^{n}is equal to 1, and so forth. In the case of countably infinite sums, questions of convergence become important. But what happens when you have a sum over uncountably many summands? It turns out that there is a simple answer to this, but it requires some care.

The typical definition for the sum of an arbitrary collection of non-negative summands is to take the supremum of sums over all finite subsets of the summands. It can be shown that if the collection S of positive summands in the sum is uncountable, then the sum must be infinite. Indeed, suppose by way of contradiction that the sum of elements in S is finite. Then the collection S

_{n}of summands s ≥ 1/n must be finite for each n--otherwise, the supremum over all finite collections of such summands would be larger than k/n for any k. But then we may write S as the countable union over all n of the S

_{n}, and this implies that S is countable, a contradiction with the starting hypothesis. Thus we must conclude that the sum is infinite.

Although a general sum is not defined for complex summands, the argument can be extended to this context to show that partial sums of an uncountable collection of complex numbers may have arbitrarily large magnitude. Certainly, using the same argument as above, an uncountable number of summands must have magnitude at least 1/n for some n. Further, an uncountable number of these must have argument within an interval of length at most π/4. The combination of these two conditions restricts either the imaginary or the real part of the summands, fixing the sign and providing a strictly positive lower bound on the magnitude. Since we have infinitely (uncountably) many such summands, the result follows.

Very nice, sir... an actually-understandable treatment!

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