Interesting sums yield rational value Let $$A_k=\sum_{n=1}^{\infty}{n^3\over e^{kn\pi}-1}$$

How can we show that $$A_1+11A_2-32A_4={1\over 12}?$$

I haven't got any idea where to begin...
 A: This is very similar to the sum in this answer (do have a look at it for getting familiar with the notation) but uses another Ramanujan function $$Q(q) = 1+240\sum_{n=1}^{\infty}\frac{n^{3}q^{n}}{1-q^{n}}\tag{1}$$ which is related to the elliptic modulus $k$ and elliptic integral $K=K(k) $ via the relations
\begin{align}
Q(q) &=\left(\frac{2K}{\pi}\right)^{4}(1+14k^{2}+k^{4})\tag{2}\\
Q(q^{2}) &= \left(\frac{2K}{\pi}\right)^{4}(1-k^{2}+k^{4})\tag{3}\\
Q(q^{4}) &= \left(\frac{2K}{\pi}\right)^{4}\left(1-k^{2}+\frac{k^{4}}{16}\right)\tag{4} 
\end{align}
Next note that if $q=e^{-\pi} $ then $k^{2}=1/2$ and hence we have
\begin{align}
1+240A_1 &= Q(e^{-\pi}) =\left(\frac{2K} {\pi} \right ) ^{4}\cdot \frac{33} {4}\tag{5} \\
1+240A_2 &= Q(e^{-2\pi}) =\left(\frac{2K}{\pi}\right) ^{4}\cdot\frac{3}{4}\tag{6}\\
1+240A_4 &= Q(e^{-4\pi})= \left(\frac{2K}{\pi}\right)^{4}\cdot\frac{33}{64}\tag{7}
\end{align}
And then the linear combination $(5)+11(6)-32(7)$ of above equations gives us $$240(A_1+11A_2-32A_4)-20=0$$ as desired.

You may also have a look at this answer which combines $A_2,A_4$ in a different manner to get another rational number. 
