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Title: Series convergence tests
Description: Series convergence tests, done rigorously and through examples
Description: Series convergence tests, done rigorously and through examples
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Applied Mathematics
Advanced Calculus and
Methods of Mathematical
Physics
Homework 2
Author: Lirik Maxhuni
Jacobs University
Problem 1
We start the problem by writing the ratio test:
lim |
k→∞
ak+1
| = r < 1,
ak
and r ∈ R+
This means that
|
ak+1
|
for k large enough, k ≥ N, where N is a arbitrary number
If we try now for:
-k = N
|
aN+1
| < R ⇒ |aN+1 | < |aN | ∙ R
aN
-k = N + 1
|
aN+2
| < R ⇒ |aN+2 | < |aN+1 | ∙ R < |aN | ∙ R2
aN+1
|
aN+3
| < R ⇒ |aN+3 | < |aN+2 | ∙ R < |aN | ∙ R3
aN+2
-k = N + 2
Let’s add the terms in the left-hand side
|aN | + |aN+1 | + |aN+2 | + |aN+3 | + |aN+4 | + ⋯
< |aN | + |aN |R + |aN |R2 + |aN |R3 + ⋯
= |aN | (1 + R + R2 + R3 + ⋯ )
This can be rearranged into sigma notation:
∞
∞
∑ |aN | = |aN | ∙ ∑ Rn
k=N
k=0
geometric series
Since above we defined R < 1, and following this statement, the right-hand
side always converges (property of geometric series)
...
We took N to be any number which fulfilled the condition needed, so it is
basically an arbitrary real number
...
Therefore, the convergence part of the ratio test is proved
...
∞
∑(−1)k
k=0
1
k+1
This is an alternating series, therefore the most convenient way to check if
it converges or not is the Liebniz alternating series test
...
e
...
|
1
1
|≤|
|
k+2
k+1
and
1
=0
k→∞ k + 1
lim
Both conditions hold, so the series is convergent
...
∞
∑
k=1
1
k2
For this kind of series, the best tool to use in order to know if it is
convergent or not is the integral test
...
∞
∫
1
1
1 ∞
dk
=
−
| =1
k2
k 1
Result of the improper integral is a finite number, so it converges
...
c)
We are going to check if the series below converges
...
ek+1
ek+1 ∙ k
ek ∙ e ∙ k
e∙k
k
+
1
lim | k | = lim |
=
lim
=
lim
|
|
|
|
|
k→∞
k→∞ (k + 1) ek
k→∞ (k + 1)ek
k→∞ (k + 1)
e
k
k
= e lim
= e∙1= e
k→∞ k + 1
ek+1
lim | k +k 1 | = e > 1
k→∞
e
k
Hence, the series diverges
...
a)
∞
∑ xk
k=0
k
lim √|x|k = lim |x| = |x| < 1
k→∞
k→∞
Thus, the radius of convergence is 𝜌: = |x| < 1
b)
∞
∑
(−1)k
k=1
x 2k
(2k)!
For this one we will run the ratio test
...
Problem 4
𝑓(𝑥) = ln(1 + 𝑥) ,
𝑓 ′ (𝑥) =
𝑐=0
1
= (1 + 𝑥)−1
1+𝑥
𝑓 ′′ (𝑥) = −(1 + 𝑥)−2
𝑓 (3) (𝑥) = 2(1 + 𝑥)−3
𝑓 (𝑘) (𝑥) = (−1)𝑘−1 (𝑘 − 1)! (
1
)
(1 + 𝑥)𝑘
So,
𝑓 (𝑘) (0) = (−1)𝑘−1 (𝑘 − 1)! ,
𝑓𝑜𝑟 𝑘 ≥ 1
If 𝑐 = 0 , then we obtain
𝑓(𝑐) = 𝑓(0) = ln(1 + 0) = ln 1 = 0
Taylor Series for the function 𝑓(𝑥) = ln(1 + 𝑥) , 𝑐 = 0 is then:
𝑛
𝑓(𝑥) = ∑
𝑘=1
(−1)𝑘−1 (𝑘 − 1)! 𝑘
𝑥
𝑘!
𝑛
(−1)𝑘−1 𝑘 (−1)𝑘−1
1
=∑
𝑥 +
∙
𝑥 𝑛+1 = 𝑙𝑛(1 + 𝑥)
𝑛+1
(1
)
𝑘
𝑛+1
+ 𝜓𝑥
𝑘=1
𝜓𝑥 ∈ (0, 𝑥) 𝑜𝑟 𝜓𝑥 ∈ (𝑥, 0)
Question:
When does the remainder go to 0?
Or, for which x does
lim 𝐸𝑛 (𝑥) ⟶ 0
𝑛→∞
𝑛+1
(−1)𝑛
𝑥
lim 𝐸𝑛 (𝑥) = lim
=0
(
)
𝑛→∞
𝑛→∞ 𝑛 + 1 1 + 𝜓𝑥
If the equation above is true
⇒0<
𝑥
< 1 ⇒ 𝑥 − 𝜓𝑥 < 1 ,
1 + 𝜓𝑥
𝑤𝑖𝑡ℎ 𝜓𝑥 ∈ (0, 𝑥)
⇒𝑥≤1
Here:
lim 𝐸𝑛 (𝑥) = 0 , 𝑖𝑓𝑓 0 < 𝑥 ≤ 1
𝑛→∞
This means the Taylor Series represents the function 𝑓(𝑥) = ln(1 + 𝑥) for 𝑥 ∈
[0,1]
...
So, indeed this series fully represents the function 𝑓(𝑥) = ln(1 + 𝑥) in the
interval 𝑥 ∈ [−1, 1]
...
So, if
g(x) =
f(x)
(1 + x)𝛼
then we have:
g(x) =
(1 + x)𝛼
=1
(1 + x)𝛼
c)
In part a) was already concluded that f(x) = (1 + x)𝛼
Title: Series convergence tests
Description: Series convergence tests, done rigorously and through examples
Description: Series convergence tests, done rigorously and through examples