional Convergence
...
2

2

Sequences and Series of Functions

9

2
...


9

2
...
11
2
...
1

Uniform convergence and boundedness
...
2
...
14

2
...
3

Uniform convergence and Riemann integration
...
2
...
17

2

2
...
20
2
...
1

Some tests for uniform convergence of series of functions
...
3
...
20

3

1

Review: Sequences and Series of Real Numbers

1
...
1
...
1
...
e
...

In other words, a sequence can be written as f (1), f (2), f (3),
...
The number an is called the n-th term of the sequence
...
2
...

n
converges to 1, that is, prove that
n+1
n
lim
=1
x→∞ n + 1

Example 1
...
Prove that the sequence

Proof
...

n+1
n+1
1
1
We first note that
< for any N ∈ N
...
It follows that if n > N , then
N
1
1
1
< <
<
n+1
n
N
and so we are done
...
1
...
1 (Uniqueness of Limits)
...
Then the limit of
the sequence is unique
...
3
...
Similarly, a sequence {an } is bounded below if there exists m ∈ R such that
an > m for all n ∈ N
...

Theorem 1
...
If {an } is a convergent sequence,
then it is bounded
...
3 (Arithmetic Properties of Convergent Sequences))
...


Suppose lim an = A and lim bn = B for some A, B ∈ R
...

n→∞

n→∞

limn→∞ an
A
an
= , provided that bn 6= 0 for all n ∈ N and B 6= 0
...
4 (Sandwich Theorem)
...
Suppose that
an ≤ b n ≤ c n
for all n ∈ N, and that lim an = lim cn = L
...

n→∞

1
...
3

n→∞

n→∞

Monotone Sequences

Definition 1
...
Let {an } be a sequence
...
We say that {an } is a monotone if it is
either increasing or decreasing
...
5
...


(b) If {an } is decreasing and bounded below, then {an } converges
...
1
...
5
...
Then the sequence
an1 , an2 , an3 · · ·
is called a subsequence of {an } and is denoted by {ank }, where k ∈ N indexes the subsequence
...
6
...

Theorem 1
...
Let {an } be a sequence
...
,
Proof
...
In other words, the term ak is dominant if ak ≥ am for all m ≥ k
...
Then we have a
infinite subsequence an1 , an2 , an3 ,
...
Similarly, we have
an1 ≥ an2 ≥ an3 ≥ · · ·
Hence we have a decreasing subsequence ank and we are done
...
(including the
case where there is no dominant term)
...
Now an1 is not dominant and so there must exist n2 > n1 such that an1 ≤ an2
...

Theorem 1
...
Let {an } be a bounded sequence
...

6

1
...
5

Cauchy’s Convergence Criterion

Definition 1
...
A sequence {an } is called a Cauchy sequence if ∀ > 0, ∃N ∈ N such that
if n, m > N , then |an − am | < 
...
9
...
Then {an } is a Cauchy sequence
...
10
...
Then {an } is a convergent sequence
...
2

Series

1
...
1

Series

Definition 1
...
Let {an } be a sequence of real numbers
...


...


n=1

The term an is called the n-th term of the series and the term sk is called the k-th partial
∞
X
sum of the series
...
We write
∞
X

an = L

n=1

7

and we refer to L as the sum of the series
...

n=1

1
...
2

Some tests for convergence

Theorem 1
...
The series

∞
X

an converges if

n=1

and only if for any  > 0 there exists N ∈ N such that if n > m > N , then
|am+1 + am+2 + · · · + an | < 
...
12 (Comparison Test)
...

(a) If

∞
X

bn , then

(b) If

an converges

n=1

n=1
∞
X

∞
X

an diverges, then

∞
X

bn diverges
...
13 (Ratio Test)
...
Suppose that
an
n→∞ an+1
lim

exists and equals r
...


n=1

1
...
3

Absolute and Conditional Convergence

Definition 1
...
Let {an } be a sequence of real numbers
...
The series

n=1

∞
X
n=1
∞
X

an is said to be
an is said to be

n=1

conditionally convergent if it converges but does not converge absolutely, that is,
verges, but

∞
X

∞
X
n=1

|an | diverges
...
In most cases, X would be a subset of R
...
In other words, if the sequence
of functions under consideration is bounded, continuous, differentiable or integrable, and if
the limit function of the sequence exists, does it carry these properties?

