   Chapter 11, Problem 42RE

Chapter
Section
Textbook Problem

# Find the radius of convergence and interval of convergence of the series.42. ∑ n = 1 ∞ 2 n ( x − 2 ) n ( n + 2 ) !

To determine

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

Explanation

Given:

The series is n=12n(x2)n(n+2)! .

Result used:

(1)Ratio test:

If limn|an+1an|=L<1 , then the series n=1an is absolutely convergent.

Calculation:

Let an=2n(x2)n(n+2)! .

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

Obtain |an+1an| to apply the Ratio test.

|an+1an|=|2n+1(x2)n+1(n+3)!2n(x2)n(n+2)!|

Take limn on both sides,

limn|an+1an|=limn|2n+1(x2)n+1(n+3)!2n(x2)n(n+2)!

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