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.
 A: Maybe you are looking for something like this: $$\begin{pmatrix}1 &-1 & 0&0&0 & 0 &\cdots \\
0 & 1 & -1& 0 & 0 & 0&\cdots\\
0 & 0 & 1 & -1 & 0 & 0&\cdots\\
0 & 0 & 0 & 1 & -1 & 0&\cdots \\
0 & 0 & 0 & 0 & 1 & -1&\cdots \\
\vdots &\vdots  &\vdots  & \vdots & \vdots & \ddots&\ddots 
\end{pmatrix}.$$
So that $x_{p,p} = 1$, $x_{p,p+1}=-1$ and $x_{p,q} = 0$ if $q \neq p,p+1$. 
Here we have $$\sum_{q=1}^{\infty} x_{p,q} = 0$$ for all $p \in \mathbb N$, and so $$\sum^\infty_{p=1} \sum^\infty_{q=1} x_{p,q} = 0.$$ However, we also have $$\sum_{p=1}^{\infty} x_{p,q} = 0 \,\,\,\, \text{ for } \,\,\,\, q \ge 2, \,\,\,\,\, \text{ but } \,\,\,\,\, \sum_{p=1}^{\infty} x_{p,1} = 1$$ and so $$\sum^\infty_{q=1} \sum^\infty_{p=1} x_{p,q} = 1.$$
As was said in the other answer, if $$\sum_{p,q} \lvert x_{p,q}\rvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_{p,q} >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+\infty$.  
A: 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,
$$\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{1}{k^2 - j^2} = \sum_{j=1}^\infty \frac{3}{4j^2} = \frac{3}{4} \frac{\pi^2}{6}$$
