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

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

• Mathematica solve sum and give answer by QPolyGamma function's. – Mariusz Iwaniuk Jul 25 '20 at 8:51
• You should have worked a lot longer with $\omega$ instead of going straight to its real and imaginary parts. Note that $\omega^3 = 1$ and $\omega^2 = \overline \omega$ – Paul Sinclair Jul 25 '20 at 16:30
• @PaulSinclair I was thinking of using the residue theorem. What do the properties of $\omega$ have to do with the series? – hwood87 Aug 30 '20 at 23:31
• When you have a problem involving an $n$-th root of unity, you better believe that this fact will play a pivotal role in solving it. They are the Checkov's guns of the mathematical world. You might try expanding the taylor series of $\csc^2 x$, and using $\omega^3 = 1$ to reduce it to a quadratic expression in $\omega$. – Paul Sinclair Aug 31 '20 at 0:21
• @PaulSinclair Sorry could you show me? – hwood87 Aug 31 '20 at 1:37

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 $$\frac{1-\omega}{3}+2(1-\omega)\sum_{n=1}^\infty\csc^2\omega\pi n=-\frac{2}{\omega\pi i}\implies\color{blue}{\sum_{n=1}^\infty\csc^2\omega\pi n=\frac{1}{\pi\sqrt3}-\frac16}.$$

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 $$G_{2k}(\tau+1)=G_{2k}(\tau)$$ and $$G_{2k}(-1/\tau)=\tau^{2k}G_{2k}(\tau)$$ (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.): $$\sum_{n=1}^\infty\csc^2n\pi\omega=\frac{1}{\pi^2}\sum_{n=1}^\infty\sum_{m\in\mathbb{Z}}(m+n\omega)^{-2}=\frac{1}{2\pi^2}\left(G_2(\omega)-\sum_{m\neq 0}\frac{1}{m^2}\right)=\frac{\color{blue}{G_2(\omega)}}{2\pi^2}-\frac16;\\G_2(\omega)=G_2(1+\omega)=G_2(-1/\omega)=\omega^2 G_2(\omega)-2i\pi\omega\implies \color{blue}{G_2(\omega)=2\pi/\sqrt{3}}.$$

• Are there any other ways to evaluate this sum? I was thinking the residue theorem would be more simple and adequate for a problem like this. – hwood87 Oct 25 '20 at 21:20
• Thank you for your time, however your answer is not exactly what I am looking for. I am looking for more of an exam or practise problem answer to this question. That is what I meant by 'simple'. Can you suggest any other ways of working it out? – hwood87 Oct 25 '20 at 23:39
• @hwood87: Posted an approach using residues. (If that's still not enough, let someone else try...) – metamorphy Oct 27 '20 at 19:07