Chapter 9.5, Problem 41E

a.

To determine

To find: The interval of convergence of the series.

Answer to Problem 41E

The interval of convergence of the series is $-3\le x\le 3$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n\sqrt{3}{3}^n}}$

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.

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\left|\frac{x^{\left(n+1\right)}}{\left(n+1\right)\sqrt{n+1}\cdot 3^{n+1}}\right|}{\left|\frac{x^n}{n\sqrt{n}\cdot 3^n}\right|}=\underset{n\to \infty }{\lim }\left(\frac{{\left|x\right|}^{n+1}}{\left(n+1\right)\sqrt{n+1}\cdot 3^{n+1}}.\frac{n\sqrt{n}\cdot 3^n}{{\left|x\right|}^n}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{n}{n+1}\cdot \sqrt{\frac{n}{n+1}}\cdot \frac{\left|x\right|}{3}\right)\\ =\frac{\left|x\right|}{3}\end{array}$

The series converges absolutely based on the previous step if $\left|x\right|/3<1$ i.e. $\left|x\right|<3$ , from which the interval of convergence is given by,

  $-3<x<3$

The left endpoint of the interval is given by,

  $\left(-3,3\right)$

Where,

  $x=-3$

  $\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-3\right)}^n}{n\sqrt{n}\cdot 3^n}}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{1}{n\sqrt{n}}}\\ ={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{1}{n^{3/2}}}\end{array}$

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 1/n^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ converges.

The right endpoint of the interval is given by,

  $\left(-3,3\right)$

Where,

  $x=3$

  $\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }\frac{3^n}{n\sqrt{n}\cdot 3^n}}={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n\sqrt{n}}}\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n^{3/2}}}\end{array}$

Since it only contains positive components, if it converges it will always be absolute convergent; it will never converge conditionally.

However, this series diverges because the general term does not approach zero, which prevents the series from completing an absolute convergent series at the desired endpoint.

From the above steps it is known that,

The interval of convergence of the series is,

  $-3\le x\le 3$

Therefore, the interval of convergence of the series is $-3\le x\le 3$ .

b.

To determine

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

Answer to Problem 41E

The value of $x$ for which the series converges absolutely is $-3\le x\le 3$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n\sqrt{3}{3}^n}}$

Calculation:

From part (a) it is known that,

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

  $\left(-3,3\right)$

Using the Ratio test,

The series diverges if,

  $x\notin \left[-3,3\right]$

Then, the series converges absolutely on $\left(-3,3\right)$ .

Therefore, the value of $x$ for which the series converges absolutely is $-3\le x\le 3$ .

c.

To determine

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

Answer to Problem 41E

The value of $x$ for which the series converges conditionally does not exist.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n\sqrt{3}{3}^n}}$

Calculation:

As identified the interval on which the series converges and converges absolutely in parts (a) and (b).

It is concluded that there is no value of $x$ for which the series conditionally converges by applying these findings and the analysis of the conditional convergence at the endpoints in part (a).

Therefore, there does not exist a value of $x$ for which the series converges conditionally.