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:

$$$S(n_1, n_2, n_3, n_4) := \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \frac{(-1)^{n_1+n_4}}{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2 + n_4^2}$$$

We note that for $$$\lambda > 0$$$, $$$\int_{0}^\infty e^{-\lambda x}dx = \frac{1}{\lambda}$$$,

therefore $$$S(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \int_{0}^\infty (-1)^{n_1+n_4} \exp((-n_1^2-(n_2+0.5)^2-(n_3+0.5)^2-n_4^2)x)dx$$$.