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: $$\begin{align}\sum_{n=1}^N (-1)^n\sin^2 n &= \sum_{n=1}^N (-1)^{n+1} \frac{e^{2in}-2+e^{-2in}}{4}\\ & = \sum_{n=1}^N \frac{(-1)^{n+1}}{4}e^{2in} -2\sum_{n=1}^N \frac{(-1)^{n+1}}{4} +\sum_{n=1}^N \frac{(-1)^{n+1}}{4}e^{-2in} \end{align}$$
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 $-e^{2i}$.
I know that can be evaluated as $\frac{e^{2i}}{4} \frac{(-1)^M e^{2iM}-1}{-e^{2i}-1}$. We see that $|\sum_{n=1}^M b_n|<2$ for all $M$. By Dirichlet's test $\sum a_n b_n$ converges. The remaining two series can be handled similarly.
BUT MY QUESTIONS :
Why: we can be evaluated as $\frac{e^{2i}}{4} \frac{(-1)^M e^{2iM}-1}{-e^{2i}-1}$ how to transcribe???? And how to show that $|\sum_{n=1}^M b_n|<2$ for all M ????????
