In my Analysis III course, we have recently proven the following result.
If the power series $$\sum_{n=1}^\infty a_nz^n$$ has radius of convergence $R$, then the series converges uniformly on $\{z\in\mathbb C:|z|<a\}$ for every $a<R$, i.e. on every closed interval $[-a,a]\subseteq(-R,R)$.
We then went on to say that although the power series converges on every closed interval subset of $(-R,R)$, it does not converge uniformly on the open interval $(-R,R)$ itself. But how can this be? If we can choose $a<R$ to be arbitrarily close to $R$, then surely this is equivalent to saying that it converges on the open interval $(-R,R)$ itself? Perhaps I've misunderstood what an open interval actually is - I've always thought of $(-2,2)$, for example, to be the interval obtained by making a closed interval $[-a,a]$ get arbitrarily close to when $a=2$, in a limiting way.
The only difference I can think of is that when we consider $[-a,a]$ where $a<R$, there is always a 'gap' between $a$ and $R$, no matter how close to $R$ we get; whereas there are no 'gaps' when considering the open interval. But I still can't see how this shows that we do not have uniform convergence.
Uniform convergence means that for any $\varepsilon>0$ there exists $n_0\in\mathbb{N}$, such that for all $n\ge n_0$ and $x\in\left[-a,a\right]$, $$\left|\sum_{k=1}^n a_k z^k-S\left(x\right)\right|<\varepsilon$$ where $S\left(x\right)$ is the limit function. The problem is that $n_0$ depends on $a$ (but not on $x$). So, theoretically, when you take $a$ close enough to $R$, the value of $n_0$ for a given $\varepsilon$ may tend to $\infty$, and thus the convergence may not be uniformly.