Chapter 10, Problem 2RE

(a)

To determine

The radius of convergence ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$

Answer to Problem 2RE

The radius of convergence for is 3. $R=3$

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$

Formula used:

Ratio test is used.

Calculation:

Check for absolute convergence using ratio test. $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(x+4\right)}^{n+1}\right|}{{\left(n+1\right)}^{3n+1}}\frac{n{3}^n}{\left|{\left(x+4\right)}^n\right|}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{n}{n+1},\frac{\left|x+4\right|}{3}\\ =\frac{\left|x+4\right|}{3}\underset{n\to \infty }{\lim }\frac{n}{n+1}\\ =\frac{\left|x+4\right|}{3}.1\\ =\frac{\left|x+4\right|}{3}\end{array}$

Hence,

  $\frac{\left|x+4\right|}{3}<1$

Therefore,

  $\left|x+4\right|<3$

That gives

  $-7<x<-\mathrm{1..............}\left(1\right)$

Checking convergence at end points At $x=-7$

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

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

Hence,

  $\frac{\left|x+4\right|}{3}<1$

Therefore,

  $\left|x+4\right|<3$ , then the series converges absolutely and hence convergence. Therefore, the radius of convergence for is 3. $R=3$

Conclusion:

The radius of convergence for is 3. $R=3$

(b)

To determine

The interval of convergence is ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$

Answer to Problem 2RE

The series always converges for interval $\left[-7,-1\right)$

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\left|x+4\right|}{3}$

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

  $-7\le x<-1$

Therefore,

The series always converges for interval $\left[-7,-1\right)$

Conclusion:

The series always converges for interval $\left[-7,-1\right)$

(c)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$ converges absolutely.

Answer to Problem 2RE

The series always converges for interval $\left[-7,-1\right)$

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\left|x+4\right|}{3}$

From equation (1) as shown below,

  $-7\le x<-1$

Therefore,

The series always converges for interval $\left[-7,-1\right)$

Conclusion:

The series always converges for interval $\left[-7,-1\right)$

(d)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$ converges conditionally.

Answer to Problem 2RE

The series converges conditionally at x = -7.

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$

Formula used:

Ratio test is used.

Calculation:

From equation (2), the series converges conditionally at x = -7. Since, the rearrangement of the terms affects the convergence of the series x = -7.

Conclusion:

The series converges conditionally at x = -7.