Q 69. - Calculus

Question

Find the values of x for which the series $\sum _{K=0}^{\infty }{\left(\sin x\right)}^k$ converges.

Step-by-Step Solution

Answer

The series $\sum _{K=0}^{\infty }{\left(\sin x\right)}^k$ converges for all real numbers except for  $x\in \left\{\left(2n+1\right)\frac{\mathrm{\pi }}{2}|n\in \mathrm{\mathbb{N}}\right\}$.

Step 1. Given information.

Given a series $\sum _{K=0}^{\infty }{\left(\sin x\right)}^k$.

Step 2. Find all values of x for which the series converges.

A geometric series is of the form $\sum _{k=0}^{\infty }c{r}^k$for some constants c and r.

Suppose r is a non-zero real number, then $\sum _{k=0}^{\infty }c{r}^k$  converges to $\frac{c}{1-r}$ if and only if $\left|r\right|<1$.

Here, the series $\sum _{K=0}^{\infty }{\left(\sin x\right)}^k$ has $r=\sin x$.

For the series to converge, $\left|\sin x\right|<1$.

Note that $-1\le \sin x\le 1$.

So, $\left|\sin x\right|=1$ if and only if $x=\left(2n+1\right)\frac{\mathrm{\pi }}{2}$ for all natural number n.

It follows that $\sum _{K=0}^{\infty }{\left(\sin x\right)}^k$ converges for all real number except for $x=\left(2n+1\right)\frac{\mathrm{\pi }}{2}$ where $n\in \mathrm{\mathbb{N}}$.