2
...
1
...
We can
define a function f by
f (x) = lim fn (x), x ∈ E
...
Sometimes we shall use a more descriptive terminology and shall
say that ”{fn } converges for every x ∈ E”, and if we define
f (x) = lim

n→∞

n
X

fi (x), x ∈ E,

i=1

then the function f is called the sum of the series

X

fn
...
1
...
For instance, if the
functions {fn } are bounded, continuous, differentiable, or integrable, is the same true of the
9

limit function? What are the relations between fn0 and f 0 , say, or between the integrals of fn
and that of f ?
To say that f is continuous at a limit point x means
lim f (t) = f (x)
...
e
...

We shall now show by means of several examples that limit processes cannot in general
be interchanged without affecting the result
...

Example 2
...
For m = 1, 2, 3, · · · , n = 1, 2, 3, · · · , let
m

...


n→∞ m→∞

On the other hand, for every fixed m,
lim sm,n = 0,

n→∞

so that
lim lim sm,n = 0
...
2
...

(1 + x2 )n

Since fn (0) = 0, we have f (0) = 0
...
Hence



0
if x = 0,
f (x) =


1 + x2 if x 6= 0,
so that a convergent series of continuous functions may have a discontinuous sum
...
2
...
1, which will enable us to arrive at positive results
...
2

Uniform convergence of sequences of functions

Definition 2
...
We say that a sequence of functions {fn }, n = 1, 2, 3, · · · , converges unif ormly
on E to a function f if for every  > 0 there exists N ∈ N such that n ≥ N implies
|fn (x) − f (x)| ≤ 
for all x ∈ E
...
1 (Cauchy Critirion ,)
...

Proof
...
Then, given  > 0, we can find N such that

n > N implies |fn (x) − f (x)| < for all x ∈ E
...
Thus it follows from the triangular inequality
2
that
|fn (x) − fm (x)| < |fn (x)| − f (x)| + |f (x) − fm (x)| < 
...
Then, for each x ∈ E, the sequence {fn (x)} converges
...
We have to prove that the
n→∞

convergence is uniform
...
, |fn (x) − fn+k (x)| <


2

for all x ∈ E
...
Hence, n > N implies |fn (x) − f (x)| <  for all x ∈ E
...

Theorem 2
...
Suppose
lim fn (x) = f (x) (x ∈ E)
...

x∈E

Then fn → f uniformly on E if and only if Mn → 0 as n → ∞
...
3
...
Prove that {fn } converges pointwise but not uniformly on R
...
Since lim fn (x) exists for all x ∈ R, according to pointwise convergence, the limit
n→∞

function f is given by



0




f (x) = 1

2




1
12

if |x| < 1
if |x| = 1
if |x| > 1

Put
Mn = sup |fn (x) − f (x)|
...
Thus, it follows from Theorem 2
...


2
...
1

Uniform convergence and boundedness

Theorem 2
...
If {fn } is a sequence of bounded functions on E, and if fn → f uniformly on
E, then f is bounded on E
...
Let  = 1
...
Particularly, consider n = N + 1
...
Since fN +1 is bounded on E, there exists B > 0 such that
|fN +1 (x)| < B
for all x ∈ E
...
This proves that f is bounded on E
...
4
...




n if x ∈ (0, 1 ]
n
Consider fn (x) =

1
1


if x ∈ ( , 1]
...

Since ∀n ∈ N, |fn (x)| < n + 1 for all x ∈ (0, 1], all the functions in the sequence are
bounded
...


2
...
2

Uniform convergence and continuity

Theorem 2
...
Suppose fn → f uniformly on a set E
...


t→x

Then {An } converges, and
lim f (t) = lim An
...
Let  > 0 be given
...

Letting t → x, we obtain
|An − Am | ≤ 
14

for n ≥ N, m ≥ N , so that {An } is a Cauchy sequence and therefore converges, say to A
...

We first choose n such that

3

|f (t) − fn (t)| ≤

for all t ∈ E (this is possible by the uniform convergence), and such that

|An − A| ≤
...
Substituting the inequalities above, we see that
|f (t) − A| ≤ ,
provided that t ∈ V ∩ E, t 6= x
...

t→x

n→∞

Theorem 2
...
If {fn } is a sequence of continuous functions on E, and if fn → f uniformly
on E, then f is continuous on E
...
3
...


x→x0 n→∞

n→∞ x→x0

If x0 is an isolated point of E, then f is automatically continuous at x = x0
...
5 follows directly from Theorem 2
...

Note 2
...
Is the converse of Theorem 2
...
e
...

In Example 2
...

15

Example 2
...
Consider {fn } defined on [0, 1] such that fn (x) = n2 x(1 − x)n for all n ∈
N, x ∈ [0, 1]
...

Obviously, each term in the sequence, as well as the limit function, is continuous on [0, 1]
...
Since lim Mn > 0, it follows from Theorem 2
...


