TITLE: Showing convergence of $\sum_{n=1}^{\infty} {(-1)^{n} \frac{\sin^{2} n}{n}}$
QUESTION [5 upvotes]: I've tried to prove the convergence of $$\sum_{n=1}^{\infty} {(-1)^{n} \frac{\sin^{2} n}{n}}$$ using three different criteria but ran into the following issues:
$1.$ I first tried the Leibnitz criterion.
Let $a_n = \displaystyle{\frac{\sin^{2} n}{n}}$. Though $$\lim_{n\to \infty} \frac{\sin^{2} n}{n} = 0$$
the sequence $a_n$ is not monotone thus the test fails. 
$2.$ Next I tried Dirichlet's test.
I set $\displaystyle{ {a_n} = \frac{1}{n}}$ which monotonically decreases and has a limit of zero. But with $\displaystyle{{b_n} = (-1)^n \sin^2{n}}$ the partial sums $\displaystyle{\left|\sum_{n=1}^{m}b_n\right|}$ are not bounded. Same issue with setting $\displaystyle{ a_n = \frac{(-1)^n}{n}}$ and $\displaystyle{ b_n = \sin^2 n }$. 
$3.$ The last criterion I know is Abel's test. I'm stuck on similar points as with Dirichlet's test.
If I set $\displaystyle{ a_n = \frac{\sin^{2} n}{n} }$, then $\displaystyle{ \sum {a_n} }$ is convergent, but what remains $\displaystyle{ (-1)^n }$ is bounded but not monotone. 
If $ \displaystyle{ a_n = \frac{(-1)^n}{n} }$ then $\displaystyle{ \sum {a_n} }$ is convergent, but $\displaystyle{ \sin^2 n }$ is not monotone.  
What other tests can I apply? Or perhaps this shows that the series is not convergent?

REPLY [3 votes]: Hint:
$$\sum_{n=1}^{m} (-1)^{n} \frac{\sin^{2} n}{n} = \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} -
\sum_{n=1}^{m}\frac{ (-1)^{n}\cos 2n}{2n} \\= \sum_{n=1}^{m} \frac{ (-1)^{n}}{2n} -
\sum_{n=1}^{m}\frac{ \,\cos (\pi+2)n}{2n} $$
The partial sums on the RHS are both convergent as $m \to \infty$.