Determining Convergence and Intervals of Power Series in Calculus

This assignment solution addresses two main problems in calculus: determining the convergence of infinite series and finding the radius and interval of convergence for power series. The first problem analyzes the convergence of various series using the comparison test, integral test, and alternating series test. The second problem focuses on finding the radius and interval of convergence for different power series, including those derived from known functions. The solution employs techniques such as the ratio test and considers conditional convergence where applicable. The assignment covers a range of series types, including those involving factorials, alternating terms, and more complex expressions, providing a comprehensive analysis of convergence properties.

Order Id : 814869
Question 1: Test for convergence:
i) ∑
n =1
∞ 1
n2 5n Comparison test
∑
n =1
∞ 1
n2 5n =∑
n=1
∞ 1
n2 ∗∑
n=1
∞ 1
5n
Using the p-series
∑
n =1
∞ 1
n2 Converges since ∑
n =1
∞ 1
np and p>1
∑
n =1
∞ 1
5n Converges
Therefore ∑
n =1
∞ 1
n2 5n converges
ii) ∑
n =1
∞ n
( n2+2 ) 3 comparison Test
∑
n =1
∞ n
( n2+2 ) 3 Written in form of ∑ an ∑ bnand are two positive sequence: if ∑ bn
converges so does ∑ an
∑
n =1
∞ n
( n2+2 ) 3 <∑
n=1
ê n
n6 And when factored, then ∑
n =1
ê 1
n5
Using the p-series ∑
n =1
ê 1
n5 converges
Thus, ∑
n =1
∞ n
( n2+2 ) 3 converges
iii) ∑
n =1
∞ n
en Integral Test

f ( x )= n
en If ∫
n=1
∞
f ( x)dx converges, so does ∑
n =1
∞ n
en
∫
n=1
∞
f (x)dx= ∫
n=1
∞
n
en dn=2
e , so it converges at 2
e
Therefore, ∑
n =1
∞ n
en converges
iv) ∑
n =1
∞ (−1 )n +1
√n Alternating series test
Suppose that for anthere exist an N so that for all n ≥ N
an is positive and monotone decreasing and lim
n → ∞
an =0 then the alternating series
∑ (−1 )n
naand ∑ (−1 )n−1 an both converges
an= 1
√ n is positive and monotone decreasing
lim
n → ∞ ( 1
√n )=0 then by Alternating series test criteria ∑
n =1
∞ (−1 )n +1
√n converges
v) ∑
n =1
∞
(−1 )n+1 n
n+1 Alternating series test
Suppose that for anthere exist an N so that for all n ≥ N
an is positive and monotone decreasing and lim
n → ∞
an =0 then the alternating series
∑ (−1 )n
naand ∑ (−1 )n−1 an both converges
an= n
n+1is positive and not-monotone decreasing
lim
n → ∞ ( n
n+1 ) ≠ 0 Then by Alternating series test criteria ∑
n =1
∞
(−1 )n+1 n
n+1 diverges
Question 2: Determining the radius and interval of convergence

i) ∑
n=0
∞ xn
7n
lim
n → ∞ (|( xn
7n ) 1
n
|) :| x
7 |
The sum converges for L<1, solving for | x
7 |<1
The radius of convergence is R=7
| x
7 |<1Thus the interval is −7< x <7
The sum may converge for L=1, therefore the convergence for | x
7 |=1
For x=7, ∑
n=0
∞ 7n
7n : diverges
For x=-7, ∑
n=0
∞ (−7)n
7n : diverges
Therefore, the convergence interval ∑
n=0
∞ xn
7n is −7<x <7
ii) ∑
n=0
∞ ( x+3)n
5n
lim
n → ∞ (|((x +3)n
5n )1
n
|): |x +3
5 |
The sum converges for L<1, solving for | x+ 3
5 |< 1
The radius of convergence is R=5
| x+ 3
5 |< 1Thus the interval is −8< x <2

