Chapter 9.5, Problem 44E

a.

To determine

To find: The interval of convergence of the series.

Answer to Problem 44E

The interval of convergence of the series is $-4<x<4$ .

Explanation of Solution

Given information:

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

Calculation:

By applying the Ratio Test and looking at the interval's endpoints, one can determine the interval where the series will converge.

Taking the absolute value of the terms in the series and looking at the following limit, only using the Ratio Test for series with non-negative terms.

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\left|\frac{\left(n+1\right)x^{n+1}}{4^{n+1}\left({\left(n+1\right)}^2+1\right)}\right|}{\left|\frac{n{x}^n}{4^n\left(n^2+1\right)}\right|}=\underset{n\to \infty }{\lim }\left(\frac{\left(n+1\right){\left|x\right|}^{n+1}}{4^{n+1}\left(n^2+2n+2\right)}.\frac{4^n\left(n^2+1\right)}{n{\left|x\right|}^n}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{n+1}{n}\cdot \frac{1}{4}\cdot \frac{n^2+1}{n^2+2n+2}\cdot \left|x\right|\right)\\ =\frac{\left|x\right|}{4}\end{array}$

The series converges absolutely based on the previous step if $\left|x\right|/4<1$ , from which the interval of convergence is given by,

  $-4<x<4$

The left endpoint of the interval is given by,

  $\left(-4,4\right)$

Where,

  $x=-4$

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

The series does not absolutely converge at the left endpoint since the $p$ -series test indicates that it diverges; even though it is absolute convergent if ${\displaystyle \sum \left|a_n\right|={\displaystyle \sum n/\left(n^2+1\right)}}$ converges.

Comparing the general term of the series to terms of $1/n$ and utilize the limit comparison test because the general term of the series

  ${\displaystyle \sum _0^{\infty }n/\left(n^2+1\right)}$ behaves like $n^2/n=1/n$ for high $n$ .

Using the Limit comparison test,

Here,

  $\begin{array}{c}{a}_n=n/n^2+1\\ b_n=1/n\end{array}$

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{n/n^2+1}{1/n}\\ =\underset{n\to \infty }{\lim }\frac{n^2}{n^2+1}\\ =1\end{array}$

Because the limit of $\frac{a_n}{b_n}$ is $1$ and the series ${\displaystyle \sum 1/n}$ diverges.

Thus, based on the limit comparison test ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{n^2+1}}$ also diverges.

Therefore, the series does not converge absolutely at the left endpoint.

Check to see if the series conditionally converges at the left endpoint. The series at the left endpoint is determined by the prior steps as follows:

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

Alternating Series Test can be used to determine whether it converges.

The series converges because it meets the requirements of the Alternating Series Test, which states that it is positive for every $n$ , decreasing, and close to zero as $n$ approaches infinity.

As a result, the original series conditionally converges at the left endpoint,

  $x=-4$

The right endpoint of the interval is given by,

  $\left(-4,4\right)$

Where,

  $x=4$

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

Since it only contains positive components, if it converges it will always be absolute convergent; it will never converge conditionally. However, the Limit Comparison Test indicates that this series diverges; hence the series does not converge at the intended endpoint.

From the above steps it is known that,

The interval of convergence of the series is,

  $-4<x<4$

Therefore, the interval of convergence of the series is $-4<x<4$ .

b.

To determine

To find: For what value of $x$ does the series converge absolutely.

Answer to Problem 44E

The value of $x$ for which the series converges absolutely is $-4<x<4$ .

Explanation of Solution

Given information:

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

Calculation:

From part (a) it is known that,

The interval on which the series converges absolutely is given by,

  $\left(-4,4\right)$

Using the Ratio test,

The series diverges if,

  $x\notin \left[-4,4\right]$

Then, the series converges absolutely on $\left(-4,4\right)$ .

Therefore, the value of $x$ for which the series converges absolutely is $-4<x<4$ .

c.

To determine

To find: For what value of $x$ does the series converge conditionally.

Answer to Problem 44E

The value of $x$ for which the series converges conditionally is $x=-4$ .

Explanation of Solution

Given information:

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

Calculation:

Identifying the intervals in parts (a) and (b) on which the series converges and on which it converges absolutely.

These findings can be used to determine where the series conditionally converges by looking at the conditional convergence of the endpoints.

From part (a) it is known that,

The interval on which the series converges is given by,

  $x=-4$

But from part (a) and part (b) it is found that it does not converge absolutely at this point.

Then, the series converges conditionally on $x=-4$ .

Therefore, the value of $x$ for which the series converges conditionally is $x=-4$ .