Chapter 9, Problem 8RE

a.

To determine

To find: The radius of convergence for the power series ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$ .

Answer to Problem 8RE

  $e$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$

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

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

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

The series is converges if $\frac{1}{e}\left|x\right|<1$ and diverges if $\frac{1}{e}\left|x\right|>1$ .

So, the radius of convergence $\left|x\right|<e\text{ or }\left|x+0\right|<e$ .

The radius of convergence is $e$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$ .

Answer to Problem 8RE

  $\left(-e,e\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$

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 $R=e$ .

So, the interval of convergence $\left|x+0\right|<e$ .

  $\begin{array}{c}-e<x+0<e\\ -e<x<e\end{array}$

Now check the end points of the interval.

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

For $x=e$ , the series ${\displaystyle \sum _{n=0}^{\infty }\frac{e^n}{n^n}}$ is not converges.

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

c.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$ converges absolutely.

Answer to Problem 8RE

  $\left(-e,e\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$

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 $R=e$ .

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

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

d.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$ converges conditionally.

Answer to Problem 8RE

  $\left(-e,e\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^n}}$

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 $R=e$ .

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

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