   Chapter 11.8, Problem 28E

Chapter
Section
Textbook Problem

Find the radius of convergence and interval of convergence of the series. ∑ n = 1 ∞ n ! x n 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ ⋅ ( 2 n − 1 )

To determine

To find:

The radius of convergence and the interval of convergence of the series

n=1n!xn1·3·5··(2n-1)

Explanation

1) Concept:

i) For a power series n=0cnx-an, there is a positive number R such that the series converges if x-a<R and diverges if x-a>R, this number R is called as a radius of convergence. From this, there are four possible cases of interval of convergence

a-R, a+R,  a-R, a+R,  a-R, a+R,  a-R, a+R

ii) The ratio test state, that if limnan+1an<1, then the series n=1an converges.

2) Given:

n=1n!xn1·3·5··(2n-1)

3) Calculation:

The given series is n=1n!xn1·3·5··(2n-1)

Therefore, the nth term is an=n!xn1·3·5··(2n-1)

Therefore,

limnan+1an=limnn+1!xn+11·3·5··(2n-1)·(2n+1) ·1·3·5··(2n-1)n!xn

limnan+1an=

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