Convergence of $\sum_n^\infty (-1)^n\frac{\sin^2 n}n$ Could anyone give a hint how to prove the convergence of the following sum?

$$\sum_n^\infty (-1)^n\frac{\sin^2 n}n$$

I tried writing it like this instead:

$$\sum_n^\infty \frac1n (-1)^n \sin^2 n.$$

From here, it is easy to see that $\frac1n$ is a bounded and strictly decreasing sequence. It would be sufficient to prove that the sequence of partial sums of $(-1)^n\sin^2 n$ is bounded.

From here, I get that $(-1)^n\sin^2 n = (-1)^n\frac{1 - \cos 2n}2 = \frac{(-1)^n}2 - \frac{(-1)^n \cos2n}2$, where the sequence of partial sums of $\frac{(-1)^n}2$ is bounded as well as the sequence of partial sums of $\frac{\cos 2n}2$. Unfortunately, I cannot tell anything about $\frac{(-1)^n\cos 2n}2$.

Thank you.
We will use Dirichlet's test: http://en.wikipedia.org/wiki/Dirichlet%27s_test

Write $\frac 1 n (-1)^n\sin(n)^2 = \frac 1 n  (-1)^{n+1} \frac{e^{2in}-2+e^{-2in}}{4}$, which follows from Euler's formulas.  

Our series naturally splits in a sum of $3$ series. Set $a_n = \frac 1 n$. Set $b_n =(-1)^{n+1} \frac{e^{2in}}{4}$. Then $\sum_{n=1}^M b_n $ is a geometric sum, with quotient $-e^{2i}$, that can be evaluated as $\frac{e^{2i}}{4} \frac{(-1)^M e^{2iM}-1}{-e^{2i}-1}$. We see that $|\sum_{n=1}^M b_n|&lt;2$ for all $M$. By Dirichlet's test $\sum a_n b_n$ converges. The remaining two series can be handled similarly.
Using Alternative convergence theorem.
As long As $b_n$ has limit when $n$ approaches infinity, and the limit equals to $0$,
the serise $a_n=(-1)^n b_n$ is convergence.