In this blog we prove a Ramanujan-type identity:

$$$\sum\limits_{n_1 \in \mathbb{Z}}(-1)^{n_1}\sum\limits_{(n_2, n_3) \in \mathbb{Z}^2}\frac{1}{\sqrt{n_1^2+ (n_2+0.5)^2+(n_3+0.5)^2}\sinh{(\pi\sqrt{n_1^2+ (n_2+0.5)^2+(n_3+0.5)^2})}} = 1$$$. (1)

First, we consider a 4D lattice sum:

$S(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^2} \frac{(-1)^{(n_1+n_4)}}{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2 + n_4^2}