The Riemann 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*}
thus 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 the Riemann 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*}
which in Gauss' notation is (1)
Now consider
\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 the Śleszyński–Pringsheim theorem we can see that $f(z)$ converges for $\Im{z}=0$ and $\Re{z}\geq 0$. This is to say that $1/\zeta(z)$ converges for real $z\geq 0$.
My question: can a bigger region of convergence be found using the theory of continued fractions?
Best Answer
(Too long for a comment.)
There's a (somewhat) simpler (Eulerian) continued fraction:
$$\sum_{k=1}^\infty \frac1{k^s}=1+\sum_{k=2}^{\infty} \prod_{j=2}^k \left(1-\frac1{j}\right)^s=\cfrac1{1-\cfrac{\left(1-\frac12\right)^s}{1+\left(1-\frac12\right)^s-\cfrac{\left(1-\frac13\right)^s}{1+\left(1-\frac13\right)^s-\cfrac{\left(1-\frac14\right)^s}{1+\left(1-\frac14\right)^s-\cdots}}}}\;\;\;\;\;\;$$
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...