Calculus 8th Edition

The given series converges for all of $x$ and the sum is $\frac{3}{3-sinx}$.
Given: $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}}$ $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \Sigma^{\infty}_{n=0} (\frac{sinx}{3})^{n}$ Here, $a=1$ and $r=\frac{sinx}{3}$ The series converges when $|r| \lt 1$ $|\frac{sinx}{3}| \lt 1$ $-3 \lt sinx \lt 3$ Since $|sinx| \leq 1$ for all $x$, we should have $-3 \lt sinx \lt 3$ for all $x$ and so the given series converges for all $x$. Sum can be calculated as follows: $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \frac{a}{1-r}$ $=\frac{1}{1-\frac{sinx}{3}}$ $=\frac{3}{3-sinx}$