# Proving convergence of $\sum \frac{\sin n}{2^n}$

Prove that the following sequence ($x_n$) is convergent:

$$x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n}$$ I have tried to use to the sequence is contractive, but am unable to do so. Any help as to which direction I should head to? Thanks!

• Since you're only interested in proving convergence, I'll confine this (well-known) result to a comment: $$\sum_{n=0}^\infty \frac{\sin n}{2^n}=\text{Im}\left[\sum_{n=0}^\infty \left(\frac{e^{i}}{2}\right)^{n}\right] =\text{Im}\left[\frac{1}{1-e^i/2}\right]=\frac{\sin 1}{5-4\cos 1}$$ which is to say, it's not only convergent but can be explicitly summed – Semiclassical Sep 25 '14 at 4:25

We have $\frac{|\sin \ n|}{2^n} \leq \frac{1}{2^n}$ for all $n$ (because $|\sin x| \leq 1$ for all $x$). The series $\sum 1/2^n$ (it is a geometric series) is convergent. By the comparison test, the series $\sum \frac{|\sin n|}{2^n}$ converges. Then $\sum \frac{\sin \ n}{2^n}$ converges.