2

In here, I succeed in proving one-dimensional cases. Here is a high dimensional example: $$\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{n+k}}{n^2k^2\sqrt{n^2+k^2}\sinh\left ( \pi\sqrt{n^2+k^2} \right ) } +\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^5\sinh(\pi n)} =-\frac{\pi^5}{11340}.$$ The above sum comes from $$\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^{n+k+j}}{n^2k^2}\frac{1}{n^2+k^2+j^2}.$$ I am sure that it can establish a relationship with high dimensional residues. But I still don't know how to do that. I will be thankful for all the help.

Theorem (Residue theorem).
If a function $f$ is meromorphic in $D\subset\mathbb{C}^n$, and $P$ is a polar-set of $f$, then for any $n$-dimensional cycle $\sigma\subset D\backslash P$, $$ \int_\sigma f(z)\mathrm{d}z =(2\pi i)^n\sum_{\mu=1}^{\rho}k_\mu R_\mu. $$

Setness Ramesory
  • 5,172
  • Using symmetry we have $$\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^{n+k+j}}{n^2k^2}\frac{1}{n^2+k^2+j^2} = \frac{1}{3}(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2})^3 $$ Now, instead using the symmetry, try to do the triple sum with respect to $k$. – pisco Nov 29 '21 at 19:11
  • @pisco You might misunderstnd what I asked. I am actually interested in evaluating the sum through some unconventional ways. As I mentioned, one of the possible ways is computing high dimensional residues. – Setness Ramesory Nov 30 '21 at 11:20

0 Answers0