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Let $(a_{n,k})_{n, k \in \mathbb N} \subset \mathbb C$ be a series satisfying

$$ \sum_{n=0}^\infty \left| \sum_{k=0}^\infty a_{n,k}\right| \lt \infty $$ and $$ \sum_{k=0}^\infty \left|a_{n,k}\right| \lt \infty \qquad \forall n\in \mathbb N. $$ Does this imply that $\sum_{k=0}^\infty \sum_{n=0}^\infty a_{n,k}$ converges with $$ \sum_{k=0}^\infty \sum_{n=0}^\infty a_{n,k} = \sum_{n=0}^\infty \sum_{k=0}^\infty a_{n,k}, $$ ie. the order of summation can be rearranged?

In a lecture a proof was given (at least I think so!) for the case that instead of the first condition above the series satisfies $$ \sum_{n=0}^\infty \sum_{k=0}^\infty |a_{n,k}| \lt \infty. $$ (Am I right that this last condition itself is enough to justify switching the order because of the Fubini theorem?)

Can the series be rearranged in general or is there a (nasty) counter example?

Keba
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2 Answers2

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Consider the double series

$$1-1+0+0+0+0 +\cdots$$ $$0+2-2+0+0+0+\cdots $$ $$0+0+3-3+0+0 +\cdots $$ $$0+0+0 +4 -4+ 0 + \cdots$$ $$0+ 0+0+0+5-5+\cdots$$

where of course we continue the rows in this pattern on out to $\infty.$ We have absolutely convergent series in each row and column. But the interated double series sums to $0$ in one order, and $\infty$ in the opposite order.

zhw.
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  • This is also a nice example. (…and probably both quicker to check and easier to remember than the accepted answer.) – Keba Nov 09 '16 at 21:22
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The condition $$ \sum_{n=0}^\infty \left| \sum_{k=0}^\infty a_{n,k}\right| \lt \infty $$ is not sufficient, see the counter-example below.

Define a serie $\{a_{n,k}\}$, where $$ a_{n,k} = \cases{\frac{1}{n^2}-1 & if $n=k$, \\ 1 & if $n \ne k, k=1$, \\ 0 & else.} $$ Then we can easily verify that $$\sum_{n=0}^\infty \left| \sum_{k=0}^\infty a_{n,k}\right| = \sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} -1 \lt \infty,$$ but $$\sum_{k=0}^\infty \left|\ \sum_{n=0}^\infty a_{n,k}\right| = \infty.$$

  • Thanks for your answer. I was able to verify the counter example (although I think the series should star with $n=1$ and $k=1$.) – Keba Nov 09 '16 at 20:21
  • You're right, I confused myself with the notations. But it's not essential, anyway. – Victor Chen Nov 09 '16 at 20:23