Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.