I try to compute the sum of the inverse distances to the fourth power of all unit cubes inside $\mathbb R^3$: $$\sum_{n=1}^\infty\frac{1}{\left\lVert\vec{r}_n\right\rVert^4}$$
with cartesian coordinates this sum becomes:
$$\sum_{n=-\infty}^\infty\sum_{k=-\infty}^\infty\sum_{j=-\infty}^\infty\frac{1}{(n^2+k^2+j^2)^2},$$
where the term $(n,k,j)=(0,0,0)$ is excluded. After some calculations (using symmetry, etc.) I arrived at the simpler sums: $$\sum_{n=-\infty}^\infty\sum_{k=-\infty}^\infty\sum_{j=-\infty}^\infty\frac{1}{(n^2+k^2+j^2)^2}=8\cdot\sum_{n=1}^\infty\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{1}{(n^2+k^2+j^2)^2}+12\cdot\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{(n^2+k^2)^2}+6\cdot\sum_{n=1}^\infty\frac{1}{n^4}$$
The last two sums are equal to $2\pi^2G-\frac{\pi^4}{15}$, where $G$ is Catalan's constant, but I don't know to analyticly compute the first sum (if even possible) or any tricks how to tackle a triple sum. Does anybody know, if the triple sum has a nice analytic value or how someone might compute it?
