This is a simple consequence of the fact that each function $S_k:x\mapsto1+x+\cdots+x^k$ is bounded while the limit function $S:x\mapsto1/(1-x)$ is not.
Hence each function $S_k-S$ is unbounded, that is, the sup-norm of $S_k-S$ is infinite, in particular the sequence of the sup-norms does not converge to zero. This last assertion is equivalent to the fact that $(S_k)$ does not converge uniformly to $S$.
A series may converge (absolutely and therefore uniformly) at all points of the boundary of convergence, consider $\displaystyle \sum \frac{z^n}{n^2}$.
However, there is at least one point along which the holomorphic function the series defines cannot be analytically continued. Intuitively, think of $\sum z^n$. This series coincides with $1/(1-z)$, which is holomorphic everywhere (except at $z=1$), even though the series only has radius of convergence 1. What this result says is that the "except at $z=1$" is unavoidable.
Let me be more precise. A point on the boundary of the disk $D$ of convergence of a power series (say, about 0) is called regular iff there is an open neighborhood $U$ of that point and an analytic function that coincides with the power series on the intersection of $D$ and $U$. Otherwise, the point is singular.
The theorem your instructor was referring to is probably the following:
Theorem. If a power series has finite radius of convergence, then the set of singular points is a nonempty closed subset of its boundary of convergence.
The idea of the proof that there is at least one singular point is easy. For details, examples, and a discussion of analytic continuation, see for example Chapter 5 of Berenstein-Gay "Complex Variables. An introduction", or a similar textbook.
The thing is, calling $f$ the function defined by the power series, if there are no singular points, we can find around each point $p$ of the boundary, a little disk $D_p$ and an analytic function $f_p$ that coincides with $f$ on $D_p\cap D$.
But, by connectedness, if $D_p$ and $D_{p'}$ intersect, then on their intersection $f_p$ and $f_{p'}$ coincide. This is because they coincide (with $f$) on $D_p\cap D_{p'}\cap D$, which is nonempty if $D_p\cap D_{p'}\ne\emptyset$ to begin with.
Using compactness, we can then use the disks $D_p$ to see that there is a disk concentric with $D$ but slightly larger where there is an analytic function extending $f$. But then, the disk of convergence of $f$ was not $D$ to begin with.
Best Answer
The power series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n^2}$ has radius of convergence $1$ and it converges uniformly on $\overline{D(0,1)}$; in particular, it converges uniformly on $D(0,1)$.