Chapter 9, Problem 6RE

a.

To determine

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

Answer to Problem 6RE

  $1$

Explanation of Solution

Given information:

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

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

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

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

The series is converges when $L<1$ .

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

The radius of convergence is $1$ .

b.

To determine

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

Answer to Problem 6RE

  $\left(-1,1\right)$

Explanation of Solution

Given information:

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

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 $R=1$ .

So, the interval of convergence $\left|x\right|<1$ .

  $-1<x<1$

Now check the end point of the interval.

For $x=-1$ , the series ${\displaystyle \sum _{n=0}^{\infty }\left(n+1\right)x^{3n}}={\displaystyle \sum _{n=0}^{\infty }\left(n+1\right){\left(-1\right)}^{3n}}$ is not converges.

For $x=1$ , the series ${\displaystyle \sum _{n=0}^{\infty }\left(n+1\right)x^{3n}}={\displaystyle \sum _{n=0}^{\infty }\left(n+1\right){\left(1\right)}^{3n}}$ is not converges.

Therefore, the interval of convergence is $\left(-1,1\right)$ .

c.

To determine

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

Answer to Problem 6RE

  $\left(-1,1\right)$

Explanation of Solution

Given information:

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

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 $R=1$ .

In part (b) calculated the interval of convergence is $\left(-1,1\right)$ .

So, the value of $x$ when series converges absolutely is $\left(-1,1\right)$ .

d.

To determine

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

Answer to Problem 6RE

  $\left(-1,1\right)$

Explanation of Solution

Given information:

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

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 $R=1$ .

In part (b) calculated the interval of convergence is $\left(-1,1\right)$ .

So, the value of $x$ when series converges conditionally is $\left(-1,1\right)$ .