# Interchanging the order of limits of a particular double sequence

I would like to interchange the iterated limits of a particular double sequence $a_{n,m}$, that is I would like that $\lim_{m\to\infty}\lim_{n\to\infty}a_{n,m}=\lim_{n\to\infty}\lim_{m\to\infty}a_{n,m}$. The sequence has the following properties.

Firstly, it is actually formed by a sum over one of the variables of a double sequence $b_{n,i}$, so that $a_{n,m}=\sum_{i=1}^mb_{n,i}$, where $0\leq b_{n,i}\leq 1$. Hence for each fixed $n$, $a_{n,m}$ is an increasing sequence in $m$.

For each fixed $n$ the limit $\lim_{m\to\infty}a_{n,m}=l_n$ exists, i.e. the series $\sum_{i=1}^\infty b_{n,i}$ converges.

The iterated limit $\lim_{n\to\infty}\lim_{m\to\infty}a_{n,m}=\lim_{n\to\infty}\sum_{i=1}^\infty b_{n,i}$ exists.

For each fixed $m$ the limit $\lim_{n\to\infty}a_{n,m}=\lim_{n\to\infty}\sum_{i=1}^m b_{n,i}=\sum_{i=0}^m\lim_{n\to\infty}b_{n,i}=p_m$ exists.

I know that if it were the case that $a_{n,m}\to l_n$ uniformly in $n$ then the double limit $\lim_{n,m\to\infty}a_{n,m}=a$ exists, and then since $\lim_{n\to\infty}a_{n,m}=p_m$ exists, the iterated limits would commute and equal $a$ (as in When can you switch the order of limits?), but I don't have uniformity.

So are the above properties sufficient for the the iterated limits to commute, or do I need something more?

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You might find this interesting: math.la.asu.edu/~jss/courses/fall06/mat472/… (It's not an answer to the question; I think John has already answered the question; but it's a nice general treatment of interchange of limits and uniform convergence.) – joriki Jul 18 '11 at 4:50
@joriki thanks for the link. – Mark Jul 18 '11 at 22:32

How about $a_{n,m}=(1-1/m)^n$? This is increasing in $m$, with limit $l_n=1$, so the iterated limit exists and equals $1$. However, as $n$ goes to infinity, $a_{n,m}$ goes to $p_m = 0$.
An even simpler counterexample would be $b_{n,i}=\delta_{n,i}$, so $a_{n,m}=1$ if $m\ge n$ and 0 otherwise. – Florian Jul 18 '11 at 8:11
The condition is not sufficient. Take $b_{n,i} = \dfrac1{(1 + 1/n)^i}$ for example.
This example doesn't satisfy that the iterated limit $\lim_{n\to\infty}\lim_{m\to\infty}a_{n,m}$ exists, since $\sum_{i=1}^\infty b_{n,i}=n$. – Mark Jul 18 '11 at 1:22