Is $\sum^\infty_{n=0}(\frac z{1+z})^n$ holomorphic on the right half plane? Since $\sum^k_{n=0}(\frac z{1+z})^n$ is holomorphic for every $k$ in the right half plane, it suffices to show that the series is uniformly convergent in every compact subset of right half plane. But how do we show this?

The geometric series $\sum_{n=0}^\infty w^n$ converges to $(1-w)^{-1}$ for $|w|<1$ and diverges otherwise.
So $\sum_{n=0}^{\infty} (\frac{z}{1+z})^n=(1-\frac{z}{1+z})^{-1}=1+z$ for $|\frac{z}{1+z}|<1$ and diverges otherwise. Of course, $1+z$ is holomorphic, so this series is holomorphic everywhere it exists (where $|\frac{z}{1+z}|<1$).
Note $|\frac{z}{1+z}|<1 \iff |z|^2<|1+z|^2$ and $|1+z|^2=1+2\mathrm{Re}(z)+|z|^2$; can you simplifty?
