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I'm trying to rewrite a double sum in the following format

$$ \sum_{l=0}^\infty \sum_{n=0}^\infty z^{n-2l} g(n,l) = \sum_{m=-\infty}^\infty z^m h(n) $$

For some $h(n)$, which will probably involve a second sum. In other words, I'm trying to pull the $z$ term out of one of the sums (but not necessarily getting rid of the second sum, which I think in general is not always possible).

Is this possible? In general, how do you solve these types of double sum problems?

smörkex
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1 Answers1

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Assuming that the double sum converges absolutely (such that we are allowed to rearrange the order of summation) collecting terms of the same degree yields

$$ h(m) = \sum_{\substack{n-2l =m \\ n,l\geq 0}} g(n,l) = \sum_{\substack{n=2l+m \\ n,l\geq 0}} g(n,l) =\sum_{l\geq \max\{0, \lceil -\frac{m}{2} \rceil \}} g(m+2l, l).$$

Where $l\geq \max\{0, \lceil -\frac{m}{2} \rceil \} $ comes from the assumption that $2l+m=n\geq 0$.

Severin Schraven
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