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What is the necessary and sufficient condition for $\sum\limits_{i=0}^\infty \sum\limits_{j=0}^\infty a_{ij}=\sum\limits_{j=0}^\infty \sum\limits_{i=0}^\infty a_{ij}$? Suppose that both sides are convergent. As we know, the absolute convergence of any side is sufficient. My textbook says there is a necessary and sufficient condition found by Markov. But I can't find it on the Internet. Can anyone tell me what is it?

Michael Hardy
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Eclipse Sun
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  • Hint: Use Fubini-Tonelli theorem. – Mustafa Said Apr 17 '14 at 07:06
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    Some related posts: http://math.stackexchange.com/questions/89814/positivity-and-interchange-of-summation, http://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits/ and http://math.stackexchange.com/questions/23057/under-what-condition-we-can-interchange-order-of-a-limit-and-a-summation – Martin Sleziak Apr 17 '14 at 08:41
  • Could you be more specific and instead of my textbook says give us a more precise reference to the book where you have seen this claim_ – Martin Sleziak Apr 17 '14 at 08:43
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    I'm reading Fichtenholz's Differential and Integral Calculus. The book may have no English translation. The claim is mentioned in Chapter 11. – Eclipse Sun Apr 17 '14 at 10:11
  • You wrote: \sum\limits_{i=0}^{\infty }{\sum\limits_{j=0}^{\infty }{{{a}{ij}}}}=\sum\limits{j=0}^{\infty }{\sum\limits_{i=0}^{\infty }{{{a}{ij}}}} I changed it to: \sum\limits{i=0}^\infty \sum\limits_{j=0}^\infty a_{ij}=\sum\limits_{j=0}^\infty \sum\limits_{i=0}^\infty a_{ij} Your overly complicated way of coding this just makes editing harder. – Michael Hardy Oct 16 '14 at 17:10
  • For example, you had {{{a}{ij}}} where a{ij} suffices. And you have {\sum} where \sum suffices. And you seem to think that whatever follows \sum must be enclosed in braces. And in your second sum you had {\sum etc. etc.} with needless braces surrounding the whole expression. – Michael Hardy Oct 16 '14 at 17:13

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After translating your reference and the writing of Markov to German, guessing which volume you are talking about, searching on Amazon inside the book for Markoff, then taking the name "Markoffsche Reihentransformation" and putting it in Google, one gets the following reference (Knopp, Theorie und Anwendung der unendlichen Reihen) giving the exact theorem of Markoff:

http://librarum.org/book/7105/261

So, basically, the criterion says that the sum of the tails of equal starting point has to converge to 0, i.e.

$$\lim_{m\to\infty}\sum_{i=0}^\infty \sum_{j=m}^\infty a_{ij} =0.$$

(The linked book by Knopp also says that the convergence of the interchanged double sum is equivalent to the mere existence of the limit.)

theobromine
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You can think of infinite sums as Lebesgue integrals over $\mathbb{N}$ with respect to the counting measure (which is $\sigma$-finite). Fubini's theorem pretty much says if $\mu$,$\nu$ are $\sigma$-finite measures, then $\int \int f(x,y) d\mu(x)d\nu(y)= \int \int f(x,y) d\nu(y)d\mu(x)$ provided either $f$ is positive or $\int \int |f(x,y)| d\mu(x) d\nu(y) <\infty$ (n.b in general the positive valued case can make this condition easy to check). This corresponds to being able to change the order of summation whenever $a_{ij}$ is either positive or the sum is absolutely convergent.

It's possible that there's a weaker condition, but I don't know of any. There's probably also proof of this not involving any measure theory, possible adapting the proof that an absolutely convergent series with just one variable can be re-arranged without changing the limit.

Xita Meyers
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Matt Rigby
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