How can I swap the order of a finite product of infinite sums?

[smart_ansatz]

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 $$\nu$$) of infinite sums and seem to wave a magic wand and arrive at an infinite sum of finite products, thus:

$$\prod_{\nu} \sum_{k_{\nu}=-\infty}^{\infty} a_{k_{\nu}} \to \sum_{k}\prod_{\nu}a_{k_{\nu}}$$

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

$$\prod_{i=1}^{I}\sum_{n=1}^{\infty}\frac{1}{n^2} \ne \sum_{n=1}^{\infty}\frac{1}{n^{2I}}$$

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

Hope someone out there can help!

 

[mathman]

Your trivial example is incorrect. i doesn't appear in the terms of the left side.

 

[smart_ansatz]

yes it does

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

anyway, that's missing the point. We know this is not general. but is/are there any occasions when it can be done?

 

[smart_ansatz]

ok I've solved it. The key was in the "sum over all possible k" on the right hand side. Cheers anyway.