An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

Quiz problem: Outline of theorem 6.5.2


One of the problems Mark wanted us to be ready for was to outline the proof of the following:

If $\sum a_n x^n$ converges absolutely at $x_0$, then it converges uniformly on $[-c,c]$, whenever $ 0\leq c<|x_0|$.

I will go ahead and write out my outline; since we're calling it an outline, I'd love not only thoughts on logic, but also how much detail I'm including.

By assumption, we know that $\sum a_n x_0^n$ converges absolutely. Since $x \in [-c,c]$ are strictly less in absolute value than $x_0$, we know that $|a_n x^n|<|a_n x_0^n|$, and so we can apply the Weierstrauss $M$ test with $M_n = a_0 x_0^n$. Therefore, $\sum a_n x^n$ converges uniformly on $[-c,c]$.

I know that didn't fully flesh things out, but do you guys think that works as a rough outline?


I think it's absolutely crucial to write something along the lines of
$$\sum |a_n x^n| \leq \sum |a_n x_0^n| \left| \frac{x^n}{x_0^n}\right|.$$
You want control over the thing on the left and you've got control over the things on the right.


After doing some more exploring in our notes, and reading your comment, here are some thoughts:

We can choose an $M$ such that $a_n x_0^n < M$ for all $n$, and then we can write that

\sum |a_n x^n| \leq \sum |a_n x_0^n| \left| \frac{x^n}{x_0^n}\right| \leq \sum M \left| \frac{x}{x_0}\right|^n,

which is a converging geometric series since $\left| \frac{x}{x_0}\right| < 1$, and this implies absolute convergence of $\sum a_n x^n$ by the comparison test. This much we covered in lecture. What I am wondering is, do we get the result of uniform convergence from now applying the Weierstrauss $M$ test?