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Product of infinite sum

  1. Mar 23, 2010 #1
    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: Mar 24, 2010
  2. jcsd
  3. Mar 23, 2010 #2

    mathman

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    Science Advisor
    Gold Member

    Your trivial example is incorrect. i doesn't appear in the terms of the left side.
     
  4. Mar 24, 2010 #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?
     
  5. Mar 25, 2010 #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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