How to evaluate $\sum_{j=0}^\infty\;\sum_{\substack{k=0 \\ k \neq j}}^\infty \frac{1}{j^2-k^2}$ I was reading an introductory text on multiple integrals and I have encountered a problem asking me to explain why 
$$
\sum_{j=0}^\infty\;\sum_{\substack{k=0 \\ k \neq j}}^\infty \frac{1}{j^2-k^2}\neq\sum_{k=0}^\infty\;\sum_{\substack{j=0 \\ j\neq k}}^\infty \frac{1}{j^2-k^2}
$$
I felt that the infinite sum would diverge because the two terms corresponding to the pair of points $(j,k)$ and $(k,j)$ are always included in the sum and the sum of the two terms $\frac{1}{j^2-k^2}+\frac{1}{k^2-j^2}$are zero. However, I have no idea how to show that by evaluating the sum in any way.
I would appreciate any help.
 A: For any natural number $j\neq 0$ we have:
$$\begin{eqnarray*}\sum_{\substack{k=0\\k\neq j}}^{+\infty}\frac{1}{j^2-k^2}&=&\frac{1}{2j}\left[\sum_{k<j}\left(\frac{1}{j-k}+\frac{1}{j+k}\right)+\sum_{k>j}\left(\frac{1}{j-k}+\frac{1}{j+k}\right)\right]\\&=&\frac{1}{2j}\left[\sum_{k=0}^{j-1}\left(\frac{1}{j-k}+\frac{1}{j+k}\right)+\sum_{\eta=1}^{+\infty}\left(\frac{1}{2j+\eta}-\frac{1}{\eta}\right)\right]\\&=&\frac{1}{2j}\left[H_j+(H_{2j-1}-H_{j-1})-H_{2j}\right]\\&=&\frac{1}{2j}\left[\frac{1}{j}-\frac{1}{2j}\right]=\color{red}{\frac{1}{4j^2}}\end{eqnarray*}$$
hence: $$\sum_{j=0}^{+\infty}\sum_{\substack{k=0\\k\neq j}}\frac{1}{j^2-k^2}=-\sum_{k=1}^{+\infty}\frac{1}{k^2}+\sum_{j=1}^{+\infty}\frac{1}{4j^2}=-\frac{3}{4}\zeta(2)=\color{red}{-\frac{\pi^2}{8}}$$
and you can compute the other sum in the same way, then check they do not match (they are opposite, indeed).
A: Your series is not absolutely convergent, if only because it includes a divergent positive sub-series: take all terms for which $k \ge 0$ and $j = k+1$; then $\frac1{j^2-k^2} = \frac1{2k+1}$, whose sum is $+\infty$.
Also, we know that in general, if $\sum_{ij} |a_{ij}| = +\infty$, we don't have $\sum_i\sum_j a_{ij} = \sum_j\sum_i a_{ij}$.
