How to prove that series $\sum (-1)^n\sin^4n /\sqrt{n}$ converges?

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.

• tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx – TomGrubb Dec 31 '15 at 14:35
• @bburGsamohT Is that function strictly decreasing? You need that for the AST. – Gregory Grant Dec 31 '15 at 14:36
• @GregoryGrant Compare it with the series $\sum(-1)^n\frac{1}{\sqrt{n}}$ – TomGrubb Dec 31 '15 at 14:37
• If $\alpha$ is not a multiple of $2\pi$, then the partial sums $\sum_{n = 1}^N e^{i\alpha n}$ are bounded. $(-1)^n\sin^4 n = \frac{e^{i\pi n}}{16}(e^{in} - e^{-in})^4$. Ask Dirichlet. – Daniel Fischer Dec 31 '15 at 14:43
• @bburGsamohT we can compare series that have only non-negative terms. – niar_q Dec 31 '15 at 14:47

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.