Consider a power series $\sum a_n(z-z_0)^n$, and assume it has radius of convergence $r$. Then we know that $\forall z\in(z_0 -r,z_0 +r)$, this power series converges absolutely by root test. Thus we can define $f(z):=\sum a_n(z-z_0)^n, z\in B_r(z_0)$.
Moreover, we know that if we have a compact set $\bar B_\epsilon(z_0) \subset B_r(z_0),$ the power series actually converges uniformly to $f(z)$ in $\bar B_\epsilon(z_0)$.
I'm trying to prove that $f(z)$ is actually a continuous function on $B_r(z_0)$. Now, if $z \in \bar B_\epsilon(z_0)$, then since $\sum a_n (z-z_0)^n$ converges to $f(z)$ uniformly, and each $a_n (z-z_0)^n$ is apparently continuous, then on that compact subset $f(z)$ is continuous. But that's how far I got. How do I extend the conclusion to the whole disk of convergence, not just a compact subset of it?
Update: I was confusing myself. To conclude that function $f(z)$ is continuous, we don't need $\sum a_n (z-z_0)^n$ to converge uniformly on the entire $B_r(z_0)$. It suffices to show that for each $x \in B_r(z_0)$ the series converges uniformly on a compact subset around it.