Chapter 9.4, Problem 29E

To determine

To find the interval of convergence of the series

The interval of convergence of the series is $S=\frac{3}{4-x^3},-2<x<2$

Answer to Problem 29E

The interval of convergence of the series is $S=\frac{3}{4-x^3},-2<x<2$

Explanation of Solution

Given:

The given series ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{\sin x}{2}\right)}^n}$

Formula used:

If the sequence of partial sum has a limit as $n\to \infty$ it is series convergence otherwise the sequence is divergence. For series to be convergence the term should approach to zero. That is common ratio should be less than $1$ . That is $\left|r\right|<1$ , where r is the common ratio.

Calculation:

The given function is ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{\sin x}{2}\right)}^n}$

It is a geometric series with common ratio $\frac{\sin x}{2}$

  $\begin{array}{l}\left(\frac{\sin x}{2}\right)<1\\ -\infty <x<\infty \end{array}$

The interval of convergence of the series is $-\infty <x<\infty$ .

The sum of the infinite geometric series whose ratio is $<1$ . $S=\frac{a}{1-r}$ , where $a$ is initial term and r is the common ratio. ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{\sin x}{2}\right)}^n}$ . It is a geometric progression with initial term is $1$ and common ratio is $\frac{\sin x}{2}$ .

Sum of the series

  $\begin{array}{l}S=\frac{1}{1-\frac{\sin x}{2}},\\ S=\frac{2}{2-\sin x},-\infty <x<\infty \end{array}$