Exchange the order of limit This is a problem I met when doing the homework, consider a positive sequence $\{a_{ij}\}$, if for each $i$, $\sum\limits_{j=1}^{\infty}a_{ij}$ exists and $\leq1$, for each $j$, $\lim\limits_{i\rightarrow\infty}a_{ij}$ exists and $\leq1$, then, do we have
$$\lim_{i\rightarrow\infty}\sum\limits_{j=1}^{\infty}a_{ij}=\sum\limits_{j=1}^{\infty}\lim_{i\rightarrow\infty}a_{ij}$$
 A: No. Let $\{a_{ij}\}$ be defined by 1's across the diagonal and zero everywhere else:*
$$a_{ij} = 
\begin{cases}
1, & \text{if $i = j$} \\
0, & \text{otherwise.} \\
\end{cases} $$
Then this sequence satisfies your hypotheses, but the limits are not equal. In fact,
$$\lim_{i \to \infty}\sum\limits_{j=1}^{\infty}a_{ij} = 
\lim_{i \to \infty} 1 = 1.$$
However,
$$\sum\limits_{j=1}^{\infty}\lim_{i\to\infty}a_{ij} =
\sum\limits_{j=1}^{\infty}0 = 0.
$$
*Of course, I'm interpreting "positive" as "non-negative". It isn't hard to make a strictly positive sequence. Let the indices $i$, and $j$ each start at 1. The following more complicated sequence works:
$$a_{ij} = 
\begin{cases}
1 - 1/2^{i}, & \text{if $i = j$} \\
1/2^{i+j}, & \text{otherwise.} \\
\end{cases} $$
A: Yes, the limit of the sum is the sum of the limits, provided each limit exists.
As Peter Tamaroff correctly states, the linked proof is not sufficient to assume you can, in fact, switch the sum and limit, but you can easily extend the method of the proof to infinite series.
