Finite Sum m−1∑k=11sin2kπm\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}

Question : Is the following true for any mN?
m1k=11sin2kπm=m213()

Motivation : I reached () by using computer. It seems true, but I can’t prove it. Can anyone help?

By the way, I’ve been able to prove n=11n2=π26 by using ().

Proof : Let
f(x)=1sin2x1x2=(xsinx)(x+sinx)x2sin2x.
We know that f(x)>0 if 0<xπ/2, and that lim. Hence, letting f(0)=1/3, we know that f(x) is continuous and positive at x=0. Hence, since f(x)\ (0\le x\le {\pi}/2) is bounded, there exists a constant C such that 0\lt f(x)\lt C. Hence, substituting x={(k\pi)}/{(2n+1)} for this, we get
0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.
Then, the sum of these from 1 to n satisfies
0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.
Here, we used (\star). Then, considering n\to\infty leads what we desired.

Answer

Note that
\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}

using this identity we can write
\begin{align}\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}&=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}\\
&=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^2}{m^2}}\\
&=m^2\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(x+k\pi+mn\pi)^2}\\
&=m^2\sum_{n\in\mathbb{Z}}\sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{m^2}{\sin^2(x)}\end{align}

and
\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}
Hence,
\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}=\lim_{x\to0}\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}=\frac{m^2-1}{3}.

Attribution
Source : Link , Question Author : mathlove , Answer Author : user91500

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