# Relative convergence of a series with sine

I have this series:

$$\sum_{n=1}^\infty \frac{2^n\sin^n{x}}{n^2}$$

and I need to find the domain of convergence and absolute convergence. Using the root test, it's easy to see that this series converges absolutely iff $|\sin{x}|<{1\over2}$. However I know nothing about relative convergence. I thought about using either Abel's or Dirichlet's test. Using Abel's test every condition is met except I need to show that: $a_n=2^n\sin^n{x}$ is bounded, or more precisely for which $x \in \mathbb R$ is $a_n$ bounded.

I'd be grateful for any advice or guidance

• if $|\sin x| = \frac 12$ there is convergence as well – mookid Nov 4 '14 at 19:06
• Are you certain the problem involved $2^n \sin^n x$ and not $2^n \sin(x^n)$? In the problem as posed, the $\sin$ is a bit of a red herring. – Mark Fischler Nov 4 '14 at 19:11
• @MarkFischler My first thought was $2^n \sin(nx)$ – user58697 Nov 4 '14 at 19:22

## 2 Answers

Let $a(x):=2\sin(x)$, so that your series is equivalent to $\sum_{n\geq} \frac{a^n}{n^2}$. Can you show the series always diverges if $|a|>1$ and converges if $|a|\leq 1$?

Hint: We have absolute convergence where $\sin x=\pm \frac{1}{2}$ because of the $n^2$ in the denominator.