Suppose
$
\displaystyle\sum_{n=0}^\infty a_n z^n
$
converges at $z = z_1$. Does the power series then converge uniformly on the open ball $B(0, |z_1|)$?
Let $R$ be the radius of convergence of the power series. If $|z_1| < R$, then $B(0,|z_1|)$ is contained in some closed ball with radius $R'$ where $|z_1| < R' < R$. We know the series converges uniformly in the closed ball $\overline{B(0,R')}$, so it must also converge uniformly in the open ball $B(0,|z_1|)$.
What happens if $|z_1| = R$? Intuitively I think this doesn't converge uniformly, since otherwise we wouldn't need the closed ball condition. But I can't seem to think of a counterexample.
Best Answer
Take $a_n=\frac 1 n$ and $z_1=-1$. $\sum \frac {(-1)^{n}} n$ conveges but $\sum \frac 1 n z^{n}$ does not converge uniformly for $|z| <|-1|$.
[ If it is uniformly convergenet then there exists $N$ such that $|\sum\limits_{k=N+1}^{N+n}a_kz^{k}|<\epsilon$ for all $n$ whenever $|z| <1$. Let $z\to 1$ to get a contradiction].