Chapter 9, Problem 2RE

a.

To determine

To find: The radius of convergence for the power series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$ .

Answer to Problem 2RE

  $3$

Explanation of Solution

Given information:

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

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 use consider $a_n=\frac{{\left(x+4\right)}^n}{n{3}^n}$ and $a_{n+1}=\frac{{\left(x+4\right)}^{n+1}}{\left(n+1\right)3^{n+1}}$ .

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

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{\frac{{\left(x+4\right)}^{n+1}}{\left(n+1\right)3^{n+1}}}{\frac{{\left(x+4\right)}^n}{\left(n\right)3^n}}\right|\left[\text{Simplify}\right]\\ L=\underset{n\to \infty }{\lim }\left|\frac{\left(x+4\right)\left(n\right)}{3\left(n+1\right)}\right|\\ L=\frac{1}{3}\left|x+4\right|\underset{n\to \infty }{\lim }\frac{n}{n+1}\\ L=\frac{1}{3}\left|x+4\right|\end{array}$

The series is converges when $\frac{1}{3}\left|x+4\right|<1\text{ or }\left|x+4\right|<3$ .

The radius of convergence is $3$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+4\right)}^n}{n{3}^n}}$ .

Answer to Problem 2RE

  $\left[-7,-1\right)$

Explanation of Solution

Given information:

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

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 $R=3$ .

So, the interval of convergence $\left|x+4\right|<3$ .

  $\begin{array}{c}-3<x+4<3\\ -3-4<x<3-4\\ -7<x<-1\end{array}$

Now check the convergent at end point of interval.

For $x=-7$ , the power series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-7+4\right)}^n}{n{3}^n}}={\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{n}}$ . Thus, the power series converges.

For $x=-1$ , the power series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1+4\right)}^n}{n{3}^n}}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ . Thus, the power series diverges.

Therefore, the interval of convergence is $\left[-7,-1\right)$ .

c.

To determine

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

Answer to Problem 2RE

  $-7\le x<-1$

Explanation of Solution

Given information:

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

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 $R=3$ .

In part (b) calculated the interval of convergence is $\left[-7,-1\right)$ .

So, the value of $x$ when series converges absolutely is $-7\le x<-1$ .

d.

To determine

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

Answer to Problem 2RE

  $-7\le x<-1$

Explanation of Solution

Given information:

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

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 $R=3$ .

In part (b) calculated the interval of convergence is $\left[-7,-1\right)$ .

So, the value of $x$ when series converges conditionally is $-7\le x<-1$ .