Finding a closed form solution for an infinite sum 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. 
 A: $\sum_{n=1}^\infty \sin^2(\frac{\pi}{n}) = \sum_{n=1}^\infty \frac{1-\cos(\frac{2\pi}{n})}{2} = \frac{1}{2}\sum_{n=1}^\infty \left[\frac{4\pi^2}{2n^2}-\frac{2^4\pi^4}{4!n^4}+\frac{2^6\pi^6}{6!n^6}+\dots\right] = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n}\pi^{2n}}{(2n)!}\zeta(2n) = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n}\pi^{2n}}{(2n)!}(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!} = \frac{1}{4}\sum_{n=1}^\infty \frac{B_{2n}(2\pi)^{4n}}{(2n)!(2n)!}$
A: 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
$$
\sum_{n=1}^\infty \sin^2\left(\frac{\pi}{n}\right)=\frac{1}{2}\sum_{n=1}^\infty \left(1-\cos\Big(\frac{2\pi}{n}\Big)\right)=\frac{1}{2}\sum_{n=1}^\infty\left(\sum_{j=1}^\infty (-1)^{j-1} \frac{(2\pi)^{2j}}{(2j)!n^{2j}}\right)\\=\frac{1}{2}\sum_{j=1}^\infty\frac{(-1)^{j-1}(2\pi)^{2j}}{(2j)!}\left(\sum_{n=1}^\infty \frac{1}{n^{2j}}\right)=\frac{1}{2}\sum_{j=1}^\infty\frac{2(-1)^{j}(2\pi)^{2j}a_{2j-1}}{(2j)!}
$$
