Chapter 9.4, Problem 9E

To determine

To find: Radius of convergence of the power series ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n{\left(4x+1\right)}^n}$ .

Answer to Problem 9E

Radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n{\left(4x+1\right)}^n}$ is $\frac{1}{4}$ .

Explanation of Solution

Given information: Series ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n{\left(4x+1\right)}^n}$ is given.

Concept used :

If ${\displaystyle \sum a_n}$ is a series with non-negative terms such that $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$ , then ${\displaystyle \sum a_n}$ converges if $L<1$ .

If the interval of convergence is $a<x<b$ , radius of convergence is given by $R=\frac{b-a}{2}$ .

Calculation:

Determine $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}$ for the given series as shown below:

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(-1\right)}^{n+1}{\left(4x+1\right)}^{n+1}\right|}{\left|{\left(-1\right)}^n{\left(4x+1\right)}^n\right|}\\ =\underset{n\to \infty }{\lim }\left|\left(-1\right)\left(4x+1\right)\right|\\ =\left|-\left(4x+1\right)\right|\end{array}$

The obtained series converges only when $\left|-\left(4x+1\right)\right|<1$ .

Apply the absolute rule, if $\left|u\right|<a$ , $a>0$ then $-a<u<a$ to $\left|x+5\right|<1$ .

  $\begin{array}{c}-1<-\left(4x+1\right)<1\\ -\left(4x+1\right)>-1\text{ and }-\left(4x+1\right)<1\\ x<0\text{and}x>-\frac{1}{2}\end{array}$

Merge overlapping intervals.

  $-\frac{1}{2}<x<0$

Thus, the series converges absolutely for $-\frac{1}{2}<x<0$ .

Determine the radius of convergence for the given problem in the interval $-\frac{1}{2}<x<0$ .

  $\begin{array}{c}R=\frac{0-\left(-\frac{1}{2}\right)}{2}\\ =\frac{1}{4}\end{array}$

As a result, radius of convergence for the given problem is $\frac{1}{4}$ .