I've come across a infinite series for which I've had difficulty finding a closed form solution:

$$\sum_{i=1}^\infty \sin^2(\pi/i).$$

I believe that the series does converge and I've tried looking at transformations to different trig functions and exponentials, however the answer remains elusive. Putting this into WolframAlpha yields a numerical result however I'm much more interested in finding a closed form solution if one exists.

Would be swell if anyone could offer some guidance, thanks.

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    $\begingroup$ @mathworker21 thanks for the quick response, I've explored going down that route however I don't know how to resolve this sum with multiple terms (having the extra 1/2 in this case) $\endgroup$ – Thuy Guevarra Aug 8 '19 at 22:02
  • $\begingroup$ Don't know if finding closed form solution is possible. In your question you seem to be in doubt if the series converges. This is readily seen to hold by applying the comparison test, since the $n$-th term is asymptotically $\frac{\pi^2}{n^2}$. $\endgroup$ – Daniel Aug 8 '19 at 22:06
  • $\begingroup$ A269611 gives some equivalent series for this, but no closed form. $\endgroup$ – Varun Vejalla Aug 8 '19 at 22:08
  • $\begingroup$ @Daniel Maybe I wasn't clear, I believe that it does converge which is why I hope there's a closed form solution $\endgroup$ – Thuy Guevarra Aug 8 '19 at 22:13
  • $\begingroup$ @automaticallyGenerated that's an interesting resource, I'll explore around a bit but yeah the equivalent functions all also involve another series of some type $\endgroup$ – Thuy Guevarra Aug 8 '19 at 22:16

$\sum_{n=1}^\infty \sin^2(\frac{\pi}{n}) = \sum_{n=1}^\infty \frac{1-\cos(\frac{2\pi}{n})}{2} = \frac{1}{2}\sum_{n=1}^\infty \left[\frac{4\pi^2}{2n^2}-\frac{2^4\pi^4}{4!n^4}+\frac{2^6\pi^6}{6!n^6}+\dots\right] = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n}\pi^{2n}}{(2n)!}\zeta(2n) = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n}\pi^{2n}}{(2n)!}(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!} = \frac{1}{4}\sum_{n=1}^\infty \frac{B_{2n}(2\pi)^{4n}}{(2n)!(2n)!}$

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  • $\begingroup$ The second result still seems to not be closed, the first is interesting through how did you find that? The answer differs from what WolframAlpha yields $\endgroup$ – Thuy Guevarra Aug 8 '19 at 22:53
  • $\begingroup$ I don't know what you just said. $\endgroup$ – mathworker21 Aug 8 '19 at 23:07
  • $\begingroup$ The second result is still an infinite sum, the first reduces to a constant but differs from the constant I get when I plug the equation into WolframAlpha. $\endgroup$ – Thuy Guevarra Aug 9 '19 at 0:13
  • $\begingroup$ I don't know what you just said. $\endgroup$ – mathworker21 Aug 21 '19 at 20:33

Using standard techniques to calculate infinite series via residue calculus one obtains that, if $$ \pi \cot (\pi z)=\frac{a_{-1}}{z}+a_1z+a_3z^3+\cdots+a_{2k-1}z^{2k-1}+\cdots $$ then $$ \sum_{n=1}^\infty \frac{1}{n^{2k}}=-2a_{2k-1}. $$ Hence $$ \sum_{n=1}^\infty \sin^2\left(\frac{\pi}{n}\right)=\frac{1}{2}\sum_{n=1}^\infty \left(1-\cos\Big(\frac{2\pi}{n}\Big)\right)=\frac{1}{2}\sum_{n=1}^\infty\left(\sum_{j=1}^\infty (-1)^{j-1} \frac{(2\pi)^{2j}}{(2j)!n^{2j}}\right)\\=\frac{1}{2}\sum_{j=1}^\infty\frac{(-1)^{j-1}(2\pi)^{2j}}{(2j)!}\left(\sum_{n=1}^\infty \frac{1}{n^{2j}}\right)=\frac{1}{2}\sum_{j=1}^\infty\frac{2(-1)^{j}(2\pi)^{2j}a_{2j-1}}{(2j)!} $$

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  • $\begingroup$ It's an interesting result, however there's still an infinite sum to be dealt with in the end $\endgroup$ – Thuy Guevarra Aug 9 '19 at 14:12

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