Question

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

Short 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-by-step solution

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.

Supposer 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 localid="1648832395698" $\sum _{K=0}^{\infty }{\left(\sin x\right)}^k$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 localid="1648832424318" $x=\left(2n+1\right)\frac{\mathrm{\pi }}{2}$ 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 localid="1648832538629" $n\in \mathrm{\mathbb{N}}$.