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How to multiply this series: $$(\sum_{t=-\infty}^{\infty}a_{t})(\sum_{k=-\infty}^{\infty}b_{k})$$

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    $\begingroup$ Any answers to this question will depend on knowing something about the rates of convergence of your series. What do you know? $\endgroup$ Feb 8, 2010 at 19:21
  • $\begingroup$ this series are absolute convergence $\endgroup$
    – WBT
    Feb 8, 2010 at 19:25
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    $\begingroup$ Check out en.wikipedia.org/wiki/Fubini%27s%5ftheorem#Corollary and replace the integrals by sums. Sums are indeed integrals with respect to counting measures, so this use of Fubini's theorem is completely kosher. Though of course, more elementary proofs can be given. (That wikipedia page needs some rewriting, though.) $\endgroup$ Feb 8, 2010 at 19:40
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    $\begingroup$ I don't think that's the right kind of question for MO... $\endgroup$ Feb 8, 2010 at 21:51
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    $\begingroup$ Right, but absolutely convergent sums have none of these subtleties. $\endgroup$ Feb 8, 2010 at 22:06

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It's not a problem to multiply the series: the product is $\sum_{(t,k)\in\mathbb Z^2} a_tb_k$. The question is how to sum the double series that we have.

For series with nonnegative terms summation is not a problem either: we take the supremum of all finite sums. And since any finite sum is contained in a sufficiently large square, it follows that $\sum_{(t,k)\in\mathbb Z^2} |a_tb_k|$ is finite whenever $\sum_{t\in\mathbb Z} |a_t|$ and $\sum_{k\in\mathbb Z} |b_k|$ are.

In general, $\sum_{(t,k)\in\mathbb Z^2} a_tb_k=S$ if for any $\epsilon>0$ there is a finite subset $A\subset \mathbb Z^2$ such that $|\sum_{(t,k)\in B}a_tb_k - S|<\epsilon$ whenever $B$ is finite and $B\supset A$. Now if both given series converge absolutely, then the contribution from outside of a large square is small, and it follows that $S$ is the product of two sums.

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