Chapter 8.5, Problem 11E

To determine

To find: The radius of convergence and the interval of convergence.

Answer to Problem 11E

The radius of convergence is 2 and the interval of convergence is -2 < x < 2.

Explanation of Solution

Given information: ${\displaystyle \sum _{n-1}^{\infty }\frac{{(-2)}^n{x}^n}{\sqrt[4]{n}}}$

Calculation:

The given series is, ${\displaystyle \sum _{n-1}^{\infty }\frac{{(-2)}^n{x}^n}{\sqrt[4]{n}}}$

Where, $a_n=\frac{{(-2)}^n}{\sqrt[4]{n}}$

  $=\frac{{(-2)}^n}{(n)\frac{1}{4}}$

  $a_{n+1}=\frac{{(-2)}^{n+1}}{(n+1)\frac{1}{4}}$

  $\underset{n\to \infty }{\lim }|\frac{a_{n+1}}{a_n}|=\underset{n\to \infty }{\lim }.\frac{{(-2)}^n.(-2)}{(n+1)\frac{1}{4}}.\frac{n\frac{1}{4}}{{(-2)}^n}$

  $\begin{array}{l}$ =\underset{n\to \infty }{\lim .}2.{(\frac{n}{n+1})}^4$$ =\underset{n\to \infty }{\lim }\mathrm{.2.}{(\frac{1}{1+\frac{1}{n}})}^4$=2\end{array}$

Thus the radius of convergence is 2 and the interval of convergence is -2 < x < 2 .