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Suppose f(m,n) is a double sequence in $\mathbb R$. Under what condition do we have $\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty f(m,n)=\sum\limits_{m=1}^\infty \lim\limits_{n\to\infty} f(m,n)$? Thanks!

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3 Answers 3

up vote 14 down vote accepted

A fairly general set of conditions, sufficient for many applications, is given by the hypotheses of dominated convergence. (Note that sums are just integrals with respect to the counting measure on $\mathbb{N}$, so dominated convergence applies with no modification.)

Without domination, the idea is that lumps of positive mass can "escape to infinity" when one attempts to interchange sum and limit. Here is a basic example: let $f_{m,n} = 1$ if $m = n$ and $0$ otherwise. Then $\sum_{m=1}^{\infty} f(m, n) = 1$ for all $n$, so the LHS is $1$, but $\lim_{n \to \infty} f(m, n) = 0$, so the RHS is $0$. The point of domination is to prevent these lumps of mass from escaping.

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Thank you, Yuan. It is a good idea to view the sum of a series as an integral w.r.t. counting measure. But due to the property of absolute integrability, only absolutely convergent series can be viewed as an integral, while conditionally convergent series can not, as a result, we have to assume that $\sum\limits_{m=1}^\infty f(m,n)$ converges absolutely for each n, right? –  zzzhhh Feb 21 '11 at 13:37
    
@zzzhhh: yes, if your series don't converge absolutely then it's not a good idea to try exchanging limits and sums anyway... –  Qiaochu Yuan Feb 21 '11 at 14:11
    
Based on this result, the method Yuan mentioned, and monotone convergence theorem, we can prove: Given a sequence $<\mu_j>$ of measures on $(X,\mathcal M)$, define measure $\mu=\sum\limits_{j=1}^\infty \mu_j$. If $f\in M^+(X,\mathcal M)$ or $f\in L^1(\mu)$, Then $\int fd\sum\limits_{j=1}^\infty\mu_j=\sum\limits_{j=1}^\infty\int fd\mu_j$. –  zzzhhh Feb 21 '11 at 17:53
    
I think uniform convergence is another weaker condition. By virtue of Weierstrass M-test, dominated convergence implies uniform convergence (in the case of discrete summation, rather than Riemann integral). –  Frank Science Jun 28 '13 at 16:43

Not an answer to your question as such, but a note which seems worth making. Again from the point of view of measure theory (as previously mentioned by Qiaochu Yuan), you can use Fatou's Lemma to show that you have:

$$ \lim_{n \rightarrow \infty} \sum_{m=1}^\infty f(m,n) \geq \sum_{m=1}^\infty \left( \lim_{n \rightarrow \infty} f(m,n) \right). $$

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Denote $h(m)=\lim_{n\to\infty}f(m,n)$, and suppose that $\sum_{m=1}^\infty h(m)<\infty$. Then my guess is that the sufficient condition is

\begin{align} \limsup_{n,m}\left|\sum_{k=1}^mf(k,n)-\sum_{k=1}^m h(k)\right|=0. \end{align}

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