It's not a problem to multiply the series: the product is sum(t,k)inmathbbZ2atbk. 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)inmathbbZ2|atbk| is finite whenever sumtinmathbbZ|at| and sumkinmathbbZ|bk| are.
In general, sum(t,k)inmathbbZ2atbk=S if for any epsilon>0 there is a finite subset AsubsetmathbbZ2 such that |sum(t,k)inBatbk−S|<epsilon whenever B is finite and BsupsetA. 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.
No comments:
Post a Comment