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?

  • 1
    $\begingroup$ Did you try the comparison test? $\endgroup$ – user99914 Sep 29 '15 at 4:00
  • $\begingroup$ The summation does in fact converge to $\frac{1}{2}(\pi-1)$ if that helps $\endgroup$ – Brevan Ellefsen Sep 29 '15 at 4:02
  • $\begingroup$ Okay, I have to know how you calculated that $\endgroup$ – William Stagner Sep 29 '15 at 4:02
  • $\begingroup$ @WilliamStagner I used Mathematica. Wolfram Alpha has the partial sums, but it is messy. I'm sure there is a more elegant proof that could be constructed, I just threw that out there as an interesting observance. I am now curious as to whether there is a simple way to prove this... I'll look into it for a bit $\endgroup$ – Brevan Ellefsen Sep 29 '15 at 4:06
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    $\begingroup$ The sum $\sum_{n\in \mathbb{Z}} \sin^2(n)/n^2 = \pi$ can be evaluated easily using Parseval's identity. The result stated by @BrevanEllefsen then follows because $\sin^2 n/n^2$ is even under $n\mapsto -n$. $\endgroup$ – Fabian Sep 29 '15 at 4:48

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.

  • $\begingroup$ We need all the terms to be non-negative to apply the comparison test. A fix is to show that the series is absolutely convergent. $\endgroup$ – user217285 Sep 29 '15 at 4:16
  • $\begingroup$ @Nitin: Thanks; that is correct; I was being sloppy... But note that it is the square of $\sin$ takes place. $\endgroup$ – Megadeth Sep 29 '15 at 4:18
  • $\begingroup$ aha you're right. Guess I was being sloppy too :~) $\endgroup$ – user217285 Sep 29 '15 at 4:29

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 $$\sum_{n\geq1}\frac{\sin^{2}\left(n\right)}{n^{2}}=\frac{1}{2}\left(\frac{\pi^{2}}{6}+\frac{\pi^{2}}{12}+\frac{1}{4}\log^{2}\left(-e^{2i}\right)\right)=\frac{1}{2}\left(\pi-1\right).$$


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