Chapter 10.5, Problem 56E

To determine

To find : the radius of convergence of the series.

Answer to Problem 56E

The interval of convergence is $(-5e,5e)$

Explanation of Solution

Given information:

Given expression is : ${\displaystyle \sum }_{n=1}^{\infty }\frac{n!{(x)}^n}{n^n{5}^n}$

Calculation:

The given series is

  ${\displaystyle \sum }_{n=1}^{\infty }\frac{n!{(x)}^n}{n^n{5}^n}$

The series of absolute values is

  ${\displaystyle \sum }_{n=1}^{\infty }\left|a_n\right|={\displaystyle \sum }_{n=1}^{\infty }\frac{n!|x{|}^n}{n^n{5}^n}$

The ratio Test: Let $\sum a_n$ be a series with positive terms, and with

  $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$

Then, the series converges if $L<1,$ the series diverges if $L>1$ and the test is inconclusive if $L=1$

Using the ratio test, we check for absolute convergence as follows:

  $\begin{array}{cc}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}& =\underset{n\to \infty }{\lim }\frac{(n+1)!|x{|}^{n+1}}{{(n+1)}^{n+1}{5}^{n+1}}\times \frac{n^n{5}^n}{n!|x{|}^n}\\ & =\frac{\left|x\right|}{5}\underset{n\to \infty }{\lim }{\left(\frac{n}{n+1}\right)}^n\\ & =\frac{\left|x\right|}{5}\underset{n\to \infty }{\lim }\frac{1}{{\left(1+\frac{1}{n}\right)}^n}=\frac{\left|x\right|}{5e}\end{array}$

The series converges absolutely for

  $\frac{\left|x\right|}{5e}<1$

  $\left|x\right|<5e$

  $-5e<x<5e$

And diverges for

  $\frac{\left|x\right|}{5e}>1$

  $\left|x\right|>5e$

And the radius of convergence is $5e$

Therefore, the interval of convergence is $(-5e,5e)$