Prove $\frac{1}{1^{5}\cosh(\frac{\pi}{2})}-\frac{1}{3^{5}\cosh(\frac{3\pi}{2})}+\frac{1}{5^{5}\cosh(\frac{5\pi}{2})}+\cdots=\frac{\pi^{5}}{768}$ 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:
$$\frac{1}{1^{5}\cosh(\frac{\pi}{2})}-\frac{1}{3^{5}\cosh(\frac{3\pi}{2})}+\frac{1}{5^{5}\cosh(\frac{5\pi}{2})}+\cdots=\frac{\pi^{5}}{768}$$
 A: 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 $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5 \cosh \left(\frac{(2n+1)\pi}{2} \right)} = \frac{\pi^5}{768} $$
