$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $

Question

$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $$ if $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ find $\frac{A^2}{B^2}$

My Attempt

$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \sum_{n=1}^\infty csch^2(i\omega\pi n)= 4\sum_{n=1}^\infty \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2} $$ $$\sum_{n=1}^\infty \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2}= \big(ie^{\pi\frac{\sqrt{3}}{2}}+ie^{-\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(-e^{\pi\sqrt{3}}+e^{-\pi\sqrt{3}}\big)^{-2} +\big(-ie^{3\pi\frac{\sqrt{3}}{2}}-ie^{-3\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(e^{2\pi\sqrt{3}}-e^{-2\pi\sqrt{3}}\big)^{-2} +...$$$$= \sum_{n=0}^\infty \big(ie^{(4n+1)\pi\frac{\sqrt{3}}{2}}+ie^{-(4n+1)\pi\frac{\sqrt{3}}{2}}\big)^{-2} +\sum_{n=0}^\infty \big(-e^{(2n+1)π√3}+e^{-(2n+1)π√3}\big)^{-2} +\sum_{n=0}^\infty \big(-ie^{(3+4n)\pi\frac{\sqrt{3}}{2}}+-ie^{-(4n+3)\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \sum_{n=0}^\infty \big(e^{(2n)π√3}-e^{-(2n)π√3}\big)^{-2} $$ $$\sum_{n=0}^\infty \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} $$ I have found the sums numerically and $\sum_{n=0}^\infty \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} \approx -0.00429$ How can I evaluate this analytically?

Answer 1

Here is an approach using the residue theorem.

The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are: $$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy to obtain using power series; we keep in mind $\omega^3=1\implies 1/\omega=-1-\omega$).

Now consider $\int_{\Gamma}f(z)\,dz$ where $\Gamma$ is the boundary of the parallelogram

$\hspace{3cm}$ $$\big\{z : \max\{|\Re z|,|\Re(\omega z)|\}\leqslant N+1/2\big\}$$ and take $N\to\infty$. On $AB$ or $CD$ we have $z=\pm\frac1\omega\left(N+\frac12+it\right)$, with "$+$" on $CD$ and "$-$" on $AB$, and $t\in\mathbb{R}$; using $\lim\limits_{y\to-\infty}\cot(x+iy)=i$ (uniformly w.r.t. $x\in\mathbb{R}$), we see that both $\int_{AB}$ and $\int_{CD}$ tend to $$-\frac\pi\omega\int_{-\infty}^\infty\frac{dt}{\cosh^2\pi t}=-\frac2\omega$$ as $N\to\infty$. The integrals along $BC$ and $DA$ both tend to $0$, hence the residue theorem gives $$\frac{1-\omega}{3}+2(1-\omega)\sum_{n=1}^\infty\csc^2\omega\pi n=-\frac{2}{\omega\pi i}\implies\color{blue}{\sum_{n=1}^\infty\csc^2\omega\pi n=\frac{1}{\pi\sqrt3}-\frac16}.$$

Answer 2

In my mind, this is most naturally linked to Eisenstein series.

For $\tau\in\mathbb{C}$ with $\Im\tau>0$, and an integer $k>1$, these are defined by $$G_{2k}(\tau)=\sum_{m,n\in\mathbb{Z}}'(m+n\tau)^{-2k}:=\sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}}(m+n\tau)^{-2k}$$ (i.e., $\sum'$ means $\sum$ over $(m,n)$ excluding $m=n=0$), and satisfy the well-known $$G_{2k}(\tau+1)=G_{2k}(\tau),\qquad G_{2k}(-1/\tau)=\tau^{2k}G_{2k}(\tau)$$ (the latter equality corresponds to the validity of $\sum_m\sum_n=\sum_n\sum_m$).

For $k=1$, the series doesn't converge in the usual sense; however, it's possible to define $$G_2(\tau)=\sum_{n\in\mathbb{Z}}\sum_{m\in\mathbb{Z}}'(m+n\tau)^{-2}$$ where, in the $\sum'$, the term with $m=0$ is excluded if $n=0$. We still have $G_2(\tau+1)=G_2(\tau)$ but $$G_2(-1/\tau)=\tau^2 G_2(\tau)-2i\pi\tau$$ (now the order of the summations is important).

To prove the last formula, $G_2(\tau)$ is "approximated" with $G(\tau)=\sum_n\sum_m'\frac{1}{(m+n\tau)(m+1+n\tau)}$ (here, the meaning of $\sum'$ is clear but obviously different!), so that $G_2(\tau)-G(\tau)$ is absolutely convergent (hence the interchange of summations is valid), and both $G(\tau)$ and its version with the summations interchanged are easy to evaluate in closed form (because of "telescoping").

The link to the present question is given by $\pi^2\csc^2\pi\tau=\sum_{m\in\mathbb{Z}}(m+\tau)^{-2}$ (see (1), (2), etc.): $$\sum_{n=1}^\infty\csc^2n\pi\omega=\frac{1}{\pi^2}\sum_{n=1}^\infty\sum_{m\in\mathbb{Z}}(m+n\omega)^{-2}=\frac{1}{2\pi^2}\left(G_2(\omega)-\sum_{m\neq 0}\frac{1}{m^2}\right)=\frac{\color{blue}{G_2(\omega)}}{2\pi^2}-\frac16;\\G_2(\omega)=G_2(1+\omega)=G_2(-1/\omega)=\omega^2 G_2(\omega)-2i\pi\omega\implies \color{blue}{G_2(\omega)=2\pi/\sqrt{3}}.$$