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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...

Simply Beautiful Art
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gymbvghjkgkjkhgfkl
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    Related: Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$ – Simply Beautiful Art Dec 15 '17 at 23:10
  • $f(z) = \sum_{n=1}^\infty \frac{n^3}{e^{n z}-1} = \sum_{n=1}^\infty n^3 \sum_{m=1}^\infty e^{-nmz} = \sum_{k=1}^\infty \sigma_3(k) e^{-kz}$ where $\sigma_3(k) = \sum_{d | k} d^3$. Then $E_4(z) = \frac{1}{240}+f(2i\pi z)$ is an Eisenstein series (a modular form) and you are asking about the special value $E_4( i/2)$, which can be evaluated from the identity $E_4 = (\theta_2^2+\theta_3^2+\theta_4^2)$ mentioned there, and the fact $\theta_l(ix/2) = \pm x^{-1/2}\theta_l(i/(2x))$ – reuns Dec 15 '17 at 23:57
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    By geometric expansion:$$\frac1{e^{kn\pi}-1}=\sum_{p=1}^\infty e^{-pkn\pi}$$and by rearranging sums:$$A_k=\sum_{p=1}^\infty\sum_{n=0}^\infty n^3e^{-pkn\pi}=\sum_{p=1}^\infty\frac{e^{-pk\pi}+4e^{-2pk\pi}+e^{-3pk\pi}}{(1-e^{-pk\pi})^4}=\sum_{p=1}^\infty\frac{\cosh(pk\pi)+2}{8\sinh^4(pk\pi/2)}$$which leaves a sum which is probably computable via some suitable application of the residue theorem. Though I haven't the mind to do this today. – Simply Beautiful Art Dec 15 '17 at 23:58
  • By the way how did you stumble upon this sum? It's difficult to come by unless you are interested in the theory of theta functions. – Paramanand Singh Dec 17 '17 at 05:26
  • Observing ramanujan equation and do trial and error and guess my way throught it @Paramanand Singh. Everything here on my site are all just luck guess I did. – gymbvghjkgkjkhgfkl Dec 17 '17 at 06:55
  • I got more of them, if you would like me to post it. – gymbvghjkgkjkhgfkl Dec 17 '17 at 06:59
  • If you have something interesting do share it. I don't know if I can prove them, but someone here will probably be able to find a proof. – Paramanand Singh Dec 17 '17 at 07:04
  • You may also have a look at the identities mentioned here: http://www.genautica.com/math/e_and_pi/e_pi_integer_multiples.html They can all be proved using Ramanujan's technique but with more effort. – Paramanand Singh Dec 17 '17 at 07:15
  • I just check my note book and may have a question to share. I don't know whether is trivial case or not. – gymbvghjkgkjkhgfkl Dec 17 '17 at 07:15

1 Answers1

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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.

Paramanand Singh
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