Uniform convergence of $\frac{1}{1+nz}$ $\bigg(\dfrac{1}{1+nz}\bigg)$ converges pointwise to $f(z) = 0$ on $\mathbb{C}- \{0\}$.

What is the region of uniform convergence of $\bigg(\dfrac{1}{1+nz}\bigg)$ to $f(z) = 0$?

For any $r>0$ , let $D_r = \{|z|>r\}$.
For any $\epsilon >0,\ $ let N be an integer larger than $\frac{1}{r\epsilon}$.
Then $\forall n\geq N, \forall z\in D_r$, 
$|\frac{1}{1+nz}| \leq |\frac{1}{nz}|<\frac{1}{nr} < \epsilon$.
Hence, the region of uniform convergence is $|z|>r$ for any $r>0$. $\ \ \ \square$
Is this the correct region of uniform convergence?
If we let $z=\frac{1}{n}$, then $\frac{1}{1+nz} = \frac{1}{2}$, which contradicts the above result.
Where did i go wrong? Help please.

Furthermore, i want to prove that $(\frac{1}{1+nz})$ converges uniformly on any compact subset of the complex plane that does not contain the origin.

Suppose that $K$ is a compact subset of $\mathbb{C}$. I want to show that there exists $\delta>0$ such that $\forall z\in K, |z| \geq \delta$.
I was given the following topological result (without proof) in the introductory course:

If $f:K\to \mathbb{C}$ is continuous, then $|f|$ is continuous and $\max_{z\in K} |f(z)|$ exists.

Let $f(z) = \frac{1}{z}$, where $f$ is continuous on $K$. Then, $\exists m = \max_{z\in K} \frac{1}{|z|}$. Hence, $\min_{z\in K} |z|= \frac{1}{m}$ and let $\delta = \min_{z\in K} |z|$.
Is this correct? 

There's not "the" region, but any of the $D_r$ sets with $r>0$ is a region of uniform convergence. That is: once you fix an $r>0$ you have $D_r$ a region of u.c. But is not true that $\bigcup_{r>0}D_r=\mathbb{C}\setminus\{0\}$ is a region of u.c.: your example actually disproves it. But your example doesn't work on a fixed $D_r$ since only finitely many points of the form $\tfrac1n,\, n\in\mathbb{N},$ are in $D_r$.
