Chapter 10.5, Problem 69E

To determine

To find : the radius of convergence of the series.

Answer to Problem 69E

The radius of convergence is : $2<x<3$ Option $(\text{A})$

Explanation of Solution

Given information:

Given expression is : ${\displaystyle \sum }_{n=0}^{\infty }\frac{n{(2x-5)}^n}{n+2}$

Calculation:

The given series is

  ${\displaystyle \sum }_{n=0}^{\infty }\frac{n{(2x-5)}^n}{n+2}$

The series of absolute values is

  ${\displaystyle \sum }_{n=0}^{\infty }\left|a_n\right|={\displaystyle \sum }_{n=0}^{\infty }\frac{n|2x-5{|}^n}{n+2}$

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.

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

The series converges absolutely for

  $|2x-5|<1$

  $\left|x-\frac{5}{2}\right|<\frac{1}{2}$

  $2<x<3$

And the series diverges at the endpoints by the $n$ th Term Test. So the interval of convergence is:

  $2<x<3$

Hence, $2<x<3$ Option $(\text{A})$ is the correct answer