Chapter 9, Problem 13RE

a.

To determine

To find: The radius of convergence for the power series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$ .

Answer to Problem 13RE

  $0$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$

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 $a_n=\frac{n!x^{2n+1}}{2^n}$ and $a_{n+1}=\frac{\left(n+1\right)!x^{2n+2}}{2^{n+1}}$ .

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

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{\frac{\left(n+1\right)!x^{2n+2}}{2^{n+1}}}{\frac{n!x^{2n}}{2^n}}\right|\left[\text{Simplify}\right]\\ L=\frac{1}{2}\left|x^2\right|\underset{n\to \infty }{\lim }\left(n+1\right)\\ L=\infty \end{array}$

The series is converges if $\infty <1$ and diverges if $\infty >1$ .

So, the radius of convergence $\left|x-0\right|<0$ .

The radius of convergence is $0$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$ .

Answer to Problem 13RE

  $x=0$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$

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 $R=0$ .

So, the interval of convergence $x=0$ .

Therefore, the interval of convergence is $x=0$ .

c.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$ converges absolutely.

Answer to Problem 13RE

  $x=0$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$

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 $R=0$ .

In part (b) calculated the interval of convergence is $x=0$ .

So, the value of $x$ when series converges absolutely is $x=0$ .

d.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$ converges conditionally.

Answer to Problem 13RE

  $x=0$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!x^{2n}}{2^n}}$

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 $R=0$ .

In part (b) calculated the interval of convergence is $x=0$ .

So, the value of $x$ when series converges conditionally is $x=0$ .