# 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.

• @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) – Thuy Guevarra Aug 8 '19 at 22:02
• 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 $\frac{\pi^2}{n^2}$. – Daniel Aug 8 '19 at 22:06
• A269611 gives some equivalent series for this, but no closed form. – Varun Vejalla Aug 8 '19 at 22:08
• @Daniel Maybe I wasn't clear, I believe that it does converge which is why I hope there's a closed form solution – Thuy Guevarra Aug 8 '19 at 22:13
• @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 – Thuy Guevarra Aug 8 '19 at 22:16

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