# cauchy product for general case [closed]

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

## closed as off-topic by Ricardo Andrade, Jack Huizenga, Lucia, David White, Daniel MoskovichDec 1 '13 at 6:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Jack Huizenga, Lucia, David White, Daniel Moskovich
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• Any answers to this question will depend on knowing something about the rates of convergence of your series. What do you know? – David E Speyer Feb 8 '10 at 19:21
• this series are absolute convergence – WBT Feb 8 '10 at 19:25
• 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.) – Harald Hanche-Olsen Feb 8 '10 at 19:40
• I don't think that's the right kind of question for MO... – Johannes Hahn Feb 8 '10 at 21:51
• Right, but absolutely convergent sums have none of these subtleties. – David E Speyer Feb 8 '10 at 22:06

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.