Chapter 9, Problem 14RE

a.

To determine

To find: The radius of convergence for the power series ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$ .

Answer to Problem 14RE

  $\frac{1}{10}$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$

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 $a_n=\frac{{\left(10x\right)}^n}{\ln x}$ and $a_{n+1}=\frac{{\left(10x\right)}^{n+1}}{\ln \left(n+1\right)}$ .

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

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{\frac{{\left(10x\right)}^{n+1}}{\ln \left(n+1\right)}}{\frac{{\left(10x\right)}^n}{\ln \left(n\right)}}\right|\left[\text{Simplify}\right]\\ L=\left|10x\right|\underset{n\to \infty }{\lim }\left(\frac{\ln n}{\ln \left(n+1\right)}\right)\\ L=\left|10x\right|\end{array}$

The series is converges if $\left|10x\right|<1$ and diverges if $\left|10x\right|>1$ .

So, the radius of convergence $\left|x\right|<\frac{1}{10}$ .

The radius of convergence is $\frac{1}{10}$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$ .

Answer to Problem 14RE

  $\left[-\frac{1}{10},\frac{1}{10}\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$

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 $R=\frac{1}{10}$ .

  $\begin{array}{c}\left|x\right|<\frac{1}{10}\\ -\frac{1}{10}<x<\frac{1}{10}\end{array}$

So, the interval of convergence $-\frac{1}{10}<x<\frac{1}{10}$ .

Now check the end points of the interval.

For $x=-\frac{1}{10}$ , the power series ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10\cdot -\frac{1}{10}\right)}^n}{\ln n}}={\displaystyle \sum _{n=2}^{\infty }\frac{{\left(-1\right)}^n}{\ln n}}$ is converges.

For $x=\frac{1}{10}$ , the power series ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10\cdot \frac{1}{10}\right)}^n}{\ln n}}={\displaystyle \sum _{n=2}^{\infty }\frac{1}{\ln n}}$ is not converges.

Therefore, the interval of convergence is $\left[-\frac{1}{10},\frac{1}{10}\right)$ .

c.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$ converges absolutely.

Answer to Problem 14RE

  $\left(-\frac{1}{10},\frac{1}{10}\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$

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 $R=\frac{1}{10}$ .

In part (b) calculated the interval of convergence is $\left[-\frac{1}{10},\frac{1}{10}\right)$ .

So, the value of $x$ when series converges absolutely is $\left(-\frac{1}{10},\frac{1}{10}\right)$ .

d.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$ converges conditionally.

Answer to Problem 14RE

  $\left[-\frac{1}{10},\frac{1}{10}\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(10x\right)}^n}{\ln x}}$

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 $R=\frac{1}{10}$ .

In part (b) calculated the interval of convergence is $\left[-\frac{1}{10},\frac{1}{10}\right)$ .

So, the value of $x$ when series converges conditionally is $\left[-\frac{1}{10},\frac{1}{10}\right)$ .