Let$\:(m,n,k)\underset{k\:\text{fixed}}{\in}\mathbf{N^3}\begin{equation} \:\:\land a_{m,n}=\begin{cases} \left( m^{-k}-n^{-k}\right)^{-1} \ &, m\neq n\\ 0 &, m=n. \end{cases} \end{equation} $
Is $\{a_{m,n}\}_{_{\large (m,n)\in\mathbf N^2}}\:\:$a summable collection?
First thought is obviously to look for $$\sum_{m\ge 0}\sum_{n\ge 0}a_{m,n}=\sum_{m\ge 0}\sum_{n\ge 0}\left({A\over{m-n}}+{B\over{\sum_{_{\large j=0}}^{k-1}}m^j\:n^{k-j-1}}\right),$$ where $A,B\:$are to be found, which is not a piece of cake to yield a fresh closed form.
Although, if we let the permutation $a_{m,n}\mapsto \ a_{n,m}$
$$\large\:\sigma\normalsize=\sum_{m\ge 0}\sum_{n\ge 0}a_{m,n}=\sum_{n\ge 0}\sum_{m\ge 0}a_{n,m}\underset{_{k\:\text{even}}}{=}-\sum_{n\ge 0}\sum_{m\ge 0}a_{m,n}=-\large \sigma$$
So$\:\{a_{m,n}\}_{_{\large (m,n)\in\mathbf N^2}}\:\:$is not a summable collection by $\text{Fubini Theorem}.$
Note: $$a_{m,n}:=a_{m,n}(k)\implies a_{m,n}(2)={1\over{2m}}\left({1\over{n-m}}-{1\over{n+m}}\right),m\ne n$$
Now, two things : Trick to yield a closed form of $\sigma$; see wether or not the collection is summable for even $k$ exponents.
Any input is appreciated.
Cheers