$f_{n}=\frac{1}{1+z^{n}}$ uniform convergence $\newcommand{\Z}{\mathbb{Z}}$ 
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I am trying to show and disprove uniform convergence in the following example: 

$$D=\{z\in \C | |z| < 1 \} \ and \ f_{n}:D\rightarrow \C : f_{n}(z)=\frac{1}{1+z^{n}}  $$

Proposition 1: 
$f_{n}$ converges uniformly in all $B(0)$ with $0<r<1$
Proof 1: With $f_{n}(z)= (1+z^n)^{-1}$ put   
$$ f'_{n}(z)=-nz^{n-1}(1+z^n)^{-2}=0 \Rightarrow z=0$$ 
(I wanted to use that $\lim sup |f_{n} - f | = 0 $ but I fail at finding the supremum of $|f_{n}-f|$) How does one find the supremum? 
So I try finding an estimate instead: $$|f_{n} - f| = |\frac{1}{1+z^{n}}-1| = |\frac{-z^{n}}{1+z^{n}}| \le \frac{r^{n}}{|1-r^{n}|} =: \epsilon$$
Is this done correctly? 
Proposition 2: $f_{n}$ does not converge uniformly in D
Proof 2: How can one show that something does not converge uniformly?? 
Thanks for suggestions.

If $|z|<1$, then $z^n \to 0$ as $n \to \infty$, so $f_n$ converges pointwise to the constant function $1$.
On the other hand $\sup_{z\in D} |f_n| = +\infty$. (To see this, for a fixed $n$, take a sequence of points in $D$ converging to a $n$:th root of $-1$.)
On $D_r = \{ z : |z| < r \}$, (if $r < 1$)
$$ \left| \frac{1}{1+z^n} - 1 \right| = \left| \frac{z^n}{1+z^n} \right| < \frac{r^n}{1-r^n} \to 0\quad\text{as $n\to\infty$}.$$
