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The question comes to me when I find there are answers on summation of some forms of trigonometric functions, i.e. $$ \sum\limits_{k=1}^{n-1} \frac{1}{\sin^2(\frac{k\pi}{n})}\\ \sum\limits_{k=0}^{n-1} \tan(\frac{k\pi}{n})\\ $$ Sum of the reciprocal of sine squared

Sum of tangent functions where arguments are in specific arithmetic series

To show the identity of $\sum\limits_{k=0}^{n-1} \frac{1}{\tan^2(\frac{k\pi}{n})}$ should be trivial as the summand can be rewritten as $\frac{1}{\sin^2(\frac{k\pi}{n})}-1$.

I am wondering what is the following summation: $$ \sum\limits_{k=1}^{n-1} \frac{1}{\sin(\frac{k\pi}{n})}? $$

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    $\begingroup$ I think the parameter $k$ starts from 1 $\endgroup$
    – DiegoMath
    Commented May 6, 2019 at 21:30
  • $\begingroup$ @DiegoMath Right. $\endgroup$
    – Ethanabc
    Commented May 6, 2019 at 21:32
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    $\begingroup$ I found that if $s_n=\sum_{k=1}^{n-1} \frac{1}{sin(kπ/n)}$, then $s_n-s_{n-1}\approx0.7+0.64\ln(n+0.6)$ fits a log graph really well. Not sure how to proceed from there but I hope it's of help to others! $\endgroup$
    – user141870
    Commented Apr 25, 2021 at 9:05

3 Answers 3

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Repeated summation of the beta function yields $$\sum_{k=1}^{n-1}\frac\pi{\sin\pi k/n}=\sum_{k=0}^{n-2}\int_0^\infty\frac{ns^k}{1+s^n}\,ds$$ which can also be derived using Ramanujan's master theorem. Equivalently, $$\sum_{k=1}^{n-1}\csc\frac{\pi k}n=\frac{2n}\pi\int_0^1\frac{s^{n-1}-1}{(s-1)(s^n+1)}\,ds.$$ For a general $n$, the integral has no closed form.

