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.
Answer
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.
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