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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These two questions seem to be related: Under what condition we can interchange order of a limit and a summation? and When can you switch the order of limits? – Martin Sleziak Dec 09 '11 at 18:44
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depends on the definition of summability you are using, everyone here assumes the classic summability. So make sure they specifically specify if they mean classic summability or not. – jimjim May 08 '20 at 02:50
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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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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 03:46
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Yes. Because either way it is equal to $\displaystyle{\sup\limits_{M,N}\;\sum_{n=1}^N\sum_{m=1}^M a_{m,n}}$.
Jonas Meyer
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@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 09 '11 at 06:46
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1One 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 09 '11 at 23:27
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1@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 01:20
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Here's an elementary proof fleshing out @JonasMeyer's answer.
Claim: If $a_{m,n}\ge 0$ for all $m,n$ then $$ \sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n}= \sup_{M,N}\sum_{m=1}^M\sum_{n=1}^N a_{m,n}.\tag1 $$ Proof. Let $S$ denote the RHS of (1). Write $$S_{M,N}:=\sum_{m=1}^M\sum_{n=1}^Na_{m,n}\quad\text{and}\quad S_M:=\lim_N S_{M,N}=\sum_{m=1}^M\sum_{n=1}^\infty a_{m,n}.$$ The limit $S_M$ exists, possibly with value infinity, since $S_{M,N}$ is monotonically increasing in $N$. By definition of sup, we have $S_{M,N}\le S\ \forall M,N$, hence $S_M\le S\,\forall M$, and $\color{green}{\lim_M S_M\le S}$. But $S_{M,N}\le S_M$ by non-negativity of $a_{m,n}$, so $S:=\sup S_{M,N}\le\sup S_M$. Since $S_M$ is nondecreasing (non-negativity again), this means $\color{green}{S\le\lim_MS_M}$. To finish, check that $\lim_M S_M$ equals the LHS of (1).
Corollary: If $a_{m,n}\ge 0$ for all $m,n$ then $$ \sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n}= \sum_{n=1}^\infty\sum_{m=1}^\infty a_{m,n}. $$ Proof. First apply the Claim to $a_{m,n}':=a_{n,m}$, then swap indices to obtain $$ \sum_{n=1}^\infty\sum_{m=1}^\infty a_{m,n}= \sup_{M,N}\sum_{n=1}^N\sum_{m=1}^M a_{m,n}.\tag2 $$ Finally note that the RHS of (1) equals the RHS of (2).
grand_chat
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