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.
 A: 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 $|\sum_{n=1}^M b_n|<2$ for all $M$. By Dirichlet's test $\sum a_n b_n$ converges. The remaining two series can be handled similarly.   
A: The series is not absolutely convergent, so the study of convergence is of interest.
We have 
$$\sin^2n=1-\cos^2n=1-\frac{1+\cos(2n)}2=\frac 12-\frac{\cos(2n)}2.$$
Since $$\sum_{n=1}^\infty\frac{(-1)^n}n\mbox{ is  convergent},$$
we only have to address the convergence of $$\sum_{n=1}^\infty (-1)^n\frac{\cos(2n)}n,$$
which can be done by a summation by parts. Indeed, we define $s_n:=\sum_{k=0}^n(-1)^k$. Then 
$$\sum_{k=M}^N(-1)^k\frac{\cos(2k)}k=\sum_{n=M}^Ns_n\frac{\cos(2n)}n-\sum_{n=M-1}^{N-1}s_n\frac{\cos(2(n+1))}{n+1}.$$
Since the series $\sum_k\frac 1{k^2}$ is convergent, we actually only have to show that the series 
$$\sum_{n=1}^\infty s_n\frac{\cos(2n)}{n}\mbox{ and }\sum_{n=1}^\infty s_n\frac{\cos(2(n+1))}{n}$$
are convergent. (Indeed, $\frac{\cos(2n)}n-\frac{\cos(2(n+1))}{n+1}=\frac{\cos(2n)-\cos(2(n+1))}n-\cos(2(n+1))\left(\frac 1n-\frac 1{n+1}\right) $.) Since $s_{2k+1}=0$, it's enough to establish the convergence of 
$$\sum_{n=1}^{\infty}\frac{\cos(4n)}n\mbox{ and }\sum_{n=1}^{\infty}\frac{\cos(2(2n+1))}n.$$
It can be done by (an other!) summation by parts.
A: 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.
A: EDIT
$$- \log(1+e^{\pm 2i}) =\sum_{n\in\mathbb{N}} (-1)^n \frac{e^{\pm 2i n}}{n}$$
And
$$\sin^2 n = -\frac{1}{4} (e^{in} - e^{-in})^2 = - \frac{1}{4} (e^{2in} - 2 + e^{-2in}) = \frac{1}{2} - \frac{1}{4} (e^{2in} + e^{-2in})$$
So with $\Sigma$ for the "target sum":
$$\Sigma = \frac{1}{2} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n \frac{1}{n}}_{=-\log(2)} - \frac{1}{4} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n\frac{e^{2in}}{n}}_{= -\log(1+e^{2in})} - \frac{1}{4} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n \frac{e^{-2in}}{n}}_{=-\log(1+e^{-2in})} = \frac{1}{4} (-\log(4) + \log(1+e^{2i}) + \log(1+e^{-2i}))$$
