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In Ramanujan lost notebook I see series like $$\sum _{k=1}^{\infty } \frac{343 \sinh \left(\sqrt{2} \pi k\right)}{32 \pi ^{11} k^{13} \left(\cos \left(\sqrt{2} \pi k\right)-\cosh \left(\sqrt{2} \pi k\right)\right)}=\frac{721 (-1)^{3/4} \pi ^2}{277992000}-\frac{721 \sqrt[4]{-1} \pi ^2}{277992000}$$ Mathematica verifies its correctness numerically, but other series cannot be dealt with Mathematica, for example: $$\sum _{k=1}^{\infty } \frac{\sinh (2 \pi k)}{2 \sqrt{2} \pi ^9 k^{11} (1-\cosh (2 \pi k))}=-\frac{1453 \pi ^2}{851350500 \sqrt{2}}$$ How can we prove this identity? Thank for helping.

Infiniticism
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user547221
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    Please, $\color{red}{\text{R}}\text{amanujan}$ – Claude Leibovici Sep 04 '20 at 13:53
  • Curious, but what can't you bring that $\pi^9$ to the R.H.S? – Mourad Sep 04 '20 at 13:58
  • Have you tried looking in George E. Andrews and Bruce C. Berndt's five volume editing of Ramanujan's Lost notebook? If it is in the lost notebook, they have a proof. – Somos Sep 04 '20 at 14:03
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    Also the second sum can be simplified to $$\sum_{k=1}^{\infty}\frac{\coth\left(\pi k\right)}{k^{11}}=\frac{1453\pi^{11}}{425675250}$$ Pretty sure Ramanujan gave a general identity for this, and such series are well known. – Mourad Sep 04 '20 at 14:04
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    Also the first sum can be simplified to $$\sum_{k=1}^{\infty}\frac{\coth\left(i^{+1/2}\pi k\right)+\coth\left(i^{-1/2}\pi k\right)}{k^{13}}=\frac{103\sqrt{2}\pi^{13}}{212837625}$$ Where $i$ is the imaginary unit. – Mourad Sep 04 '20 at 14:39

1 Answers1

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Both of your sums can be brought under same context. Let $\zeta \notin \mathbb{R}$, integrate $$f(z)=\frac{{\cot \pi \zeta z\cot \pi z}}{{{z^n}}}$$ around a big circle centered at origin. When $n\geq 2$, the integral around big circle $\to 0$. $f(z)$ has poles at $z=k, k\zeta^{-1}$ for $k\in \mathbb{Z}$, if $n$ is moreover odd, then $$\tag{*}\sum_{k = 1}^\infty {\left( {\frac{{\cot \pi \zeta k}}{{{k^n}}} + {\zeta ^{n - 1}}\frac{{\cot \pi {\zeta ^{ - 1}}k}}{{{k^n}}}} \right)} = - \frac{\pi }{2}{\mathop{\rm Res}\nolimits} [\frac{{\cot \pi \zeta z\cot \pi z}}{{{z^n}}},z = 0]$$

This is essentially the functional equation here.

  • When $\zeta = i$, $n\equiv 1\pmod{4}$, LHS of $(*)$ becomes $$-2i\sum_{k = 1}^\infty \frac{\coth \pi k}{k^n}$$ so this sum can be explicitly calculated. As pointed out by a comment, when $n=11$, this is your first sum.

  • When $\zeta = e^{\pi i /4}$, $n\equiv 5\pmod{8}$, LHS of $(*)$ becomes $$2i\sum\limits_{k = 1}^\infty {\frac{{\sinh \sqrt 2 \pi k}}{{\cos (\sqrt 2 \pi k) - \cosh (\sqrt 2 \pi k)}}\frac{1}{{{k^n}}}} $$ this is your first sum when $n=13$.

  • When $\zeta = e^{\pi i /4}$, $n\equiv 1\pmod{8}$, LHS of $(*)$ becomes $$-2\sum\limits_{k = 1}^\infty {\frac{{\sin \sqrt 2 \pi k}}{{\cos (\sqrt 2 \pi k) - \cosh (\sqrt 2 \pi k)}}\frac{1}{{{k^n}}}} $$ for example when $n=17$, this equals $-\frac{41 \pi ^{17}}{181976169375 \sqrt{2}}$.

  • When $\zeta=e^{\pi i/3}$, $\cot \pi {\zeta ^{ \pm 1}}k = \frac{{ \pm i\sinh \sqrt 3 \pi k}}{{{{( - 1)}^k} - \cosh \sqrt 3 \pi k}}$, so LHS of $(*)$ becomes $$i(1 - {\zeta ^{n - 1}})\sum\limits_{k = 1}^\infty {\frac{{\sinh \sqrt 3 \pi k}}{{{{( - 1)}^k} - \cosh \sqrt 3 \pi k}}\frac{1}{{{k^n}}}} $$ For instance, when $n=11$, we have $$\sum\limits_{k = 1}^\infty {\frac{{\sinh \sqrt 3 \pi k}}{{{{( - 1)}^k} - \cosh \sqrt 3 \pi k}}\frac{1}{{{k^{11}}}}} = - \frac{{7457{\pi ^{11}}}}{{1277025750\sqrt 3 }}$$

  • There are also formulas of comparable simplicity when $\zeta = e^{\pi i/6}$.

pisco
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