I need to prove that the complex power series $\sum\limits_{n=0}^{\infty}a_nz^n$ converges uniformly on the compact disc $|z| \leq r|z_0|,$ assuming that the series converges for some $z_0 \neq 0.$
I know that the series converges absolutely for every $z,$ such that $|z|<|z_0|.$ Since $\sum\limits_{n=0}^{\infty}a_nz_0^n$ converges, by definition of convergence this means that the tail part of the series can be made arbitrarily small. So $\lim\limits_{n \rightarrow \infty}a_nz_0^n = 0$ and this means that $\exists M \in \mathbb{R}^{+}$ so that $|a_nz_0^n| \leq M$ for $n \geq M.$ So it follows that $$|a_nz^n| \leq M \left|\dfrac{z}{z_0}\right|^n.$$ Thus $$\sum\limits_{n=0}^{\infty}|a_nz^n| \leq \sum\limits_{n=0}^{\infty}M \left|\dfrac{z}{z_0}\right|^n = M\sum\limits_{n=0}^{\infty}r^n,$$ where $r<1$ (because $|z|<|z_0| \implies \left|\dfrac{z}{z_0}\right|<1.$)
But how do I get started on this uniform convergence?
Best Answer
You must assume from the beginning that $r<1$. Then your argument proves that $$ |z|\le r|z_0|\implies |a_n\,z^n|\le M\,r^n. $$ The Weierstrass $M$-test shows that the convergence is uniform.