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.

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    $\begingroup$ Continuity is a local property. Every point of $B_r(z_0)$ is contained in some $B_\rho(z_0)$ with $0 < \rho < r$. $\endgroup$ – Daniel Fischer Dec 14 '14 at 21:32
  • $\begingroup$ @DanielFischer Indeed that's what my book says. It says each point belong to some compact subset of $B(z_0,r)$. But I can only see that it would belong to some open ball... $\endgroup$ – 3x89g2 Dec 14 '14 at 21:35
  • $\begingroup$ If $\lvert z-z_0\rvert < \rho < r$, then $B_\rho(z_0)$ is an open neighbourhood of $z$. And $\overline{B_\rho(z_0)}$ is a compact subset of $B_r(z_0)$. Therefore - what? $\endgroup$ – Daniel Fischer Dec 14 '14 at 21:37

Hint : define $f_n = \sum_{i=0}^{n}a_n(z-z_n)^i$, $\quad \lim_{n \to \infty}f_n =f$, $f_n$ is uniformly converges in $\overline{B_r(z_0)}$ ,and

Since ${f_n}$ is continuous then $f$ is contiuous


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