Chapter 10.5, Problem 44E

To determine

the interval of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{x}^n}{4^n\left(n^2+1\right)}}$

Answer to Problem 44E

The interval of convergence and the series converges absolutely on $\left[-4,4\right)$

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{n=0}^{\infty }\frac{n{x}^n}{4^n\left(n^2+1\right)}}$

Formula used:

The ratio test is used.

Calculation:

The series of absolute values is

  ${\displaystyle \sum _{n=0}^{\infty }\left|a_n\right|}=\frac{n{\left|x\right|}^n}{4^n\left(n^2+1\right)}$

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,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left(n+1\right){\left|x\right|}^{n+1}}{4^{n+1}\left[{\left(n+1\right)}^2+1\right]}\times \frac{4^n\left(n^2+1\right)}{n{\left|x\right|}^n}\\ =\frac{\left|x\right|}{4}\underset{n\to \infty }{\lim }\left(\frac{n^3+n^2+n+1}{n^3+2n^2+2n}\right)\\ \underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\left|x\right|}{4}\end{array}$

The series converges absolutely for

  $\begin{array}{l}\frac{\left|x\right|}{4}<1\\ \left|x\right|<4\\ -4<x<4\end{array}$

And diverges for

  $\begin{array}{l}\frac{\left|x\right|}{4}>1\\ \left|x\right|>4\\ x<-4,x>4\end{array}$

And when x = 5, the series is

  ${\displaystyle \sum _{n=0}^{\infty }\frac{n{x}^n}{4^n\left(n^2+1\right)}=}{\displaystyle \sum _{n=0}^{\infty }\frac{n}{n^2+1}}$

This diverges by the limit comparison test (Use ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ which converges absolutely by the p-test)

And when x = -4, the series is

  ${\displaystyle \sum _{n=0}^{\infty }\frac{n{x}^n}{4^n\left(n^2+1\right)}=}{\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}n}{n^2+1}}$

This does not converge absolutely, but it converges conditionally by the alternating series test.

(a) Interval of convergence: $\left[-4,4\right)$

(b) Series converges absolutely on $\left[-4,4\right)$

(c) Series converges conditionally at $x=-4$

Conclusion:

The interval of convergence and the series converges absolutely on $\left[-4,4\right)$