Chapter 9, Problem 5RE

a.

To determine

To find: The radius of convergence for the power series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$ .

Answer to Problem 5RE

  $\frac{1}{3}$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

Let us consider $a_n=\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}$ and $a_{n+1}=\frac{{\left(-1\right)}^n{\left(3x-1\right)}^{n+1}}{{\left(n+1\right)}^2}$ .

Using ratio test find the limit of the ratio of $\left|\frac{a_{n+1}}{a_n}\right|$ as $n\to \infty$ .

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{\frac{{\left(-1\right)}^n{\left(3x-1\right)}^{n+1}}{{\left(n+1\right)}^2}}{\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}\right|\left[\text{Simplify}\right]\\ L=\underset{n\to \infty }{\lim }\left|\frac{\left(3x-1\right)n^2}{{\left(n+1\right)}^2}\right|\\ L=\left|3x-1\right|\underset{n\to \infty }{\lim }\frac{n^2}{{\left(n+1\right)}^2}\\ L=\left|3x-1\right|\end{array}$

So, the radius of convergence $\left|3x-1\right|<1\text{ or }\left|x-\frac{1}{3}\right|<\frac{1}{3}$ .

The radius of convergence is $\frac{1}{3}$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$ .

Answer to Problem 5RE

  $\left(0,\frac{1}{3}\right]$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

In part (a) calculated the radius of convergence is $R=\frac{1}{3}$ .

So, the interval of convergence $\left|x-\frac{1}{3}\right|<\frac{1}{3}$ .

  $\begin{array}{c}\left|x-\frac{1}{3}\right|<\frac{1}{3}\\ -\frac{1}{3}<x-\frac{1}{3}<\frac{1}{3}\\ 0<x<\frac{2}{3}\end{array}$

Now check the end point of the interval.

For $x=0$ , the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(0-1\right)}^n}{n^2}}={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{2n-1}}{n^2}}$ is not converges.

For $x=\frac{1}{3}$ , the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(1-1\right)}^n}{n^2}}=0$ is converges.

Therefore, the interval of convergence is $\left(0,\frac{1}{3}\right]$ .

c.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$ converges absolutely.

Answer to Problem 5RE

  $\left(0,\frac{1}{3}\right]$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

In part (a) calculated the radius of convergence is $R=\frac{1}{3}$ .

In part (b) calculated the interval of convergence is $\left(0,\frac{1}{3}\right]$ .

So, the value of $x$ when series converges absolutely is $\left(0,\frac{1}{3}\right]$ .

d.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$ converges conditionally.

Answer to Problem 5RE

  $\left(0,\frac{1}{3}\right]$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n-1}{\left(3x-1\right)}^n}{n^2}}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

In part (a) calculated the radius of convergence is $R=\frac{1}{3}$ .

In part (b) calculated the interval of convergence is $\left(0,\frac{1}{3}\right]$ .

So, the value of $x$ when series converges conditionally is $\left(0,\frac{1}{3}\right]$ .