Please, consider the following series \begin{equation} f(z)=1+\sum_{n=1}^{\infty}2^{-\sum_{k=1}^{n}\frac{2s}{k}} =1+ \sum_{n=1}^{\infty}\left( \prod_{k=1}^{n}2^{-\frac{2z}{k}} \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^{-\frac{2z}{1}}}{ 1+2^{-\frac{2z}{1}}- \cfrac{2^{-\frac{2z}{2}}}{ 1+2^{-\frac{2z}{2}}- \cfrac{2^{-\frac{2z}{3}}}{ 1+2^{-\frac{2z}{3}} - \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^{-\frac{2z}{k}}}{1+2^{-\frac{2z}{k}}}\right)^{-1} \end{align*}

I'd like to know the region of convergence of this series using only the theory of continued fractions?

P.S. I've asked a similar question at mse.



EDITED ANSWER: First of all it follows directly from the Śleszyński–Pringsheim theorem that this converges for all values of $z\in\mathbb{R}$.

This takes more work, but using continued fraction machinery one can also prove that the continued fraction converges for all $z\in\mathbb{C}$ which satisfy $$1/4<\mathrm{frac}((\log 2)^2\Im z/\pi)<3/4,$$ where $\mathrm{frac}$ denotes the fractional part. Here is the argument:

Let $a_0=1,~$ $a_n=2^{-2z/n}$ and $b_n=1+a_n$ for $n\ge 1.$ Then the continued fraction you are interested in is equal to $$\left(1-\mathrm{K}_1^\infty \frac{1}{b_n'}\right)^{-1}$$ with $$b_n'=\frac{b_na_0a_2\cdots a_{n-1}}{a_1a_3\cdots a_n}$$ if $n$ is odd and $$b_n'=\frac{b_na_1a_3\cdots a_{n-1}}{a_0a_2\cdots a_n}$$ if $n$ is even. Now if $n=2m+1$ is odd then $$\frac{b_na_0a_2\cdots a_{n-1}}{a_1a_3\cdots a_n}=-(1+a_n)2^{-2z(-1+\sum_{k=1}^m(1/2k-1/(2k+1)))}.$$ Therefore as $n\rightarrow \infty$ through odd numbers the argument of $b_n'$ approaches $$\lim_{m\rightarrow\infty}\mathrm{arg} (b_{2m+1}')=\pi+2(\log 2)^2\Im z \mod 2\pi.$$ By our hypothesis above this means that the limit of the argument falls in the interval $(-\pi/2+\epsilon,\pi/2-\epsilon)$ for some $\epsilon>0$. Furthermore the limit along the even integers is the inverse of the limit along the odd integers, so the same comments apply. Finally by essentially the same argument the limit of the modulus exists and is nonzero, therefore $$\sum |b_n'|=\infty$$ and the continued fraction converges by van Vleck's theorem.

Final comments: this is probably not the `complete' solution to the problem, especially since the van Vleck's argument doesn't seem to pick up the nearly trivial cases when $\Im z=0$. Also in retrospect, the rearrangement in the continued fraction at the beginning of the argument is justified by the absolute convergence at the end of the argument... just rearrange the final product to get back to the original. Hope this helps.

  • $\begingroup$ Here $a_n=2^{-\frac{2z}{n}}$ and $b_n=1+2^{-\frac{2z}{n}}$ are complex numbers but the theorem only guarantees convergence for real numbers. $\endgroup$ – A.Neves Dec 22 '11 at 12:04
  • $\begingroup$ @A.Neves The theorem works for complex numbers too... for example see Theorem V.1 of the paper arxiv.org/PS_cache/math/pdf/0010/0010240v2.pdf. Nevertheless, what I said above is not so obvious to me anymore, so I will try to edit it to make it correct. $\endgroup$ – Alan Haynes Dec 22 '11 at 12:30
  • $\begingroup$ @AH, see here: en.wikipedia.org/wiki/… $\endgroup$ – A.Neves Dec 22 '11 at 12:37
  • $\begingroup$ @AH, note that $a_n=2^{-\frac{2z}{n}}$ and $b_n=1+a_n$. $\endgroup$ – A.Neves Dec 22 '11 at 12:39
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    $\begingroup$ @A.Neves Wikipedia is not the greatest reference. Check out some books or mathscinet references to the theorem and you will see that it is stated for all complex numbers. Also, coincidentally, on the wikipedia page about convergence of continued fractions, the theorem is also stated for all complex numbers. $\endgroup$ – Alan Haynes Dec 22 '11 at 12:45

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