Product of infinite sum

  • #1

Main Question or Discussion Point

Hi this the first time I've got completely stuck and need some advice. I'm trying repeat a (supposedly simple) derivation that appeared in a recently published paper. The details are not important, but I am stuck on a part of that calculation that they skip over.

They have a finite product (over [tex]\nu[/tex]) of infinite sums and seem to wave a magic wand and arrive at an infinite sum of finite products, thus:

[tex]
\prod_{\nu} \sum_{k_{\nu}=-\infty}^{\infty} a_{k_{\nu}} \to \sum_{k}\prod_{\nu}a_{k_{\nu}}
[/tex]

This is clearly not a general statement as the trivial example below shows.

[tex]
\prod_{i=1}^{I}\sum_{n=1}^{\infty}\frac{1}{n^2} \ne \sum_{n=1}^{\infty}\frac{1}{n^{2I}}
[/tex]

So does anyone have any idea how to swap the order of a finite product of an infinite sum? maybe there are some conditions [tex]a_{k_{\nu}}[/tex] has to obey for it to be valid.

Hope someone out there can help!
 
Last edited:

Answers and Replies

  • #2
mathman
Science Advisor
7,761
415
Your trivial example is incorrect. i doesn't appear in the terms of the left side.
 
  • #3
yes it does

[tex]
\prod_{i=1}^{I}\sum_{n}\frac{1}{n^2} = \left(\sum_{n}\frac{1}{n^2}\right)(\ldots) = \left(\frac{\pi^2}{6}\right)^I \ne \sum_{n}\left(\frac{1}{n^2}\right)^I
[/tex]

anyway, that's missing the point. We know this is not general. but is/are there any occasions when it can be done?
 
  • #4
ok I've solved it. The key was in the "sum over all possible k" on the right hand side. Cheers anyway.
 

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