Tricky Sum 3.0

Using complex analysis, we can prove the following famous sum (proofs can be found here ) $\displaystyle \sum_{n=1}^\infty\dfrac{1}{x^2+n^2}=\dfrac{\pi x\coth(\pi x)-1}{2x^2}$ which rearranges to $\dfrac{1}{2\pi x}-\dfrac{1}{2}+\dfrac{x}{\pi} \displaystyle \sum_{n=1}^\infty\dfrac{1}{x^2+n^2}=\dfrac{1}{e^{2\pi x}-1}$ $\implies\dfrac{1}{2\pi x}-\dfrac{1}{2}+\dfrac{1}{2\pi} \displaystyle \sum_{n=1}^\infty\left(\dfrac{1}{x+in}+\dfrac{1}{x-in}\right)=\dfrac{1}{e^{2\pi x}-1}$ Note that $e^{-2\pi x}+e^{-4\pi x}+e^{-6\pi x}+\cdots=\dfrac{e^{-2\pi x}}{1-e^{-2\pi x}}=\dfrac{1}{e^{2\pi x}-1}$ $\implies e^{-2\pi x}+e^{-4\pi x}+e^{-6\pi x}+\cdots=\dfrac{1}{2\pi x}-\dfrac{1}{2}+\dfrac{1}{2\pi} \displaystyle \sum_{n=1}^\infty\left(\dfrac{1}{x+in}+\dfrac{1}{x-in}\right)$ Let $k=4p+1$ where $p\in\mathbb{N}$ . Differentiating both sides $k$ times w.r.t. $x$ gives $1^{k}e^{-2\pi x}+2^{k}e^{-4\pi x}+3^{k}e^{-6\pi x}+\cdots=\dfrac{k!}{(2\pi x)^{k+1}}+\dfrac{k!}{(2\pi)^{k+1}} \displaystyle \sum_{n=1}^\infty\left(\dfrac{1}{(x+in)^{k+1}}+\dfrac{1}{(x-in)^{k+1}}\right)$ Summing both sides from $x=1$ to $\infty$ gives $\dfrac{1^{k}}{e^{2\pi}-1}+\dfrac{2^{k}}{e^{4\pi}-1}+\dfrac{3^{k}}{e^{6\pi}-1}+\cdots=\dfrac{k!}{(2\pi)^{k+1}}\zeta(k+1)+\dfrac{k!}{(2\pi)^{k+1}} \displaystyle \sum_{x=1}^{\infty}\sum_{n=1}^\infty\left(\dfrac{1}{(x+in)^{k+1}}+\dfrac{1}{(x-in)^{k+1}}\right)$ For $k=4p+1$ where $p\in\mathbb{N}$ , the double sum on the right changes sign when interchanging $x$ and $n$ . Hence, the double sum is $0$ for such $k$ . $\implies S(k)=\sum_{n=1}^\infty \dfrac{n^k}{e^{2\pi n}-1}=\dfrac{k!}{(2\pi)^{k+1}}\zeta(k+1)=\boxed{\dfrac{B_{k+1}}{2(k+1)}}$ where $\zeta(s)$ is the Riemann zeta function and $B_n$ is the $n^\text{th}$ Bernoulli number .

and hence $S(5)=\dfrac{1}{504},S(9)=\dfrac{1}{264},$ and $S(13)=\dfrac{1}{24}$ .

giving $\dfrac{1}{S(5)}+\dfrac{1}{S(9)}+\dfrac{1}{S(13)}=504+264+24=\boxed{792}$ .