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!
Under what condition we can interchange order of a limit and a summation?
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1see https://www.jstor.org/stable/1967602?seq=1#page_scan_tab_contents – 2016-09-05
4 Answers
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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2I 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). – 2013-06-28
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). $
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}