Determine whether $\sum_{n=1}^\infty \frac{\sin^2 n}{n^2}$ converges or diverges I've been given this interesting problem but I'm not too sure how to go about it.
My first thought was to find the limit of $\frac{\sin^2 n}{n^2}$ by using the squeeze theorem and then applying the divergence test to determine if the sum converges or diverges.
After finding the limit of the expression, I realized that I cannot apply the divergence test since the limit is $0$, and this doesn't tell me anything.
How else could I go about solving this problem?
 A: Maybe it's interesting to see a way to find the sum of the series. Using the well-known trigonometric identity $$\sin^{2}\left(x\right)=\frac{1}{2}\left(1-\cos\left(2x\right)\right)$$ we have $$\sum_{n\geq1}\frac{\sin^{2}\left(n\right)}{n^{2}}=\frac{1}{2}\left(\sum_{n\geq1}\frac{1}{n^{2}}-\sum_{n\geq1}\frac{\cos\left(2n\right)}{n^{2}}\right)$$ and note, using the representation $$\cos\left(2n\right)=\frac{e^{2in}+e^{-2in}}{2}
 $$ that the last series is $$\sum_{n\geq1}\frac{\cos\left(2n\right)}{n^{2}}=\frac{1}{2}\left(\textrm{Li}_{2}\left(e^{2i}\right)+\textrm{Li}_{2}\left(e^{-2i}\right)\right)$$ which is, using the dilogarithm identity, $$\textrm{Li}_{2}\left(z\right)+\textrm{Li}_{2}\left(z^{-1}\right)=-\frac{\pi^{2}}{6}-\frac{1}{2}\log^{2}\left(-z\right)$$ hence $$\sum_{n\geq1}\frac{\sin^{2}\left(n\right)}{n^{2}}=\frac{1}{2}\left(\frac{\pi^{2}}{6}+\frac{\pi^{2}}{12}+\frac{1}{4}\log^{2}\left(-e^{2i}\right)\right)=\frac{1}{2}\left(\pi-1\right).$$
A: Using the natural bound for $\sin$ gets easier: 
We have
$$
\frac{\sin^{2}n}{n^{2}} \leq \frac{1}{n^{2}}
$$
for all $n \geq 1$;
the series $\sum_{n\geq 1}\frac{1}{n^{2}}$ converges;
so by the comparison test the desired series converges.
