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## 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!

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!

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