# Sum of reciprocal sine function $\sum\limits_{k=1}^{n-1} \frac{1}{\sin(\frac{k\pi}{n})}=?$

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: $$\sum\limits_{k=1}^{n-1} \frac{1}{\sin(\frac{k\pi}{n})}?$$

• I think the parameter $k$ starts from 1 Commented May 6, 2019 at 21:30
• @DiegoMath Right. Commented May 6, 2019 at 21:32
• 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!
– user141870
Commented Apr 25, 2021 at 9:05

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.

• 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. Commented Apr 27, 2021 at 8:48
• Yes @EwanDelanoy Commented Apr 27, 2021 at 10:31

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 $$N$$.

• (+1) Thanks Gary. I was too lazy to figure out the asymptotics when I posted my answer. Commented Aug 13, 2021 at 20:41
• @TheSimpliFire Thanks. I simplified the coefficients and improved on the presentation. It seems that $2n/\pi$ is the natural large variable.
– Gary
Commented Aug 13, 2021 at 21:01
• (+1). Great answer ! Thanks :-) Commented Aug 14, 2021 at 1:13

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

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