Question

The Riemann's zeta function can be expressed as a continued fraction as follows \begin{align*} \zeta(z)=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}\left(1-\bigk_{k=1}^{\infty }\frac{-e^{-2\cdot (coth^{-1}(2k+1))\cdot z}}{1+e^{-2\cdot (coth^{-1}(2k+1))\cdot z}}\right)^{-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1) \end{align*} and its reciprocal as \begin{align*} \frac{1}{{\zeta(z)}}=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}1-\bigk_{k=1}^{\infty }\frac{-e^{-2\cdot (coth^{-1}(2k+1)) \cdot z}}{1+e^{-2\cdot(coth^{-1}(2k+1))\cdot z}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2) \end{align*} Proof: Note that \begin{align*} ln(n)=2 \sum_{m=1}^{n-1} \sum_{k=0}^{\infty}\frac{1}{2k+1}\left( \frac{1}{2\cdot m+1} \right)^{2k+1} \end{align*} and that \begin{align*} coth^{-1}(2 n +1)=\sum_{k=0}^{\infty}\frac{1}{2k+1}\left( \frac{1}{2\cdot n+1} \right)^{2k+1} \end{align*} so \begin{align*} ln(n)=2 \sum_{m=1}^{n-1} coth^{-1}(2 m +1) \end{align*} so teh Riemann's zeta function is expressed as \begin{align*} \zeta(z)=1+\sum_{n=1}^{\infty}\prod_{k=1}^{n-1}e^{-2\cdot(coth^{-1}(2k+1))\cdot z} \end{align*} and using Euler's continued fraction formula the result follows. \begin{equation*} \zeta(z)= \cfrac{1}{ 1- \cfrac{e^{-2(coth^{-1}(3))z}}{ 1+e^{-2(coth^{-1}(3))z}- \cfrac{e^{-2(coth^{-1}(5))z}}{ 1+e^{-2(coth^{-1}(5))z}- \cfrac{e^{-2(coth^{-1}(7))z}}{ 1+e^{-2(coth^{-1}(7))z} - \ddots}}}} \end{equation*} wich in Gauss' notation is (1)

Now considere \begin{align*} f(z)=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}\bigk_{k=1}^{\infty }\frac{-e^{-2\cdot (coth^{-1}(2k+1))\cdot z}}{1+e^{-2\cdot(coth^{-1}(2k+1))\cdot z}} \end{align*}

Using Śleszyński–Pringsheim theorem we can see that $f(z)$ converges for $\Im{z}=0$ and $\Re{z}\geq 0$. This is saying that $1/\zeta(z)$ converges for $x\geq 0$.

My question: can a bigger region of convergence be found using the theory of continued fractions?

Answer 1

(Too long for a comment.)

There's a (somewhat) simpler (Eulerian) continued fraction:

$$\sum_{k=1}^\infty \frac1{k^s}=\sum_{k=1}^{\infty} \prod_{j=2}^k \left(1-\frac1{j}\right)^s=\cfrac1{1-\cfrac{\left(1-\frac12\right)^s}{1+\left(1-\frac12\right)^s-\cfrac{\left(1-\frac13\right)^s}{1+\left(1-\frac13\right)^s-\cfrac{\left(1-\frac14\right)^s}{1+\left(1-\frac14\right)^s-\cdots}}}}$$

but as you can see from comparing successive convergents of this continued fraction and the successive partial sums of the Dirichlet series, it's not terribly useful.

Also,

$$e^{-2z\,\mathrm{arcoth}(2k+1)}=\left(\frac{k}{k+1}\right)^z$$

so your CF could certainly be simplified a fair bit...