Could this conjecture be proved ? (sum of even powers of cotangents in arithmetic progression )

Having tried (in vain) to answer this question, I worked the explicit formulae of $$\color{blue}{S_k=\sum _{n=1}^m \Big[\cot \left(\frac{n \,\pi }{2 m+1}\right)\Big]^k}$$ where $k$ is an even integer.

To my surprise, the CAS I used were able to produce explicit results only for $k=2$ and $k=4$. All the remaining was done by hand using pen and paper.

What I noticed is that apparently $$\color{red}{S_{2k}=\frac{m (2 m-1)}{a_k} \left( \sum_{n=0}^{2(k-1)} b_n\, m^n \right)}$$ where the $a_k$'s $$\color{blue}{\{3,45,945,14175,467775\}}$$ are the first terms of sequence $A171080$ in $OEIS$: $$a_k = \prod_{q\;\text{odd prime}\\ \;\;\leq 2k+1}q^{\lfloor 2k/(q-1) \rfloor}$$

Concerning the coefficients $b_n$, they are given in the following table $$\color{blue}{\left( \begin{array}{cc} k & \text{list of the } b_n \\ 2 & \{1\} \\ 4 & \{-9,10,4\} \\ 6 & \{135,-252,8,112,32\} \\ 8 & \{-1575,3834,-1388,-2248,496,864,192\} \\ 10 & \{42525,-122760,77040,70400,-57920,-38720,14720,14080,2560\} \end{array} \right)}$$

My questions are :

• Why cannot we compute explicit forms for odd values of $k$ ?
• Could this conjecture be proved ?
• What are these polynomials ?
• Could we find their generic formula ?

Solution 1:

In an article by Ejsmont and Lehner a method using operator trace is applied to evaluate finite sums involving the cotangent. The paper gives numerous references concerning finite trigonometric sums which seem to have a rich recent history. We use here a result obtained by Berndt and Yeap by a contour integration technique. It reads \begin{equation} \sum_{n=1}^{q-1}\cot^{2k}\frac{n\pi}{q}=(-1)^kq-(-1)^k2^{2k}\sum_{\substack{j_0,j_1,\ldots,j_{2k}\ge0 \\j_0+j_1+\ldots+j_{2k}=k}}q^{2j_0}\prod_{p=0}^{2k}\frac{B_{2j_p}}{(2j_p)!}\tag{BY}\label{BY} \end{equation} where $B_s$ are the Bernoulli numbers. It seems that no similar results with odd powers of the function were found.

To evaluate \begin{equation} S_{2k}=\sum _{n=1}^m \left[\cot \left(\frac{n \,\pi }{2 m+1}\right)\right]^{2k} \end{equation} we remark that, by symmetry, \begin{equation} S_{2k}=\frac{1}{2}\sum _{n=1}^{2m} \left[\cot \left(\frac{n \,\pi }{2 m+1}\right)\right]^{2k} \end{equation} then, by (\ref{BY}), \begin{equation} (-1)^kS_{2k}=m+\frac{1}{2}-2^{2k-1}\sum_{\substack{j_0,j_1,\ldots,j_{2k}\ge0 \\j_0+j_1+\ldots+j_{2k}=k}}(2m+1)^{2j_0}\prod_{p=0}^{2k}\frac{B_{2j_p}}{(2j_p)!} \end{equation} Under this form, it appears that the sum is polynomial in $m$.

The summation is however difficult to perform. It can be transformed by the use of the series expansion of $\coth Z$ \begin{equation} Z\coth Z=1+\sum_{s=1}^\infty\frac{B_{2s}}{(2s)!}(2Z)^{2s} \end{equation} One has then \begin{align} S_{2k}&=(-1)^k\left( m+\frac{1}{2}\right)\left[ 1-\left[Z^{-1}\right]\left( \coth\left( (2m+1)Z \right)\coth^{2k}(Z) \right) \right]\\ &=(-1)^k\left( m+\frac{1}{2}\right)\left[ 1-\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( (2m+1)z \right)\coth^{2k}(z)\,dz \right]\label{eq:Cauchy} \end{align} where $\left[z^r\right]\left( F(z) \right)$ represents the coefficient of $z^r$ in the Laurent series of $F(z)$ at $z=0$.

The latter expression for $S_{2k}$ vanishes when $m=0$, as $\operatorname{Res}\left( \coth^{2k+1}z,z=0 \right)=1$. Moreover, from the identity $\coth 2z=(\coth z+(\coth z)^{-1})/2$, it also vanishes for $m=1/2$ as proposed. Then, polynomials $Q_k(m)$ exist such that \begin{equation} S_{2k}=m(2m-1)Q_k(m) \end{equation} where \begin{align} Q_k(m)&=\frac{(-1)^k(2m+1)}{2m(2m-1)}\left[ 1-\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( (2m+1)z \right)\coth^{2k}(z)\,dz \right]\\ &=\frac{(-1)^k(2m+1)}{2m(2m-1)}\left[ 1-\left[Z^{-1}\right]\left( \coth\left( (2m+1)Z \right)\coth^{2k}(Z) \right) \right] \end{align}

Evaluated with a CAS, the latter formula gives the results which are given in the OP up to $k=5$. Higher values of $k$ were numerically tested up to $S_{20}$ for several values of $m$.

Some values of $m$ are remarkable. For instance, it is not difficult to show that $Q_k(-1)=(-1)^{k+1}/3$ and $Q_k\left( -\frac{3}{2} \right)=\frac{(-1)^{k+1}}{3}$. Moreover $Q_k\left( \frac{3}{2} \right)=\frac{1}{3}$. Indeed, for $m=3/2$, \begin{align} \coth\left( (2m+1)z \right)&=\coth(4z)\\ &=\frac{1}{2}\left( \coth 2z+\frac{1}{\coth 2z} \right)\\ &=\frac{1}{4}\left( \coth z+\frac{1}{\coth z} \right)+\frac{\coth z}{1+\coth^2z} \end{align}

the integral reads \begin{align} \frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( 4z \right)\coth^{2k}(z)\,dz &=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac14\left( \coth^{2k+1}(z)+\coth^{2k-1}(z) \right)\,dz+I_k\\ &=\frac{1}{2}+I_k \end{align} where \begin{equation} I_k=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac{\coth^{2k+1}(z)}{1+\coth^2z}\,dz \end{equation} We have the recurrence relation \begin{equation} I_k=1-I_{k-1} \end{equation} and \begin{align} I_0&=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac{\coth(z)}{1+\coth^2z}\,dz\\ &=\frac{1}{2}\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon}\tanh(2z)\,dz\\ &=0 \end{align} Thus $I_{2s}=0$ and $I_{2s+1}=1$. Finally, \begin{equation} Q_k\left( \frac32 \right)= \begin{cases} \frac23\left( 1-\frac{1}{2}-0\right)=\frac{1}{3}&\text{ when } k=2s\\ -\frac23\left( 1-\frac{1}{2}-1 \right)=\frac{1}{3}&\text{ when } k=2s+1 \end{cases} \end{equation} Thus $Q_k\left( \frac{3}{2} \right)=\frac{1}{3}$. This also shows, by adapting the signs, that, for $m=-5/2$, the values of the polynomials oscillate between $1/3$ and $-1/5$. Adapting to the decomposition of $\coth 3z$, should also make it possible to show (as suggested by @Blue in a comment) that $Q_k(1)=3^{-k}$, which is numerically observed.