Convergence and Radius of Convergence of Series

This text explains the convergence and radius of convergence of different series using various tests such as comparison test, alternating series test, ratio test, and integral test. It also provides solved examples for each test. The text covers series with polynomial, exponential, and trigonometric functions.

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