TITLE: On the general form of the family $\sum_{n=1}^\infty \frac{n^{k}}{e^{2n\pi}-1} $
QUESTION [18 upvotes]: I. $k=4n+3.\;$ From this post, one knows that $$\sum_{n=1}^\infty \frac{n^{3}}{e^{2n\pi}-1} = \frac{\Gamma\big(\tfrac{1}{4}\big)^8}{2^{10}\cdot5\,\pi^6}-\frac{1}{240}$$ and a Mathematica session reveals$$\sum_{n=1}^\infty  \frac{n^{7}}{e^{2n\pi}-1} =\frac{3\,\Gamma\big(\tfrac{1}{4}\big)^{16}}{2^{17}\cdot5\,\pi^{12}}-\frac{1}{480}$$
$$\sum_{n=1}^\infty  \frac{n^{11}}{e^{2n\pi}-1} =\frac{189\,\Gamma\big(\tfrac{1}{4}\big)^{24}}{2^{22}\cdot5\cdot13\,\pi^{18}}-\frac{691}{65520}$$
and so on. The $691$ is a clue that Bernoulli numbers are involved. 
II. $k=4n+1.\;$ It evaluates to a rational number,
$$\sum_{n=1}^\infty  \frac{n^{5}}{e^{2n\pi}-1} =\frac{1}{504}$$
$$\sum_{n=1}^\infty  \frac{n^{9}}{e^{2n\pi}-1} =\frac{1}{264}$$
$$\sum_{n=1}^\infty  \frac{n^{13}}{e^{2n\pi}-1} =\frac{1}{24}$$ 
etc, with the last mentioned in this post.
Q: What are the general forms of I and II in terms of the Bernoulli numbers? (And a reference to Ramanujan's Notebooks, if possible.)


Edit added June 2019.

Since the gamma function/pi ratio involved has a simple form in terms of the elliptic integral singular value $K(k_n)$,
$$\beta_1=\frac{\Gamma\big(\tfrac14\big)^8}{2^8\pi^6}=\left(\frac{K(k_1)}{\pi}\right)^4$$
hence was the case $\tau=\sqrt{-1}$. So by analogy, the case  $\tau=\sqrt{-3}$,
$$\beta_3=\frac{3\Gamma\big(\tfrac13\big)^{12}}{2^9\cdot2^{1/3}\,\pi^8}=\left(\frac{K(k_3)}{\pi}\right)^4$$
And a quick test showed,
$$\sum_{k=1}^\infty \frac{k^{3}}{e^{2\pi\sqrt3\,k}-1} = \frac{1}{16}\color{red}{\beta_3}-\frac{1}{240}$$ 
$$\sum_{k=1}^\infty \frac{k^{7}}{e^{2\pi\sqrt3\,k}-1} = \frac{17}{32}\color{red}{{\beta_3}^2}-\frac{1}{480}$$
and so on.

REPLY [4 votes]: Let
$$
G_{2k}(\tau)=\sum_{m,n=-\infty,(m,n)\ne(0,0)}^{\infty}
\frac{1}{(m+n\tau)^{2k}}
$$
Where $\tau$ is not real.And we have
$$
(2k+1)(2k-1)(k-3)G_{2k}
=3\sum_{n=2}^{k-2}(2n-1)(2k-1-2n)G_{2n}G_{2k-2n}
$$
Where $k>3$,which can calculate $G_8,G_{10},...$ .
Note that
$$G_4(i)=\frac{\Gamma\left ( \frac{1}{4}  \right )^8 }{960\pi^2},G_6(i)=0$$
To prove them,consider the elliptic curve $y^2=4x^3-x$ with $g_2=1,g_3=0$.Then
$$
\sum_{m,n = -\infty,(m,n)\ne0}^{\infty}
\frac{1}{(\omega_1 m+\omega_2n)^4}=\frac{1}{60},
\sum_{m,n = -\infty,(m,n)\ne0}^{\infty}
\frac{1}{(\omega_1 m+\omega_2n)^6}=0
$$
Where $\omega_1$ and $\omega_2$ are the periods of $\wp(z)$.Easily see that,
$$
\omega_1=\frac{\Gamma\left ( \frac{1}{4}  \right )^2}{2\sqrt{\pi} },\omega_2=i\omega_1
$$
Thus,
$$
G_4(i)=\frac{\omega_1^4}{60},G_6(i)=0.
$$
And
$$
G_{2k}(\tau)
=2\zeta(2k)\left ( 1-\frac{4k}{B_{2k}}
\sum_{n=1}^{\infty}\frac{n^{2k-1}e^{2\pi in\tau}}{1-e^{2\pi in\tau}}   \right )
$$

$$
\begin{aligned}
&\sum_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1} =\frac{1}{24}-\frac{1}{8\pi}\\
&\sum_{n=1}^{\infty} \frac{n^3}{e^{2\pi n}-1}
=\frac{\Gamma\left ( \frac{1}{4}  \right )^8 }{5120\pi^6}-\frac{1}{240} \\
&\sum_{n=1}^{\infty} \frac{n^5}{e^{2\pi n}-1}
=\frac{1}{504}\\
&\sum_{n=1}^{\infty} \frac{n^7}{e^{2\pi n}-1}
=\frac{3\Gamma\left ( \frac{1}{4}  \right )^{16} }{655360\pi^{12}}
-\frac{1}{480}\\
&\sum_{n=1}^{\infty} \frac{n^9}{e^{2\pi n}-1}
=\frac{1}{264}\\
&\sum_{n=1}^{\infty} \frac{n^{11}}{e^{2\pi n}-1}
=\frac{189\Gamma\left ( \frac{1}{4}  \right )^{24} }{272629760\pi^{18}}
-\frac{691}{65520}\\
&\sum_{n=1}^{\infty} \frac{n^{13}}{e^{2\pi n}-1}
=\frac{1}{24}\\
&\sum_{n=1}^{\infty} \frac{n^{15}}{e^{2\pi n}-1}
=\frac{43659\Gamma\left ( \frac{1}{4}  \right )^{32} }{91268055040\pi^{24}} -\frac{3617}{16320}
\end{aligned}
$$

These sums related to the third type sums:
$$
\begin{aligned}
&\sum_{n = 1}^{\infty} \frac{n^3}{(-1)^ne^{\pi n\sqrt{3}}-1}
=-\frac{1}{240}\\
&\sum_{n = 1}^{\infty} \frac{n^5}{(-1)^ne^{\pi n\sqrt{3}}-1}
=\frac{3\Gamma\left ( \frac{1}{3}  \right )^{18} }{28672\pi^{12}}-\frac{1}{504}\\
&\sum_{n = 1}^{\infty} \frac{n^7}{(-1)^ne^{\pi n\sqrt{3}}-1}
=-\frac{1}{480}\\
&\sum_{n = 1}^{\infty} \frac{n^9}{(-1)^ne^{\pi n\sqrt{3}}-1}
=\frac{1}{264}\\
&\sum_{n = 1}^{\infty} \frac{n^{11}}{(-1)^ne^{\pi n\sqrt{3}}-1}
=\frac{2025\Gamma\left ( \frac{1}{3}  \right )^{36}}{190840832\pi^{24}} -\frac{691}{65520}\\
&\sum_{n = 1}^{\infty} \frac{n^{13}}{(-1)^ne^{\pi n\sqrt{3}}-1}
=\frac{1}{24} 
\end{aligned}
$$