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I managed to compute some cotangent series, such as $~\displaystyle\sum\limits_{{\substack{i=1\\10~\nmid~i}}}^{\infty}\frac{\cot\left(\dfrac{9\pi}{10}\cdot i\right)}{i}=-\frac{6\pi}{5},~$ and $\displaystyle\sum\limits_{{\substack{i=1\\11~\nmid~i}}}^{\infty}\frac{\cot\left(\dfrac{4\pi}{11}\cdot i\right)}{i}=\frac{3\pi}{11}.~$ I am interested in the general case

$$S_{k,~n}~=~\sum\limits_{{\substack{i=1\\n~\nmid~i}}}^{\infty}\frac{\cot\left(\dfrac{k\pi}{n}\cdot i\right)}{i}$$

where $k,~n$ are positive integers with $k

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    Are you saying that the approach$($es$)$ you used to evaluate the first two series do not lend themselves over to a generalization ? And what were these approaches ? – 2017-01-06
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    Yes, the computation of the particular cases was computer based. – 2017-01-06
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    Have you ever tried evaluating infinite series of this form in a *systematic* manner, say for $0 – 2017-01-06
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    [Here](http://i.stack.imgur.com/8z6Bd.png) are the values of $~n^2\cdot\dfrac S\pi,~$ for $0$n~\nmid~ik.$ – 2017-01-06
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    And [here](http://i.stack.imgur.com/ef5RD.png) are the values of $~n^2\cdot\dfrac S\pi,~$ for ***n*** up to $20.$ – 2017-01-06
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    [Trigonometric functions](http://en.wikipedia.org/wiki/Trigonometric_functions) are [periodic](http://en.wikipedia.org/wiki/Periodic_function), so each individual series can be rewritten as a [linear combination](http://en.wikipedia.org/wiki/Linear_combination) of [harmonic numbers](http://en.wikipedia.org/wiki/Harmonic_number) of [fractional index](http://en.wikipedia.org/wiki/Harmonic_number#Calculation). Since the latter are related to the [digamma function](http://en.wikipedia.org/wiki/Digamma_function), the [result follows](http://en.wikipedia.org/wiki/Reflection_formula). – 2017-01-06
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    It makes sense to use $\enspace\displaystyle\pi\cot(\pi x)=\frac{1}{x}+\sum\limits_{j=1}^\infty\frac{2x}{x^2-j^2}\enspace$ for $\enspace x\in\mathbb{C}\setminus\mathbb{Z}$ . $\enspace$ In your case: $\enspace\displaystyle x:=\frac{k}{n}i$ . – 2017-01-06

1 Answers 1

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This answer is only a hint, the formula can still be simplified.

It’s given $ \enspace\displaystyle S_{k,n}|_{gcd(k,n)=1}=\sum\limits_{{\substack{j=1\\n~\nmid~j}}}^\infty\frac{1}{j}\cot\left(\frac{k\pi}{n}j\right) \enspace $ which means $\enspace\displaystyle S_{k,n}=\sum\limits_{m=1}^\infty\frac{a_m}{m}\enspace $ with $\enspace\displaystyle a_{m+ln}=a_m:=\cot\left(\frac{k\pi}{n}m\right)\enspace$ and $\enspace a_{ln}=a_n:=0\enspace $ , $\enspace l\in\mathbb{N}_0$.

Using $ \enspace\displaystyle E_1(x):=\sum\limits_{k=1}^\infty \frac{1}{k}e^{i2\pi kx}=i\pi\left(\frac{1}{2}-x\right)-\ln(2\sin(\pi x)) \enspace $ for $\enspace 0https://www.fernuni-hagen.de/analysis/download/bachelorarbeit_aschauer.pdf, page 4 (2.1),

one gets $\enspace\displaystyle b_m |_{n\nmid m}=\frac{1}{n}\sum\limits_{v=1}^{n-1}a_v e^{-i2\pi \frac{m}{n}v}\enspace$ and additional $\enspace\displaystyle b_n=\frac{1}{n}\sum\limits_{v=1}^{n-1}a_v =0\enspace$ and therefore $\enspace\displaystyle \sum\limits_{m=1}^\infty\frac{a_m}{m}=\sum\limits_{m=1}^{n-1}b_m E_1\left(\frac{m}{n}\right)\enspace $.

Result:

$$S_{k,n}=\frac{1}{n}\sum\limits_{m=1}^{n-1} \left(i\pi\left(\frac{1}{2}-\frac{m}{n}\right)-\ln\left(2\sin\frac{m\pi}{n}\right)\right)\sum\limits_{v=1}^{n-1}\cot\left(\frac{k\pi}{n}v\right) e^{-i2\pi \frac{m}{n}v}$$

Note:

A simple numerical test with WolframAlpha gives $11 S_{4,11}\approx 9.42478$ which means $3\pi$ ; input:

sum (i pi (0.5-m/11)-ln(2 sin((m pi)/11)))*(sum (cot(4 pi v/11) e^(-i 2 pi m v/11)) for v=1 to 10) for m=1 to 10

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    That seems to be another interesting approach. Thank you. I' try to see how much it simplifies. Since $S_{k,n} \in \mathbb{R}$, getting rid of the imaginary part could be a first step. – 2017-01-10
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    I haven't check if I have calculated right (for time reasons) but at least in principle the formula must be correct. Because of symmetrie several terms disappear. :-) – 2017-01-11
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    @MathChat : Using WolframAlpha for a numerical test gave the right result for $S_{4,11}$ so that one can assume that the formula above is right. – 2017-01-17