Chapter 10.4, Problem 27E

To determine

To find: Interval of convergence and sum as a function of x for the given series.

Answer to Problem 27E

Interval of convergence is $-2<x<2$ and sum is $S=\frac{3}{4-x^2}$

Explanation of Solution

Given:

The series given is,

  $S={\displaystyle \sum _{n=0}^{\infty }{\left(\frac{x^2-1}{3}\right)}^n}$

Calculation:

A series is convergence when the limit of partial sum tends to infinity i.e. $n\to \infty$ , otherwise it is divergence.

Also, common ratio should be less than one i.e.

  $\left|r\right|<1$ where r is common ratio.

Now, the given series is

  $S={\displaystyle \sum _{n=0}^{\infty }{\left(\frac{x^2-1}{3}\right)}^n}$

Here, the above series is geometric series having common ratio $r=\frac{x^2-1}{3}$

So,

  $\begin{array}{l}\left|\frac{x^2-1}{3}\right|<1\\ \Rightarrow -2<x<2\end{array}$

i.e. the interval of convergence is $-2<x<2$

Now, determining the sum of the geometric series.

Here in $S={\displaystyle \sum _{n=0}^{\infty }{\left(\frac{x^2-1}{3}\right)}^n}$ , the initial term is $a=1$ and common ratio $r=\frac{x^2-1}{3}$

So, sum is

  $\begin{array}{l}S=\frac{a}{1-r}=\frac{1}{1-\left(\frac{x^2-1}{3}\right)}=\frac{3}{4-x^2}\\ \Rightarrow S=\frac{3}{4-x^2}\end{array}$

Therefore, interval of convergence is $-2<x<2$ and sum is $S=\frac{3}{4-x^2}$