   Chapter 11.8, Problem 26E

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.26. ∑ n = 2 ∞ x 2 n n ( ln n ) 2

To determine

To find: The radius of convergence and interval of convergence of the series

Explanation

Given:

The series is n=2x2nn(lnn)2 .

Result used:

(1) Ratio test: If limn|an+1an|=L<1 , then the series n=1an is absolutely convergent.

(2) Definition used: The improper integral abf(x)dx is convergent if the corresponding limit exists.

Calculation:

Let an=x2nn(lnn)2

Then, an+1=x2(n+1)(n+1)(ln(n+1))2 .

Obtain |an+1an| .

|an+1an|=|x2(n+1)(n+1)(ln(n+1))2x2nn(lnn)2|

Take limn on both sides,

limn|an+1an|=limn|x2(n+1)(n+1)(ln(n+1))2x2nn(lnn)2|=limn|x2(n+1)(n+1)(ln(n+1))2n(lnn)2x2n|=limn|x(2n+2)(n+1)(ln(n+1))2n(lnn)2x2n|=limnx2nn+1[lnnln(n+1)]2

On further simplification, obtain the limit

limn|an+1an|=x2limnnn+1[limnlnnln(n+1)]2=x2limnnn+1limn[1n1n+1]2    (by L'Hospital's rule)=x2limnnn+1limn[n+1n]2=x2limn[n+1n]

Apply the limit as shown below

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