TITLE: How to prove that series $\sum (-1)^n\sin^4n /\sqrt{n}$ converges?
QUESTION [5 upvotes]: I have a series:
$$\sum_{n=1}^\infty(-1)^n\frac{\sin^4n}{\sqrt n}.$$
How can we prove that it converges?

Usually, with $\sin^4n$ we would use Comparison Test, but it only applies when the terms are nonnegative.

REPLY [3 votes]: Hint: Noting that
$$ \sin^4n=\frac{1}{8}(3-4\cos(2n)+\cos(4n)) $$ 
you have
\begin{eqnarray}
\sum_{n=1}^\infty(-1)^n\frac{\sin^4n}{\sqrt n}&=&\frac{3}{8}\sum_{n=1}^\infty(-1)^n\frac{1}{\sqrt n}-\frac{1}{2}\sum_{n=1}^\infty(-1)^n\frac{\cos(2n)}{\sqrt n}+\frac{1}{8}\sum_{n=1}^\infty(-1)^n\frac{\cos(4n)}{\sqrt n}.
\end{eqnarray}
Now you can do the rest to show that $\sum_{n=1}^\infty(-1)^n\frac{\cos(2n)}{\sqrt n}$ and $\sum_{n=1}^\infty(-1)^n\frac{\cos(4n)}{\sqrt n}$ are convergent.