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
![+a_N x^N R_0(x)].](https://s0.wp.com/latex.php?latex=%2Ba_N+x%5EN+R_0%28x%29%5D.&bg=ffffff&fg=333333&s=0&c=20201002)
Now,
clearly converges to
as
, and the convergence is uniform in the interval ![{}[-r,r].](https://s0.wp.com/latex.php?latex=%7B%7D%5B-r%2Cr%5D.&bg=ffffff&fg=333333&s=0&c=20201002)
We want to show that the remaining term
converges to 0 and does so uniformly in the interval ![{}[-r,r].](https://s0.wp.com/latex.php?latex=%7B%7D%5B-r%2Cr%5D.&bg=ffffff&fg=333333&s=0&c=20201002)
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 ![x\in{}[-r,r].](https://s0.wp.com/latex.php?latex=x%5Cin%7B%7D%5B-r%2Cr%5D.&bg=ffffff&fg=333333&s=0&c=20201002)
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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