Chapter 9, Problem 9RE

a.

To determine

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

Answer to Problem 9RE

  $1$

Explanation of Solution

Given information:

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

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 $a_n=\frac{x^n}{\sqrt{n}}$ and $a_{n+1}=\frac{x^{n+1}}{\sqrt{n+1}}$ .

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

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

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

So, the radius of convergence $\left|x\right|<1\text{ or }\left|x+0\right|<1$ .

The radius of convergence is $1$ .

b.

To determine

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

Answer to Problem 9RE

  $\left(-1,1\right)$

Explanation of Solution

Given information:

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

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 $R=1$ .

So, the interval of convergence $\left|x+0\right|<1$ .

  $\begin{array}{c}-1<x+0<1\\ -1<x<1\end{array}$

Now check the end points of the interval.

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

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

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

c.

To determine

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

Answer to Problem 9RE

  $\left(-1,1\right)$

Explanation of Solution

Given information:

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

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 $R=1$ .

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

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

d.

To determine

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

Answer to Problem 9RE

  $\left(-1,1\right)$

Explanation of Solution

Given information:

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

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 $R=1$ .

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

So, the value of $x$ when series converges conditionally is $\left(-1,1\right)$ .