Chapter 9.5, Problem 55E

To determine

To find: To find the intervals of converges.

Answer to Problem 55E

The interval of convergence is $\left(-2-\frac{3}{e},-2+\frac{3}{e}\right)$ .

Explanation of Solution

Given:

The given series ${\displaystyle \sum _{n=1}^{\infty }\frac{n^n{(x+2)}^n}{3^{n}n!}}$ .

Formula used:

The ratio test Let ${\displaystyle \sum a_n}$ be a series with positive terms and with $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$ .

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

Calculation:

Given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{n^n{(x+2)}^n}{3^{n}n!}}$ .

The series of absolute values is,

  ${\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|={\displaystyle \sum _{n=1}^{\infty }\frac{n^n{(x+2)}^n}{3^{n}n!}}}$

Using the ratio test,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{\lim }\frac{n^{n+1}{(x+2)}^{n+1}}{3^{n+1}(n+1)!}\frac{3^n(n)!}{n^n{(x+2)}^n}\\ \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\frac{\left|x+2\right|}{3}\underset{n\to \infty }{\lim }\frac{{(n+1)}^n}{n^n}\\ \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\frac{\left|x+2\right|}{3}\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n\\ \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\frac{\left|x+2\right|}{3}\times e\end{array}$

The series converges absolutely for,

  $\begin{array}{l}\frac{\left|x+2\right|}{3}\times e<1\\ \left|x+2\right|<\frac{e}{3}\end{array}$

The diverges for ,

  $\begin{array}{l}\frac{\left|x+2\right|}{3}\times e>1\\ \left|x+2\right|>\frac{e}{3}\end{array}$

The radius of convergence is $\frac{3}{e}$ .

The interval of convergence is $\left(-2-\frac{3}{e},-2+\frac{3}{e}\right)$