Consider the following series \begin{equation} f(z)=1+\sum_{k=1}^{\infty}\frac{1}{2^{k z}} =1+ \sum_{n=1}^{\infty}\left( \prod_{k=1}^{n}\frac{1}{2^{z}} \right) \end{equation} Using Euler's continued fraction formula we can express this as a continued fraction \begin{equation*} f(z)= \cfrac{1}{ 1- \cfrac{2^{-z}}{ 1+2^{-z}- \cfrac{2^{-z}}{ 1+2^{-z}- \cfrac{2^{-z}}{ 1+2^{-z} - \ddots}}}} \end{equation*} or more succinctly \begin{align*} f(z)=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}\left(1-\bigk_{k=1}^{\infty }\frac{-2^{-z}}{1+2^{-z}}\right)^{-1} \end{align*} Note that $z \in \mathbb{C}$.
We know that $f(z)$ converges for $\Re{z}>0$. How can we prove this using only the theory of continued fractions? Which theorems guarantee this?
Thanks.
NOTE: There is now a similar question at MO.