Series that evaluates to different values I am trying to find a sequence $x_{pq}$ such that $\sum_{p=1}^\infty\sum_{q=1}^\infty x_{pq}$ and $\sum_{q=1}^\infty\sum_{p=1}^\infty x_{pq}$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.

The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence. 
Consider, on the other hand, $a_{jk} = 1/(k^2-j^2)$ if $j\neq k$ and $a_{jk} = 0$ if $j=k$.
Here we have,
$$\frac{\pi^2}{8}=\tag{*}\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} \neq  \sum_{k=1}^\infty \sum_{j=1\\j \neq k}^\infty \frac{1}{k^2-j^2} = -\frac{\pi^2}{8} $$
To obtain the sum, note that
$$\sum_{k=1\\j\neq k}^K \frac{1}{k^2-j^2} = \frac{1}{2j}\sum_{k=1\\j\neq k}^K \left(\frac{1}{k-j} - \frac{1}{k+j} \right) \\= \frac{1}{2j}\left(-\sum_{k=1}^{j-1}\frac{1}{k} + \sum_{k=1}^{K-j}\frac{1}{k} -  \sum_{k=j+1}^{K+j}\frac{1}{k} + \frac{1}{2j} \right)\\=  \frac{1}{2j}\left(\frac{1}{j} +\sum_{k=j+1}^{K-j}\frac{1}{k}  -  \sum_{k=j+1}^{K+j}\frac{1}{k} + \frac{1}{2j} \right)\\ = \frac{1}{2j}\left(\frac{3}{2j} - \sum_{k=-j+1}^{j} \frac{1}{K+k}\right) $$
and,
$$\sum_{k=1\\j\neq k}^\infty \frac{1}{k^2-j^2} = \lim_{K \to \infty}\frac{1}{2j}\left(\frac{3}{2j} - \sum_{k=-n+1}^{n} \frac{1}{K+k}\right)  = \frac{3}{4j^2}$$
Thus,
$$\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} = \sum_{j=1}^\infty \frac{3}{4j^2} = \frac{3}{4} \frac{\pi^2}{6}$$
