414/514 – Multiplying power series

In lecture today we argued that under reasonable circumstances, the product of two power series is the series given by their “formal” product. I think I unnecessarily made the argument more complicated than it is, so I am writing it here so we have a clean reference.

Let and . Suppose that both and converge absolutely and uniformly in the interval . We want to prove that for , we also have

and this latter series converges absolutely and uniformly in .

To see this, let and denote the partial sums of and , and denote by the partial sums of the “formal product” series.

We want to show that

Let us call the “remainder” or “tail” series of , i.e.,

We have that Expanding this last expression, we find that it equals

Now, clearly converges to as , and the convergence is uniform in the interval

We want to show that the remaining term converges to 0 and does so uniformly in the interval

First, since uniformly as , then for any there is an such that if then for all .  If , we can split the sum above as , where

and

Note that , where is the constant This shows that uniformly for

To bound , note that it is a sum of a constant number of terms, namely . The functions are continuous and bounded in the interval , so there is a constant that bounds all of them (and, of course, does not depend on ). Hence

Since converges, there is an such that if , then

Pick so that . Then . We have now shown that uniformly for , and this completes the proof.

(We also claimed that the convergence is absolute. To see this, replace all the terms above by their absolute values. The same argument shows convergence of the relevant series, so we indeed have absolute convergence.)

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