Trigonometric identity needed for sums involving secants

Question

I am looking for a closed-form formula for the following sum:

$\displaystyle \sum_{k=0}^{N}{\frac{\sin^{2}(\frac{k\pi}{N})}{a \cdot \sin^{2}(\frac{k\pi}{N})+1}}=\sum_{k=0}^{N}{\frac{1}{a+\csc^{2}(\frac{k\pi}{N})}}$.

Is such a formula known?

Answer 1

Two other references to similar sums are

Bruce C. Berndt and Boon Pin Yeap, Explicit evaluations and reciprocity theorems for finite trigonometric sums, Advances in Applied Mathematics Volume 29, Issue 3, October 2002, Pages 358--385

and

Ira Gessel, Generating Functions and Generalized Dedekind Sums, Electronic J. Combinatorics, Volume 4, Issue 2 (1997) (The Wilf Festschrift volume), R11.

The paper of Berndt and Yeap uses contour integration and has an extensive list of references. My paper uses elementary methods, including partial fractions.

Here are the details of the partial fraction approach:

First we convert the trigonometric sum to a sum over roots of unity. Let $\eta_k=e^{k\pi i /N}$ and let $\zeta_k=\eta_k^2 = e^{2k\pi i/N}$. Then \begin{equation*} \csc^2(k\pi/N) = \left(\frac{2i}{\eta_k -\eta_k^{-1}}\right)^2 =\frac{-4\eta_k^2}{(\eta_k^2-1)^2} =\frac{-4\zeta_k}{(\zeta_k-1)^2}. \end{equation*} Thus (since the summand vanishes for $k=0$) the sum is \begin{equation*} \sum_{\zeta^N=1} \frac{1} {a-4\zeta/(\zeta-1)^2} =\sum_{\zeta^N=1} \frac{(\zeta-1)^2}{a(\zeta-1)^2 - 4\zeta}. \end{equation*} To apply the partial fraction method, we need to find the partial fraction expansion of \begin{equation*} F(z)=\frac{(z-1)^2}{a(z-1)^2 - 4z} \end{equation*} Factoring the denominator shows that we can simplify things if we make the substitution $a=4c/(c-1)^2$, so that \begin{equation*} c = \frac{a+2+2\sqrt{a+1}}{a}. \end{equation*} Then we have \begin{equation*} F(z) =\frac{(c-1)^2}{4c} +\frac{(c-1)^3}{4(c+1)}\left(\frac{1}{z-c} -\frac{1}{c(cz-1)}\right) \end{equation*} We have \begin{equation*} \sum_{\zeta^N=1} (\zeta-c)^{-1} = - \frac{Nc^{N-1}}{c^N-1} \end{equation*} and \begin{equation*} \sum_{\zeta^N=1} (c\zeta-1)^{-1} = \frac{N}{c^N-1} \end{equation*} So the sum is \begin{equation*} \sum_{\zeta^N=1} F(\zeta) = N\frac{(c-1)^2}{4c} \left(1-\frac{(c-1)}{(c+1)}\frac{(c^N+1)}{(c^N-1)}\right). \end{equation*} where $c=(a+2+2\sqrt{a+1})/a$. In terms of $a$, we can simplify this a little to \begin{equation*} \frac{N}{a} \left(1-\frac{1}{\sqrt{a+1}}\frac{(c^N+1)}{(c^N-1)}\right). \end{equation*} If you really want an expression which is rational in $a$, it's possible to write this as a quotient of polynomials in $a$ that are given by generating functions.

Answer 2

I think that the following article of our very own Roberto Bosch Cabrera might come in handy:

https://www.awesomemath.org/wp-content/uploads/reflections/2008_5/article_2.pdf

Specifically, you should take a look at pages 1 & 2 of that note.

Answer 3

This may not be of much (any!) help, but Mathematica 7 gives a closed-form solution in terms of QPolyGamma functions:

$\frac{\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(1-\frac{\log \left(\frac{a-2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i \pi }{n}}\right)}\right)-\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(n-\frac{\log \left(\frac{a-2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i \pi }{n}}\right)}+1\right)+\sqrt{a+1} n \log \left(e^{-\frac{2 i \pi }{n}}\right)}{a \sqrt{a+1} \log \left(e^{-\frac{2 i \pi }{n}}\right)}$

$+$

$\frac{\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(n-\frac{\log \left(\frac{a+2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i \pi }{n}}\right)}+1\right)-\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(1-\frac{\log \left(\frac{a+2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i \pi }{n}}\right)}\right)}{a \sqrt{a+1} \log \left(e^{-\frac{2 i \pi }{n}}\right)}$

$\psi^{(0)}_q$ is the q-PolyGamma function.