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This is an identity from Ramanujan's letter, I am just curious. How do you prove this. My math level knowledge is still very basic so a simplified proof is preferred:

$$\frac{1}{1^{5}\cosh(\frac{\pi}{2})}-\frac{1}{3^{5}\cosh(\frac{3\pi}{2})}+\frac{1}{5^{5}\cosh(\frac{5\pi}{2})}+\cdots=\frac{\pi^{5}}{768}$$

StubbornAtom
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James Warthington
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1 Answers1

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This sum can be found using the partial fraction expansion of secant, with a similar approach to the one found here. Beginning with the formula $$ \frac{\pi}{4} \frac{1}{\cosh \left(\tfrac{\pi}{2}x \right)} = \sum_{k=0}^{\infty} \frac{(-1)^k (2k+1)}{(2k+1)^2+x^2} $$ valid for all real $x$ (see this post) set $x=2n+1$, divide through by $(2n+1)^5$, and sum both sides to get $$ \frac{\pi}{4} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5 \cosh \left(\frac{(2n+1)\pi}{2} \right)} = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^n (-1)^k (2k+1)}{(2n+1)^5 ((2k+1)^2+(2n+1)^2)} $$ The fraction inside the double sum may be rewritten as $$ \frac{(-1)^n (-1)^k}{(2n+1)^5(2k+1)} - \frac{(-1)^n (-1)^k}{(2n+1)^3(2k+1)^3}+\frac{(-1)^n(-1)^k}{(2n+1)(2k+1)^5}-\frac{(-1)^n(-1)^k(2n+1)}{(2k+1)^5((2n+1)^2+(2k+1)^2)} $$ Notice that the rightmost fraction is the same as the one we started with if we switch the dummy variables. Therefore we have $$ \hspace{-10cm} 2\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^n (-1)^k (2k+1)}{(2n+1)^5 ((2k+1)^2+(2n+1)^2)} $$ $$ \hspace{5cm} = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \left(\frac{(-1)^n (-1)^k}{(2n+1)^5(2k+1)} - \frac{(-1)^n (-1)^k}{(2n+1)^3(2k+1)^3}+\frac{(-1)^n(-1)^k}{(2n+1)(2k+1)^5} \right) $$ Using the Dirichlet beta function values $$ \beta(1) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4} $$ $$ \beta(3) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32} $$ $$ \beta(5) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5} = \frac{5\pi^5}{1536} $$ we find that $$ \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^n (-1)^k (2k+1)}{(2n+1)^5 ((2k+1)^2+(2n+1)^2)} = \beta(1)\beta(5)-\frac{1}{2}\beta^2(3) = \frac{\pi^6}{3072} $$ and so $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5 \cosh \left(\frac{(2n+1)\pi}{2} \right)} = \frac{\pi^5}{768} $$

Dave
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    Nice and very simple proof. Ramanujan would have loved it. +1 – Paramanand Singh May 04 '20 at 15:58
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    Thanks! This method generalizes to give the formula $$ \sum_{n=0}^{\infty} \frac{(-1)^n sech \left(\frac{(2n+1)\pi}{2} \right)}{(2n+1)^{4m+1}} = \frac{4}{\pi} \sum_{k=0}^{m-1} (-1)^k \beta(4m-2k+1) \beta(2k+1) + \frac{(-1)^m 2}{\pi} \beta^2(2m+1) $$ for any positive integer $m$. It can also be used with the other three trig partial fractions to gives formulas for $ \sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^{4m+3}} $, $ \sum_{n=0}^{\infty} \frac{\tanh \left(\frac{(2n+1)\pi}{2} \right)}{n^{4m+3}} $, and $ \sum_{n=1}^{\infty} \frac{(-1)^n csch(\pi n)}{n^{4m+3}} $. – Dave May 05 '20 at 14:31
  • I think you should post such sums here as question and also provide an answer. This may lead to other interesting approaches by other people. – Paramanand Singh May 05 '20 at 14:34