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I encountered an interesting identity when doing physics homework, that is, $ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $

How is this identity derived? Are there any more related identities?

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    [Also related](http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-n-1-infty-frac1n2/8353#8353).2012-03-21

2 Answers 2

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From my answer here: Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$

Convert this to the problem of finding

$\sum_{n=1}^N \frac{1}{1 - \cos \frac{2\pi n}{N}}$

Convert this to the problem of getting a Chebyshev polynomial of which the roots are $\cos \frac{2\pi n}{N}$ (perhaps after using $\cos \frac{2 \pi (N-n)}{N} = \cos \frac{2 \pi n}{N}$ and getting a polynomial for a subset)

And using the fact the for any polynomial $P(x)$ with roots $r_i$ we have that

$ \sum_{j=1}^{n} \frac{1}{x - r_j} = \frac{P'(x)}{P(x)}$

Another (very similar approach) would be to start with

$\frac{2}{\sin^2 x} = \frac{1}{1+\cos x} + \frac{1}{1-\cos x}$ Note: I haven't worked out the details, but I am pretty sure this would work.

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    Where are the promised more details?2012-05-24
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Let's denote $ S_N=\sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } $

Consider an equality:

$\frac{\pi^2}{\sin^2(\pi s)}=\int_0^{\infty}\frac{x^{s-1}}{1-x}\ln\frac{1}{x}dx;0

Because of $0<\frac{n}{N}<1$, the integral form applies. Thus:

$S_N=\frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\sum_{n=1}^{N-1}x^{\frac{n}{N}-1}dx=$

$= \frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{x^{\frac{1}{N}-1}-1}{1-x^{\frac{1}{N}}}dx=$

$=\frac{N^2}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx=$

$=2\frac{N^2}{{\pi}^2} \int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{3} $

because of $\int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{N^2}\frac{{\pi}^2}{6}$