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  • $\begingroup$ When you say "the integral has no closed form", you mean it has no known closed form to this day, right ? +1, by the way. $\endgroup$ Commented Apr 27, 2021 at 8:48
  • $\begingroup$ Yes @EwanDelanoy $\endgroup$
    – TheSimpliFire
    Commented Apr 27, 2021 at 10:31
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An asymptotic formula may be derived as follows. Starting with TheSimpliFire's answer, \begin{align*} \sum\limits_{k = 1}^{n - 1} {\csc \left( {\frac{{\pi k}}{n}} \right)} & = \frac{{2n}}{\pi }\int_0^1 {\frac{{s^{n - 1} - 1}}{{(s - 1)(s^n + 1)}}\mathrm{d}s} \\ & = \frac{{2n}}{\pi }\int_0^1 {\frac{{s^n - 1}}{{(s - 1)(s^n + 1)}}\mathrm{d}s} - \frac{{2n}}{\pi }\int_0^1 {\frac{{s^{n - 1} }}{{s^n + 1}}\mathrm{d}s} \\ & = \frac{{2n}}{\pi }\int_0^1 {\frac{{s^n - 1}}{{(s - 1)(s^n + 1)}}\mathrm{d}s} - \frac{{2\log 2}}{\pi } \\ & = \frac{{2n}}{\pi }\int_0^1 {\frac{{s^n - 1}}{{s - 1}}\mathrm{d}s} - \frac{{2n}}{\pi }\int_0^1 {\frac{{s^n - 1}}{{s - 1}}\frac{{s^n }}{{s^n + 1}}\mathrm{d}s} - \frac{{2\log 2}}{\pi } \\ & = \frac{{2n}}{\pi }\left( {H_n - \int_0^1 {\frac{{s^n - 1}}{{s - 1}}\frac{{s^n }}{{s^n + 1}}\mathrm{d}s} } \right) - \frac{{2\log 2}}{\pi } \end{align*} where $H_n$ is the $n$th Harmonic number. It is known that for any $x>0$ and $N\geq 1$, $$ \frac{x}{{\mathrm{e}^x - 1}} = 1-\frac{x}{2} + \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{(2k)!}}x^{2k} } + \Theta _N \frac{{B_{2N} }}{{(2N)!}}x^{2N} $$ with a suitable $0<\Theta _N<1$ that may depend on $x$, and with $B_k$ being the Bernoulli numbers. Accordingly, using the mean value theorem for improper integrals, \begin{align*} & \int_0^1 {\frac{{s^n - 1}}{{s - 1}}\frac{{s^n }}{{s^n + 1}}\mathrm{d}s} = \int_0^{ + \infty } {\frac{{1 - \mathrm{e}^{ - t} }}{t}\frac{1}{{1 + \mathrm{e}^t }}\frac{{t/n}}{{\mathrm{e}^{t/n} - 1}}\mathrm{d}t} \\ & = \int_0^{ + \infty } {\frac{{1 - \mathrm{e}^{ - t} }}{t}\frac{1}{{1 + \mathrm{e}^t }}\mathrm{d}t} - \frac{1}{{2n}}\int_0^{ + \infty } {\frac{{1 - \mathrm{e}^{ - t} }}{{1 + \mathrm{e}^t }}\mathrm{d}t} \\ & \quad + \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{(2k)!}}\frac{1}{{n^{2k} }}\int_0^{ + \infty } {\frac{{1 - \mathrm{e}^{ - t} }}{{1 + \mathrm{e}^t }}t^{2k - 1} \mathrm{d}t} } + \theta _N \frac{{B_{2N} }}{{(2N)!}}\frac{1}{{n^{2N} }}\int_0^{ + \infty } {\frac{{1 - \mathrm{e}^{ - t} }}{{1 + \mathrm{e}^t }}t^{2N - 1} \mathrm{d}t} \\ & = \log \left( {\frac{\pi }{2}} \right) - \frac{{2\log 2 - 1}}{{2n}}+ \sum\limits_{k = 1}^{N - 1} { \frac{{B_{2k} }}{{2k}}\left( {(2 - 2^{2 - 2k} )\zeta (2k) - 1} \right)\frac{1}{{n^{2k} }}} \\ & \quad + \theta _N \frac{{B_{2N} }}{{2N}}\left( {(2 - 2^{2 - 2N} )\zeta (2N) - 1} \right)\frac{1}{{n^{2N} }} \end{align*} with a suitable $0<\theta _N<1$ that depends on $n$ and $N$. Here $\zeta(s)$ denotes the Riemann zeta function. It is well known that for any $n\geq 1$ and $N\geq 1$, $$ H_n = \log n + \gamma + \frac{1}{{2n}} - \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{2k}}\frac{1}{{n^{2k} }}} - \sigma _N \frac{{B_{2N} }}{{2N}}\frac{1}{{n^{2N} }} $$ with an appropriate $0<\sigma_N<1$ that depends on $n$ and $N$. Therefore, taking into account the sign of the remainder terms, $$ H_n - \int_0^1 {\frac{{s^n - 1}}{{s - 1}}\frac{{s^n }}{{s^n + 1}}\mathrm{d}s} = \log \left( {\frac{2n}{\pi }} \right) + \gamma + \frac{\log 2}{{n}} + \sum\limits_{k = 1}^{N - 1} {\frac{{a_{2k} }}{{(2n/\pi)^{2k} }}} + \omega _N \frac{{a_{2N} }}{{(2n/\pi)^{2N} }} $$ with $$ a_{2k}= - \frac{{B_{2k} (2 - 2^{2 - 2k} )\zeta (2k)}}{{2k}}\left( {\frac{2}{\pi }} \right)^{2k} = ( - 1)^k \frac{{2^{2k} (2^{2k} - 2)B_{2k}^2 }}{{2k \cdot (2k)!}} $$ and with a suitable $0<\omega_N<1$ that depends on $n$ and $N$. In summary, $$ \sum\limits_{k = 1}^{n - 1} {\csc \left( {\frac{{\pi k}}{n}} \right)} = \frac{{2n}}{\pi }\log \left( {\frac{2n}{\pi }} \right) + \gamma \frac{{2n}}{\pi } + \sum\limits_{k = 1}^{N - 1} {\frac{{a_{2k} }}{{(2n/\pi)^{2k - 1} }}} + \omega _N \frac{{a_{2N} }}{{(2n/\pi)^{2N - 1} }} $$ with the coefficients $a_{2k}$ as given above and $0<\omega_N<1$ being an appropriate number that depends on $n$ and $N$.

