A triple series evaluating to $\sqrt{3}$

How would you suggest me to approach the following triple series? I miss a starting point.

$$\sum_{i=-\infty}^{\infty} \sum_{j=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} (-1)^{i+j+k}\frac{1}{\displaystyle \sqrt{\left(i-\frac{1}{6}\right)^2+\left(j-\frac{1}{6}\right)^2+\left(k-\frac{1}{6}\right)^2}}=\sqrt{3}$$

Solution 1:

The main idea is to use the Mellin transform and a Jacobi product identity: $$ \left( \sum_{m\in\mathbb{Z}} (-1)^m q^{(3m^2+m)/2} \right)^3 = \sum_{n\geq 0} (-1)^n(2n+1) q^{(n^2+n)/2}. $$ This result is due to Forrester and Glasser (1982; Some new lattice sums including an exact result for the electrostatic potential within the NaCl lattice).

Set $q=e^{-2t/3}$ to get $$ \sum_{m\in\mathbb{Z}^3} (-1)^{|m|} e^{-\|m+\frac16\|^2t} = \sum_{n\geq0}(-1)^n(2n+1)e^{-\frac13(n+\frac12)^2t}, $$ where $|m|$ is the 1-norm and $\|m\|$ is the 2-norm. If you multiply this by an arbitrary function $f$ whose Laplace transforms $\varphi(s)$ exists for all $s>0$, and integrate over $t\in[0,\infty)$, you get an identity that holds for every suitable Laplace transform $\varphi$: $$ \sum_{m\in\mathbb{Z}^3} (-1)^{|m|} \varphi\big(\|m+\tfrac16\|^2\big) = \sum_{n\geq0}(-1)^n(2n+1)\varphi\big(\tfrac1{3}(n+\tfrac12)^2\big). $$

Substituting $\varphi(s) = s^{-1/2} e^{-a\sqrt{s}}$, and performing the r.h.s. sum gives $$ \sum_{m\in\mathbb{Z}^3} \frac{(-1)^{|m|}e^{-a\|m+\frac16\|}}{\|m+\frac16\|} = \frac{\sqrt{3}}{\cosh \frac{a}{2\sqrt3}}. $$ The case $a=0$ is the one in your question. (Flip the sign $(i,j,k)\mapsto(-i,-j,-k)$ to get the right sign of $\frac16$.)