Double summation switch if one index is infinite and the other finite? Is the following equation generally true?
$$\sum_{i=1}^n \sum_{j=1}^\infty\left(a_{i,j}\right)=\sum_{j=1}^\infty \sum_{i=1}^n\left(a_{i,j}\right)$$ If true, how would you prove it? 
 A: You can not switch the summation in general. An example was given in comments if $a_{1,j}=-a_{2,j}=\cos(j)$ then $$\sum_{i=1}^2\sum_{j=1}^\infty a_{i,j} $$ does not converge while $$\sum_{j=1}^\infty\sum_{i=1}^2 a_{i,j} $$ converges trivially. If all the terms are nonnegative then you can switch the order that's the Toneli theorem. Using the Fubini-Toneli theorem you can get a slightly stronger result and that is that you can swap whenever the sums converge in absolute value. I don't think you can do any better then that.
A: For each finite $m$, we have 
$$ \sum_{i=1}^n \sum_{j=1}^m a_{ij} = \sum_{j=1}^m \sum_{i=1}^n a_{ij}$$
If the limits $$L_i = \sum_{j=1}^\infty a_{ij} = \lim_{m \to \infty} \sum_{j=1}^m a_{ij}$$ all exist,
then 
$$ \sum_{j=1}^\infty \sum_{i=1}^n a_{ij} = \lim_{m \to \infty} \sum_{j=1}^m \sum_{i=1}^n a_{ij} = \lim_{m \to \infty} \sum_{i=1}^n \sum_{j=1}^m a_{ij}
= \sum_{i=1}^n L_i = \sum_{i=1}^n \sum_{j=1}^\infty a_{ij}  $$ 
However, it is possible for $\sum_{j=1}^\infty \sum_{i=1}^n a_{ij}$ to exist
but some $L_i$ not to exist, in which case $\sum_{i=1}^n \sum_{j=1}^\infty a_{ij}$ doesn't exist.  For example, take $n=2$, all $a_{1j} = 1$, all $a_{2j} = -1$. 
