When is the series converges? 
Let the series $$\sum_{n=1}^\infty \frac{2^n \sin^n x}{n^2}$$. For $x\in (-\pi/2, \pi/2)$, when is the series converges?

By the root-test:
$$\sqrt[n]{a_n} = \sqrt[n]{\frac{2^n\sin^n x}{n^2}} = \frac{2\sin x}{n^{2/n}} \to 2\sin x$$
Thus, the series converges $\iff 2\sin x < 1 \iff \sin x < \frac{1}{2}$
Is that right?
 A: I like the ratio test here:
$$\begin{split}
L &= \lim_{n\to\infty} \left|\frac{2^{n+1} \sin^{n+1} x}{(n+1)^2} \frac{n^2}{2^n\sin^n{x}} \right| \\
&=\lim_{n\to\infty} \left|\frac{2n^2\sin{x}}{(n+1)^2}\right| \\
&=\lim_{n\to\infty} \left|\frac{2\sin{x}\cdot n^2}{n^2}\right| \\
&= |2\sin{x}|
\end{split}$$
$L < 1 \iff |2\sin x| < 1 \iff |\sin x| < \frac12 \iff x \in (-\frac{\pi}6,\frac{\pi}6)$
We can test the boundary separately. If $\sin x = \pm\frac12$, then the series becomes $\sum_{n=0}^{\infty}\frac{(\pm1)^n}{n^2}$, which clearly converges. Thus, the solution is the closed interval, $x \in [-\frac{\pi}6,\frac{\pi}6]$
A: It's almost done. Remember that the root test has absolute values inside the root, so you're looking at $\lim_{n\to\infty}|a_{n}|^{\frac{1}{n}}$. As you noticed, this gives us the convergence in $-1<2\sin x<1$. For the boundary value $2\sin x=1$ you have the convergent series $\sum n^{-2}$, and for the boundary value $2\sin x =-1$ you have the series $\sum (-1)^{n}n^{-2}$, which converges by the alternating series test. So your sum converges for all $-1\leq 2\sin x \leq 1$, i.e for all $x\in [-\frac{\pi}{6},\frac{\pi}{6}]$.
