QUAM Review Question 2

I think I have a good solution to Question 2 on the QUAM review sheet based off what I understood from a brief discussion with Tim.

Suppose the series $\sum\limits_na_n(z-c)^n$ converges absolutely at $z_0\not=c$ and suppose that $0<R<|z_0-c|$, where $R$ is the radius of convergence of the series.

Let $G=\{z\in\mathbb{C}:|z-c|\leq R\}$ be a region.

This implies
$$|z-c|\leq R<|z_0-c| \ \forall \ z\in G,$$
so it must be the case that
$$|a_n(z-c)^n|<|a_n(z_0-c)^n| \ \forall \ z\in G.$$
Now let $f_n(z)=a_n(z-c)^n$ and let $M_n=|a_n(z_0-c)^n|$.

Recall $\sum\limits_n M_n$ converges by hypothesis.

Also recall
$$|f_n(z)|=|a_n(z-c)^n|<M_n \ \forall \ z\in G.$$
Therefore, $\sum\limits_n f_n=\sum\limits_n a_n(z-c)^n$ converges uniformly on $G$ by the Weierstrass M-test.

That looks great - @dgallimo!!

Don't forget that $$\sum_{i=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}!!!$$

just in time! don't ask why I received this update....