Chapter 10, Problem 14RE

(a)

To determine

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

Answer to Problem 14RE

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

Explanation of Solution

Given information:

The given series is,

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

Formula used:

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(10x\right)}^n\right|}{\ln n}.\frac{\ln \left(n-1\right)}{\left|{\left(10x\right)}^{n-1}\right|}$

Calculation:

Check for absolute convergence using ratio test. $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(10x\right)}^n\right|}{\ln n}.\frac{\ln \left(n-1\right)}{\left|{\left(10x\right)}^{n-1}\right|}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{\ln \left(n-1\right)}{\ln n}\left|10x\right|\\ =\left|10x\right|\end{array}$

Hence,

  $\left|\left(10x\right)\right|<1$

Therefore,

  $\left|x\right|<\frac{1}{10}$

That gives

  $-\frac{1}{10}<x<\frac{1}{10}\mathrm{..............}\left(1\right)$

Checking convergence at end points At $x=-\frac{1}{10}$

  ${\displaystyle \sum _{n=2}^{\infty }\frac{{\left(-1\right)}^n}{\ln n}}$ converges and at $x=\frac{1}{10}$

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{\ln n}}\text{diverges}\mathrm{........}\left(2\right)$

Since,

From equation (1) $\left|x\right|<\frac{1}{10}\text{if}\left|x\right|<\frac{1}{10}$ then the series converges absolutely and hence convergence. Therefore, the radius of convergence for is $\frac{1}{10}$ . $R=\frac{1}{10}$

Conclusion: The radius of convergence for is $\frac{1}{10}$ . $R=\frac{1}{10}$

(b)

To determine

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

Answer to Problem 14RE

The series always converges for interval $\left[-\frac{1}{10},\frac{1}{10}\right]$

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\left|10x\right|$

From equation (1) and (2) as shown below,

  $-\frac{1}{10}\le x<\frac{1}{10}$

Therefore,

The series always converges for interval $\left[-\frac{1}{10},\frac{1}{10}\right]$

Conclusion:

The series always converges for interval $\left[-\frac{1}{10},\frac{1}{10}\right]$

(c)

To determine

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

Answer to Problem 14RE

The series always converges absolutely for interval $\left[-\frac{1}{10},\frac{1}{10}\right]$

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\left|10x\right|$

From equation (1) as shown below,

  $-\frac{1}{10}\le x<\frac{1}{10}$

Therefore,

The series always converges absolutely for interval $\left[-\frac{1}{10},\frac{1}{10}\right]$

Conclusion:

The series always converges absolutely for interval $\left[-\frac{1}{10},\frac{1}{10}\right]$

(d)

To determine

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

Answer to Problem 14RE

The point of conditional convergence is at $x=-\frac{1}{10}$

Explanation of Solution

Given information:

The given series is,

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

Formula used:

Ratio test is used.

Calculation:

The point of conditional convergence is at $x=-\frac{1}{10}$ .Since, the rearrangement of the termsaffects the convergence of the series.

Conclusion:

The point of conditional convergence is at $x=-\frac{1}{10}$