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I'm told that the following is true when $m,n\in\Bbb Z$: $$\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n} \alpha_{ij}=\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{m} \alpha _{ij}$$ When $m,n\to\infty$,

  1. If both sides above have the same divergence or convergence?
  2. If the above equation is true?
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(a) What do you mean when you write summation $\Sigma$ with $m,n$ possibly not integers? Or do you mean $m,n\in \mathbb{Z}$ instead? (b) What do you mean by $m,n\to\infty$? – Willie Wong Oct 2 '12 at 15:05
Sorry, I mean $m,n\in\Bbb Z$. – Michael Li Oct 3 '12 at 12:43
up vote 5 down vote accepted

For any finite integers $m,n$, the expression $$ \sum_{i = 1}^m\sum_{j = 1}^n a_{ij} = \sum_{j = 1}^n \sum_{i = 1}^m a_{ij} $$ is true.

Your question should be interpreted as asking whether the "exchanging of limits" is allowed. I.e.

$$ \lim_{m\to\infty} \lim_{n\to\infty} \sum_{i = 1}^m\sum_{j = 1}^n a_{ij} \overset{?}{=} \lim_{n\to\infty}\lim_{m\to\infty} \sum_{j = 1}^n \sum_{i = 1}^m a_{ij} \tag{*}$$

Note that the left and right hand sides are often written, respectively, as $\sum_{i = 1}^\infty\sum_{j = 1}^\infty a_{ij}$ and $\sum_{j = 1}^\infty \sum_{i = 1}^\infty a_{ij}$.

(*) is true when $a_{ij}$ is absolutely summable. But in general there can be counterexamples. A particular one that comes to mind is

$$ a_{ij} = \begin{cases} 1 & i = j \\ -1 & i = j-1 \\ 0 & \text{otherwise}\end{cases} $$

Observe that $\sum_{j = 1}^\infty a_{ij} = 1 + (-1) = 0$ for any $i$. So the LHS of (*) evaluates to 0. On the other hand,

$$\sum_{i = 1}^\infty a_{ij} = \begin{cases} 1 & j = 1 \\ 0 & j > 1\end{cases} $$ so that the RHS of (*) evaluates to 1.

Now, for a more spectacular example, let $$ a_{ij} = \begin{cases} i & i = j \\ -i & i = j-1 \\ 0 & \text{otherwise}\end{cases}$$ The LHS still evaluates to 0. For the RHS however, notice that $\sum_{i = 1}^\infty a_{ij} = 1$ for every $j$, and so the RHS does not even converge!

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The problem basically boils down to that limits of double sequences do not commute in general. See also…… – Willie Wong Oct 2 '12 at 15:24

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