# 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 $m\in\mathbb N$?

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

By the way, I’ve been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.

Proof : Let

We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. 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

Then, the sum of these from $1$ to $n$ satisfies

Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.