Showing that the sequence $z^n$ is normally but not uniformly convergent I was able to show that $z^n$ is normally convergent on the unit disk centred at the origin, but I am not sure how to show that it is not uniformly convergent on the unit disk centred at the origin.  I've read in a few places that this is true, but I have not seen a proof.
My first idea was to fix an $\epsilon$ such that $\epsilon=1/2$.  Then $|z^n|<1/2$.  However, I haven't been able to get this to work out.
 A: HINT: if $(f_n)_n$ is a sequence of continuous functions, what can you say about its limit if the sequence converges uniformly?
A: Hint: The "tail" of the series (when you truncate immediately after the $z^{n-1}$ term) is equal to $\frac{z^n}{1-z}$. For $z$ with $|z|\lt 1$ but fairly close to $1$, this has norm $\gt \frac{1}{2}|z|^n$. Given $\epsilon \gt 0$, is there an $N$ independent of $|z|$ such that $\frac{1}{2}|z|^n\lt \epsilon$ for all $n\gt N$ and all $|z|\lt 1$?  
A: Normal convergence means it should have a subsequence which converges uniformly on compact subsets.  Here our domain is the unit disk.  If you take a compact set $K\subset\mathbb{D}$ then we can find a closed disk $B[0, r]$ where $0<r<1$ such that $K\subset B[0, r]$. Now, if we consider the original sequence on $K$, then we have $\sup_{z\in K}|z|^n \leq r^n$, which goes to zero. Hence the given sequence converges normally in $\mathbb{D}$. As $\sup_{z\in \mathbb{D}}|z|^n = 1$ which does not goes to zero, the sequence does not converge uniformly in
$\mathbb{D}$.
