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I would like to rewrite the sum

$$\sum_{i=1}^K \sum_{l=-\infty}^\infty \sum_{j=-\infty}^\infty f(i+lK;j-l)$$ In the form

$$ \dots\sum_{s=-\infty}^\infty \sum_{w=-\infty}^\infty f(s,w)$$ where $s=i+lK$, $w=j-l $. How do I do it?

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In what sense does the original sum converge? (I can imagine three possibilities: absolutely, conditionally, as a principal value). –  user31373 Jun 20 '12 at 17:57
    
The sum convergence absolutely –  Katja Jun 20 '12 at 18:02
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1 Answer 1

up vote 2 down vote accepted

I think that the sum is exactly $$\sum_{s=-\infty}^\infty \sum_{w=-\infty}^\infty f(s,w)$$ Indeed, consider an ordered pair $(s,w)\in\mathbb Z\times\mathbb Z$. The integer $s\in \mathbb Z$ has a unique representation $s=i+lK$ for some $i\in\{1,\dots,K\}$ and $l\in\mathbb Z$. Since $l$ is already determined, from $w=j-l$ we get $j=w+l$. Thus, $(s,w)$ appears in the original sum exactly once.

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