Let $\mathbb{E}$ denote the open right half-plane, that is $\mathbb{E}=-i\mathbb{H}.$
There are at least two answers on MSE showing ordinary convergence in $-\frac{1}{2}+\mathbb{E}=\{z:\Re z>-\frac{1}{2}\}.$ In this domain the function is actually a polynomial; see the nice solution here: https://math.stackexchange.com/a/3910774 .
My interest however is in following up on the hint given in the first comment to that same question (Is $\sum^\infty_{n=0}(\frac z{1+z})^n$ holomorphic on the right half plane?), in order to practice working with uniform convergence. I was able to alter the idea of the hint to show that when $|z|\leq M$ and $\Re z>0:$
$$\frac {|z|}{|1+z|}\,=\,\frac {|z|}{\sqrt{(1+\Re z)^2+(\Im z)^2}}\,=\,\frac {|z|}{\sqrt{1+|z|^2+2\Re z}}\,\leq\,\frac {|z|}{\sqrt{1+|z|^2}}\,\leq\,\frac {M}{\sqrt{1+M^2}}<1.$$
(This gives uniform convergence for compacts in $\mathbb{E}.$) For the final "$\leq$" here, I am appealing to the strict increase of $f(x)=\frac{x}{\sqrt{1+x^2}}.$ The first thing I'd like to know is how to implement the original hint, using the value $r$? Clearly it would be sufficient, and I am supposing doing it that way, we avoid calculating a derivative.
Secondly, can the hint be modified to show directly uniform convergence in compact subsets of $-\frac{1}{2}+\mathbb{E}$? I think such convergence holds, if only because $z \mapsto \frac{z}{1+z}=w$ is Möbius/linear fractional, so this map will take a compact to a compact in the unit disc $\mathbb{D}$, where we have a power series $\sum w^n$.