Analogs of $\sum_{n\ge1}\frac{n^{13}}{e^{2\pi n}-1}=\frac{1}{24}$

Question

Context:

As I and others were able to show in the answers to this question, we have that $$\sum_{n\ge1}\frac{n^{13}}{e^{2\pi n}-1}=\frac{1}{24}.$$ In my answer, I let $$E_{2k}(\tau)=1+c_{2k}\sum_{n\ge1}\frac{n^{2k-1}q^n}{1-q^n},$$ with $\tau\in\Bbb H$ and $q=e^{2\pi i\tau}$ and $c_{2k}=\frac{(2\pi i)^{2k}}{(2k-1)!\zeta(2k)}$, be the Eisenstein series of weight $2k$, where $k\in\Bbb Z_{>2}$. Using the well known property that $$E_{2k}(-1/\tau)=\tau^{2k}E_{2k}(\tau),\tag1$$ it is easy to see that $$S_{2k-1}(e^{2i\pi/\tau})-\tau^{2k}S_{2k-1}(e^{-2i\pi\tau})=\frac{\tau^{2k}-1}{c_{2k}},\tag 2$$ where $$S_{\ell}(q)=\sum_{n\ge1}\frac{n^\ell}{q^n-1},$$ since $$E_{2k}(\tau)=1+c_{2k}S_{2k-1}(e^{-2i\pi\tau}).\tag3$$ Using the values $k=7$ and $\tau=i$ in $(2)$, we get $$S_{13}(e^{2\pi})-i^{14}S_{13}(e^{2\pi})=\frac{i^{14}-1}{c_{14}},$$ which reduces to $$S_{13}(e^{2\pi})=\sum_{n\ge1}\frac{n^{13}}{e^{2\pi n}-1}=\frac{1}{24}.$$

My Problem:

I am trying to find identities analogous to $S_{13}(e^{2\pi})=1/24$ using the same general method.

It is well known that $(1)$ is a the special case of $$E_{2k}\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^{2k}E_{2k}(\tau),\tag4$$ corresponding to the choice $\begin{pmatrix}a & b \\ c & d\end{pmatrix}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$. The formula $(4)$ is true for all $\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in\text{SL}_2(\Bbb Z)$.

I was hoping of using $(4)$ instead of $(1)$ to generate identities analogous to the one in the title of this question. To this end, I denote $\gamma=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ and $\gamma(\tau)=\frac{a\tau+b}{c\tau+d}$. Then using $(3)$, we have $$S_{2k-1}(e^{-2i\pi\gamma(\tau)})-(c\tau+d)^{2k}S_{2k-1}(e^{-2i\pi\tau})=\frac{(c\tau+d)^{2k}-1}{c_{2k}}.$$ Then suppose we find $\tau\in\Bbb H$ such that $\gamma(\tau)=\tau$. This would give $S_{2k-1}(e^{-2i\pi\gamma(\tau)})=S_{2k-1}(e^{-2i\pi\tau})$ and thus $$S_{2k-1}(e^{-2i\pi\tau})=-\frac{1}{c_{2k}}.\tag{*}$$ There is an infinite family for such $\tau\in\Bbb H$, each corresponding to a unique element of $\text{SL}_2(\Bbb Z)$. Using these I am able to find things like $$\sum_{n\ge1}\frac{n^{13}}{\zeta^ne^{n\pi\sqrt3/7}-1}=\frac1{24},\qquad \zeta=(-1)^{-5/7}, \gamma=\begin{pmatrix}-2 & 1 \\ -7 & 3\end{pmatrix},\tag5$$ $$\sum_{n\ge1}\frac{n^{13}}{\zeta^ne^{2n\pi/5}-1}=\frac{1}{24},\qquad \zeta=(-1)^{-2/5},\gamma=\begin{pmatrix}2 & 1 \\ -5 & -2\end{pmatrix},\tag6$$ and others.

NOTE: I am unable to numerically test identities $(5),(6)$ because I only have desmos, which doesn't do complex numbers. So I am relying on the theory which I have laid out.

My Question:

The keen among you may have noticed that the formula $(*)$ does not hold for all values of $k\in\Bbb Z_{>2}$. Indeed, we have all the values listed here which show that the values of $S_{2k-1}(e^{-2i\pi\tau})$ for certain fixed $\tau$ differ from $-1/c_{2k}$ for certain values of $k$ but not for others.

My question is, why does this happen.

For example, if we plug in $k=6$ and $\tau=\frac{1+i\sqrt3}{2}$ into $(*)$ we should get $$\sum_{n\ge1}\frac{n^{11}}{(-1)^ne^{n\pi\sqrt3}-1}=-\frac{691}{65520},$$ but actually, $$\sum_{n\ge1}\frac{n^{11}}{(-1)^ne^{n\pi\sqrt3}-1}=\frac{189\Gamma\left ( \frac{1}{4} \right )^{24} }{272629760\pi^{18}}-\frac{691}{65520}.$$ Why does this happen? Where does that $\Gamma$ term come from? Why doesn't $(*)$ work for all $k\in\Bbb Z_{>2}$ and $\gamma\in\text{SL}_2(\Bbb Z)$ when there is a solution $z=\tau\in\Bbb H$ to $\gamma(z)=z$? I figure it has something to do with elliptic curves and elliptic integrals, but I don't know enough of the theory to see it.

Answer 1

I believe equation $(*)$ breaks down since the formula cannot hold whenever $c\tau+d$ is a $2k$th root of unity; otherwise you will be dividing both sides by $(c\tau+d)^{2k}-1=0$.

In your example you have $\tau=(1+i\sqrt3)/2$ and $k=6$ so we have $a-d=1$, $b=-1$ and $c=1$. For the matrix to have unit determinant, we must have $d=0,-1$.

If $d=0$ then we have $(c\tau+d)^{2k}-1=(e^{i\pi/3})^{12}-1=0$.

If $d=-1$ then we have $(c\tau+d)^{2k}-1=(-e^{-i\pi/3})^{12}-1=0$.

Both cases mean the identity does not follow.

By the same argument this can be extended to your linked list to understand that series of the form $$\sum_{n\ge1}\frac{n^{4k-1}}{e^{2\pi n}-1},\quad\sum_{n\ge1}\frac{n^{6k-1}}{(-1)^ne^{n\pi\sqrt3}-1},\cdots$$ do not obey the pattern. I think $\Gamma(\cdot)$ appears due to the particular values of the complete elliptic integral of the first kind.

Answer 2

The Eisenstein series $E_{2k}$ can be expressed as polynomials in $E_4,E_6$ with rational coefficients for $k\geq 2$. I have discussed this using Ramanujan's approach in this answer.

Some of these polynomials have $E_4$ as a factor and some others have $E_6$ as a factor. So knowing zeroes of $E_4,E_6$ does our job. You should prove that $E_4$ vanishes at $\tau=(-1+i(\sqrt{3})^{\pm 1})/2$ and $E_6$ vanishes at $i, (-2+i)/4$.

In particular this allows us to evaluate the sums involving $e^{2n\pi},(-e^{\pi})^n,(-e^{\pi\sqrt{3} })^n$ in closed form.

Also the expression of $E_4,E_6$ in terms of elliptic integrals and moduli is necessary to get a closed form for the cases when $E_{2k}\neq 0$.

Further note that $E_4,E_6$ can be evaluated in closed form using gamma function values if $\tau$ is a root of a quadratic equation with rational coefficients.