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Consider $\displaystyle{S = \sum_{k = 1}^{n - 1}\csc\left(k\,{\pi \over n}\right) = \frac{1}{\sin\left(\pi/n\right)} + \frac{1}{\sin\left(2\pi/n\right)} + \frac{1}{\sin\left(3\pi/n\right)} + \cdots + \frac{1}{\sin\left(\left[n - 1\right]\pi/n\right)}}$

How can we find a general formula for $S$ using trigonometry identities or complex numbers ?.

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Let we assume for first that $n$ is an odd number, $n=2N+1$. In such a case the given sum is $$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = 2\sum_{k=1}^{N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}=\frac{4N+2}{\pi}H_N+2\sum_{k=1}^{N}\left[\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}-\frac{1}{\frac{\pi k}{2N+1}}\right] $$ and $\frac{1}{\sin(x)}-\frac{1}{x}$ is an integrable function on the interval $\left(0,\frac{\pi}{2}\right)$, whose integral equals $\log\frac{4}{\pi}$.$^{(*)}$
By Riemann sums is follows that: $$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = \frac{4N}{\pi}\left[H_N+\log\frac{4}{\pi}\right]+O(\log N).$$ In the general case we get that the given sum behaves like $Cn\log n$.


$^{(*)}$ Since $\text{Res}\left(\frac{1}{\sin x},x=k\pi\right)=(-1)^k$, by Herglotz' trick we have $$\frac{1}{\sin x}-\frac{1}{x}=\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi}\right)(-1)^k $$ and by termwise integration $$ \int_{0}^{\pi/2}\left(\frac{1}{\sin x}-\frac{1}{x}\right)\,dx = \sum_{k\geq 1}(-1)^k \log\left(1-\frac{1}{4k^2}\right) $$ so $I=\log\frac{4}{\pi}$ by simplifying the partial sums of the last series and recalling Wallis product.

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  • $\begingroup$ Does your sum not blow up? $\endgroup$ – Ron Gordon Jan 18 '17 at 18:20
  • $\begingroup$ @RonGordon: the given sum behaves like $n\log n$ hence yes, it does blow up. $\endgroup$ – Jack D'Aurizio Jan 18 '17 at 18:21
  • $\begingroup$ No, I mean actually blow up. Look at the first sum when $k=2 N$. $\endgroup$ – Ron Gordon Jan 18 '17 at 18:22
  • $\begingroup$ @RonGordon: oh, sorry, wrong summation indices. Now fixing. $\endgroup$ – Jack D'Aurizio Jan 18 '17 at 18:22
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    $\begingroup$ @S.H.W it won't get any simpler $\endgroup$ – tired Jan 18 '17 at 20:14

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