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    $\begingroup$ (+1) Thanks Gary. I was too lazy to figure out the asymptotics when I posted my answer. $\endgroup$
    – TheSimpliFire
    Commented Aug 13, 2021 at 20:41
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    $\begingroup$ @TheSimpliFire Thanks. I simplified the coefficients and improved on the presentation. It seems that $2n/\pi$ is the natural large variable. $\endgroup$
    – Gary
    Commented Aug 13, 2021 at 21:01
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    $\begingroup$ (+1). Great answer ! Thanks :-) $\endgroup$ Commented Aug 14, 2021 at 1:13
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Starting with the infinite series representation $$\csc(x)=\frac 1 x+\sum_{p=1}^\infty (-1)^p\,\frac{ 2^{2 p}\, B_{2 p}\left(\frac{1}{2}\right) }{(2 p)!} x^{2p-1}$$ let $x=\frac k n \pi$ to make $$\csc \left(\frac{\pi k}{n}\right)=\frac{n}{\pi k}+\sum_{p=1}^\infty (-1)^p\,\frac{2^{2p}\,\pi ^{2 p-1} B_{2 p}\left(\frac{1}{2}\right) }{n^{2p-1}(2 p)!}\,k^{2 p-1}$$ The first summation does not make any problem $$\sum_{k=1}^{n-1}\frac{n}{\pi k}=\frac{n}{\pi } H_{n-1}$$ $$\sum_{k=1}^{n-1}\csc \left(\frac{\pi k}{n}\right)=\frac{n}{\pi } H_{n-1}+S_n$$ These $S_n$ seem to be complicated and I did not find any reasonable approximation.

At this point, I give up hoping that some users could continue.

What we can notice is that the sum is "not very far" from the integral $$I_n=\int_{1}^{n-1} \csc \left(\frac{\pi k}{n}\right)\,dk=\frac {2}\pi\,n\,\log \Big[\cot \left(\frac{\pi }{2 n}\right)\Big]$$ the asymptotics of which being $$I_n=\frac{2 n \log \left(\frac{2 n}{\pi }\right)}{\pi }-\frac{\pi }{6 n}+O\left(\frac{1}{n^3}\right)$$ So, to provide some results, for $1 \leq n \leq 500$, I pefromed some data anlysis.

What I found is that a model such $$\sum_{k=1}^{n-1}\csc \left(\frac{\pi k}{n}\right)=\frac{2 n \log \left(\frac{2 n}{\pi }\right)}{\pi }+ an$$ could be quite good.

Adjusted, $a=0.3674659$ $(\sigma_a=7.33\times 10^{-7})$ and making it rational, the proposed model is $$\sum_{k=1}^{n-1}\csc \left(\frac{\pi k}{n}\right)\sim\frac{2 n \log \left(\frac{2 n}{\pi }\right)}{\pi }+ \frac{445}{1211}n+ O\left(\frac{1}{n}\right)$$

Just a few results $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 50 & 128.5224811 & 128.5208358 \\ 100 & 301.1720822 & 301.1714095 \\ 150 & 490.4771890 & 490.4769072 \\ 200 & 690.5984044 & 690.5983681 \\ 250 & 898.7624047 & 898.7625556 \\ 300 & 1113.335738 & 1113.336047 \\ 350 & 1333.239074 & 1333.239525 \\ 400 & 1557.705289 & 1557.705871 \\ 450 & 1786.160804 & 1786.161510 \\ 500 & 2018.160409 & 2018.161235 \end{array} \right)$$

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    $\begingroup$ The explicit value of your $a$ should be $2\gamma/\pi=0.3674669\ldots$. See my answer. $\endgroup$
    – Gary
    Commented Aug 13, 2021 at 20:20

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