2
...
3

Uniform convergence and Riemann integration

Theorem 2
...
Suppose fn ∈ R([a, b]) for n = 1, 2, 3,
...
Then f ∈ R([a, b]), and
b

Z

Z
f = lim

n→∞

a

b

fn
...
)
Proof
...
Put
n = sup |fn (x) − f (x)|,
the supremum being taken over a ≤ x ≤ b
...

(fn − n ) ≤ f ≤ f ≤
a

a

Hence
Z
0≤

Z
f −

f ≤ 2n (b − a)
...

Thus f ∈ R([a, b])
...

f −
|
a

Z

b

Z
f = lim

This implies

n→∞

a

a

b

fn
...
5 (Riemann integrability)
...
The definition is as follows:
Let f : [a, b] −→ R be a bounded (not necessary continuous) function on [a, b]
...

In this case, the common value of U (f ) and L(f ) is called the Riemann integral of f on [a, b],
Z b
and is denoted by
f (x)dx
...
2
...
3, 2
...
6, one might expect the following result to hold:
Theorem 2
...
Suppose that {fn }is a sequence of functions, differentiable on [a, b] and
such that {fn (xo )} converges for some point x0 on [a, b]
...

n→∞

Actually, in Example 2
...

Example 2
...
A sequence of differentiable functions {fn } with limit 0 for which {fn0 } diverges
...

n
Then lim fn (x) = 0 for all x ∈ R
...

n→∞

Theorem 2
...
Assume that each term of {fn } is a real-valued function having a finite
derivative at each point of an open interval (a, b)
...
Assume further that there exists a function g such
that fn0 → g uniformly on (a, b)
...

b) For each x in (a, b) the derivative f 0 (x) exists and equals g(x)
...
Assume that c ∈ (a, b) and define a new sequence {gn } as follows:


f (x) − fn (c)

 n
if x 6= c,
x−c
gn (x) =


f 0 (c)
if x = c
...
Convergence of {gn (c)} follows
from the hypothesis, since gn (c) = fn0 (c)
...
If x 6= c, we have
gn (x) − gm (x) =

hn (x) − hm (x)
,
x−c

0
where h(x) = fn (x)−fm (x)
...


Applying the Mean Value Theorem, we get
0
(x1 )
gn (x) − gm (x) = fn0 (x1 ) − fm

where x1 lies between x and c
...

Now we can show that {fn } converges uniformly on (a, b)
...

We can write
fn (x) = gn (x)(x − x0 ) + fn (x0 )
an equation which holds for every x ∈ (a, b)
...

This equation, with the help of the Cauchy condition, establishes the convergence of {fn } on
(a, b)
...

To prove (b), return to the sequence {gn } defined for an arbitrary point c in (a, b) and let
G(x) = lim gn (x)
...
In other
n→∞

x→c

words, each gn is continuous at c
...
This means that
G( c) = lim G(x),
x→c

the existence of the limit being part of the conclusion
...

n→∞
x−c
x−c

G(x) = lim gn (x) = lim
n→∞

Hence, the derivative f 0 (c) exists and equals G(c)
...
Since c is an arbitrary point of (a, b), this proves (b)
...
6 (Mean Value Theorem)
...
There exists c ∈ (a, b) such that
f 0 (c) =

f (b) − f (a)

...
3

Uniform convergence of series of functions

Definition 2
...
Given a sequence {fn } of functions defined on a set S
...


k=1

If there exists a function f such that sn → f uniformly on S, we say the series

X

fn (x)

converges uniformly on S and we write
∞
X

fn (x) = f (x) (unif ormly on S)
...
3
...
9 (Cauchy Condition for uniform convergence of series)
...
, and f or every x in S
...
10 (Weierstrass M-test)
...
3
...
and f or every x in S
...


the Power Series

In this section, we shall derive some properties of functions which are represented by power
series, i
...
, functions of the form
f (x) =

∞
X
n=0

20

cn x n

or, more generally,
f (x) =

∞
X

cn (x − a)n
...

We shall restrict ourselves to real values of x
...

If (*) converges for |x − a|, R, f is said to be expanded in a power series about the point
x = a
...

1
...
11
...


x=0

Then (*) converges uniformly on [−R + , R − ], no matter which  > 0 is chosen
...


n=0

Theorem 2
...
Suppose

X

cn converges
...


n=0

Then
lim f (x) =

x→1

∞
X

cn
...
13
...
If −R < a < R, then f can be expanded in a power
series about the point x = a which converges in |x − a| < R − |a|, and
f (x) =

∞
X
f (n) (a)
n=0

n!

(x − a)n (|x − a| < R − |a|)
...
7
...

2

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