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On my previous page on Ramanujan's-type identity Jack D'Aurizio and Paramanand Singh independently offered their own method of proving that beautiful identity. I am greatly appreciated for their efforts in proving the identity.

Here we offered two Ramanujan-type identities with its closed form; we have found them during own intensive search using the sum calculator. We need anyone to verify it correctness, thank you.

(1)

$$\sum_{n=1}^{\infty}\frac{n^3}{e^{2^{-k}n\pi}-1}=\sum_{n=0}^{k}16^{n-1}$$

(2)

$$2\sum_{n=1}^{\infty}\frac{n^7}{e^{2^{-k}n\pi}-1}=\sum_{n=0}^{2k+1}16^{n-1}$$

I just wonder is there one for the 11?

$$f(n,k)=\sum_{n=0}^{f(k)}11^{n-1}$$

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    $(1)$ does not hold. For instance, $$\sum_{n\geq 1}\frac{n^3}{e^{n\pi/2}-1}=1.0625000031133585\ldots\neq 1+\frac{1}{16}.$$ – Jack D'Aurizio May 14 '16 at 22:16
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    By Abel-Plana's formula your RHS is just the main term of the series, but a small remainder depending on the square of an elliptic integral is still there. – Jack D'Aurizio May 14 '16 at 22:17
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    Also see (23) and (24) on http://mathworld.wolfram.com/EisensteinSeries.html . – Jack D'Aurizio May 14 '16 at 22:32
  • Thank you @JackD'Aurizio I am going to investigate to see if I can modify. –  May 15 '16 at 07:45
  • Maybe it could help $$\sum _{k=1}^{\infty } \frac{k^3}{e^{\pi k 1/2}-1}=-\frac{-128 \pi ^2 \left(\psi _{e^{4 \pi }}^{(1)}(1)-\psi _{e^{8 \pi }}^{(1)}(1)+\psi _{e^{4 \pi }}^{(1)}\left(1-\frac{i}{4}\right)\right)+2 \psi _{e^{4 \pi }}^{(3)}(1)+\psi _{e^{8 \pi }}^{(3)}(1)+2 \psi _{e^{4 \pi }}^{(3)}\left(1-\frac{i}{4}\right)+51 \pi ^4}{48 \pi ^4}$$ –  May 21 '16 at 23:43

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This is not an answer but it is too long for a comment.

The problem to find a closed form for this type of series is related to your previous post since holds the following

Proposition: Let $\alpha,\beta $ be positive numbers such that $\alpha\beta=\pi^{2} $. If $n>1 $ is an integer, then $$\alpha^{n}\sum_{k\geq1}\frac{k^{2n-1}}{e^{2\alpha k}-1}-\left(-\beta\right)^{n}\sum_{k\geq1}\frac{k^{2n-1}}{e^{2\beta k}-1}=\frac{\left(\alpha^{n}-\left(-\beta\right)^{n}\right)B_{2n}}{4} $$ where $B_{2n} $ are the Bernoulli numbers (for a proof see here, proposition $2.6$). So for example if we want a closed form of $$\sum_{k\geq1}\frac{k^{3}}{e^{2^{-1}\pi k}-1} $$ we can study the closed form of $$\sum_{k\geq1}\frac{k^{3}}{e^{8\pi k}-1}. $$

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Marco Cantarini
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