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Suppose I have $\sum\limits_{n = 1}^{\infty}\sum\limits_{m = 1}^{\infty} a_{m, n}$ where $a_{m, n} \geq 0$ for all $m$ and $n$. Can I interchange the two summations? If so why?

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up vote 3 down vote accepted

It's worth knowing that rearrangements can change the value of sums or integrals only if the positive and negative parts both diverge to infinity.

Fubini's theorem says rearrangements are fine if both parts are finite.

Tonelli's theorem says rearrangements are fine if what's being summed or integrated is everywhere non-negative. (It follows that it also works if it's everywhere non-positive, since the minus sign pulls out.)

Putting the two together gives you what I said in the first paragraph.

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+1 for connecting this to more general results. –  Jonas Meyer Dec 9 '11 at 18:53
Basically, if the series is conditionally convergent then the order of the elements being added matter. However, that is not to say that you will necessarily get a different sum for any two orders. –  CogitoErgoCogitoSum Feb 22 '13 at 3:46
what about the formal series? –  Val Aug 7 '13 at 6:46
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Yes. Because either way it is equal to $\displaystyle{\sup\limits_{M,N}\;\sum_{n=1}^N\sum_{m=1}^M a_{m,n}}$.

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Hmm, what about convergence issues? –  Gottfried Helms Dec 9 '11 at 6:12
@GottfriedHelms: If you show that convergence of one of the iterated series implies convergence of the other, then by contraposition you also have that divergence of one of the interated series implies divergence of the other. –  Jonas Meyer Dec 9 '11 at 6:46
One comment about convergence issues is that (with all terms non-negative) the interchange is valid even if the sum does not converge. –  Michael Hardy Dec 9 '11 at 23:27
@Michael: Yes, that is the point I meant to express in my comment, along with a way to see that it is so. (And one could allow $\sup$ to take on the value $+\infty$ to apply the suggestion in my answer.) –  Jonas Meyer Dec 10 '11 at 1:20
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