The sum may converge for L=1, therefore the convergence for | x+3
5 |=1
For x=2, ∑
n=0
∞ (2+ 3)n
5n : diverges
For x=-8, ∑
n=0
∞ ((−8)+3)n
5n : diverges
Therefore, the convergence interval ( x +3)n
5n is −8< x <2
iii) ∑
n =1
∞ ( −1 ) n
n ( x−π ) n
Using the ratio test
an +1
an
=
( −1 ) n+1
n+1 ( x−π ) n +1
( −1 ) n
n ( x−π ) n
Computing the limit
lim
n → ∞ (| an+1
an |)=lim
n → ∞
(| ( −1 ) n+1
n+ 1 ( x−π ) n+1
( −1 ) n
n ( x−π ) n |) :|x−π |
The sum converges for L<1, solving for |x−π |<1
The radius of convergence is R=1
|x−π |<1 Thus the interval is π−1< x <π +1
For x=π +1 : ∑
n =1
∞ ( −1 ) n
n ( (π +1)−π ) n
converges (conditionally)
For x=π−1 : ∑
n =1
∞ (−1 )n
n ( ( π−1)−π )n diverges

Therefore, the convergence interval ( x +3)n
5n is π−1< x ≤ π +1
iv) 5
2+ 8 ( x−1 ) 2
5
2+ 8 ( x−1 )2 Can be written as ∑
n=0
∞
1
2 ( xn
5n )
Now using the ratio test:
an +1
an
=
1
2 ( xn+1
5n+1 )
1
2 ( xn
5n )
Computing the limit
lim
n → ∞ (| an+1
an |)=lim
n → ∞
(| 1
2 ( xn+1
5n+1 )
1
2 ( xn
5n ) |) :| x
5|
The sum converges for L<1, solving for | x
5 |<1
The radius of convergence is R=5
| x
5 |<1 Thus the interval is −5< x <5
Forx=5 : ∑
n=0
∞
∑
n=0
∞
1
2 ( 5n
5n ) diverges
For x=−5 : ∑
n=0
∞ 1
2 ((−5)n
5n )diverges
Therefore, the convergence interval of 5
2+ 8 ( x−1 ) 2 is −5< x <5

v) tan−1
( x
3 )
We know that tan−1 (u )=∑
n=0
∞ (−1 )n
2 n+1 u2 n+1
Replace u with x
3 gives ∑
n=0
∞ ( −1 ) n
2 n+1 ( x
3 )
2 n+1
Now using the ratio test:
an +1
an
=
(−1 )n+ 1
2(n+1)+ 1 ( x
3 )
2(n+1 )+1
(−1 )n
2n+1 ( x
3 )
2 n+1
Computing the limit
lim
n → ∞ (| an+1
an |)=lim
n → ∞
(| ( −1 ) n +1
2(n+1)+1 ( x
3 )
2(n+ 1)+1
( −1 ) n
2 n+1 ( x
3 )
2 n+1
|) :| x2
9 |
The sum converges for L<1, solving for | x2
9 |<1
The radius of convergence is R=3
|32
9 |< 1 Thus the interval is −3< x <3
Forx=3 : ∑
n=0
∞ ( −1 ) n
2 n+1 (3
3 )
2 n+1
converges (conditionally)
For x=−3 : ∑
n=0
∞ (−1 )n
2 n+1 (−3
3 )
2 n+1
converges (conditionally)
Therefore, the convergence interval of tan−1
( x
3 )is −3 ≤ x ≤ 3

vi) 2 x
( 1−x2 ) 2
2 x
( 1−x2 )2 Can be written as ∑
n=0
∞
( 2n+2 ) x2 n+1
Now using the ratio test:
an +1
an
= ( 2( n+ 1)+2 ) x2(n+1)+1
( 2 n+2 ) x2 n +1
Computing the limit
lim
n → ∞ (|an+1
an |)=lim
n → ∞ (| ( 2( n+1)+2 ) x2(n+1 )+1
( 2 n+2 ) x2 n+1 |):|x|2
The sum converges for L<1, solving for |x|2 <1
The radius of convergence is R=1
|x|
2 <1 Thus the interval is −1< x <1
For x=1 : ∑
n=0
∞
( 2n+2 ) 12 n+1 diverges
For x=−1 : ∑
n=0
∞
( 2n+2 ) (−1)2 n+ 1diverges
Therefore, the convergence interval of 2 x
( 1−x2 ) 2 is −1< x <1