Chapter 10.5, Problem 43E

To determine

The interval of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$

Answer to Problem 43E

The interval of convergence and the series converges absolutely on $\left[-8,2\right]$

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$

Formula used:

The ratio test is used.

Calculation:

The series of absolute values is

  ${\displaystyle \sum _{n=0}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=0}^{\infty }\frac{n{\left|x+3\right|}^n}{5^n}}$

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, the series converges if L < I, 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,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left(n+1\right){\left|x+3\right|}^{n+1}}{5^{n+1}}\times \frac{5^n}{n{\left|x+3\right|}^n}\\ =\frac{\left|x+3\right|}{5}{x}^2\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\\ \underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\left|x+3\right|}{5}\end{array}$

The series converges absolutely for

  $\begin{array}{l}\frac{\left|x+3\right|}{5}<1\\ \left|x+3\right|<5\\ -5<x+3<5\\ -8<x<2\end{array}$

And diverges for

  $\begin{array}{l}\frac{\left|x+3\right|}{5}>1\\ \left|x+3\right|>5\\ x<-8,x>2\end{array}$

And when |x + 3| = 5, the series diverges by the nth-term test

(a) Interval of convergence: $\left[-8,2\right]$

(b) Series converges absolutely on $\left[-8,2\right]$

(c) None

Conclusion:

The interval of convergence and the series converges absolutely on $\left[-8,2\right]$