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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.

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  • $\begingroup$ $\sum\limits_{k=1}^{\infty}\frac{1}{2^{z}}$ trivially diverges. $\endgroup$ – J. M. is a poor mathematician Dec 20 '11 at 10:44
  • $\begingroup$ @J.M.: It appears from the $\sum\prod$ form that it should be $\sum\limits_{k=1}^\infty\frac1{2^{kz}}$. $\endgroup$ – Brian M. Scott Dec 20 '11 at 10:46
  • $\begingroup$ @J.M. forgot the $k$, now corrected. $\endgroup$ – Neves Dec 20 '11 at 10:48
  • $\begingroup$ In that case, the convergence region ought to be $\Re z > 0$, no? $\endgroup$ – J. M. is a poor mathematician Dec 20 '11 at 10:50
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    $\begingroup$ Śleszyński–Pringsheim theorem? $\endgroup$ – Noud Dec 20 '11 at 11:01

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