most recent 30 from http://mathoverflow.net 2013-05-24T20:22:10Z http://mathoverflow.net/feeds/question/14679 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14679/cauchy-product-for-general-case WBT 2010-02-08T19:18:18Z 2010-02-08T21:30:24Z

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

http://mathoverflow.net/questions/14679/cauchy-product-for-general-case/14700#14700 002 2010-02-08T21:30:24Z 2010-02-08T21:30:24Z

<p>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. </p> <p>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. </p> <p>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 $|\sum_{(t,k)\in B}a_tb_k - S|&lt;\epsilon$ 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.</p>