Convergence of $\sum_{n=1}^{\infty} \frac{\sin (n x)}{2^n}$

Question

I can't seem to solve this one. I've tried Dirichlet's test but to no avail. Any help is greatly appreciated.

Answer 1

If you want, you can actually calculate what it converges to by appealing to complex numbers. The sum is the imaginary part of $$ \sum \frac{e^{nix}}{2^n} = \sum \left(\frac{e^{ix}}{2}\right)^n = \frac{2}{2-e^{ix}}, $$ so your sum converges to $\frac{2 \sin x}{5 - 4\cos x}$.

Answer 2

HINT: To show absolute convergence use that $\sin$ is bounded by 1 (I assume that you want $x$ to take real values).