Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You might find this interesting:… (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
up vote 2 down vote accepted

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 $p_m = 0$.

share|cite|improve this answer
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
@Florian - I think that's a great example. It gets to the heart of the matter. – John M Jul 18 '11 at 13:24
Thank you both, two nice counterexamples. – Mark Jul 18 '11 at 22:31

The condition is not sufficient. Take $b_{n,i} = \dfrac1{(1 + 1/n)^i}$ for example.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.