The Riemann's Zeta Function represented as a continued fraction and a question of convergence


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?

• Is this $\mathop{\large{\bf K}}_{k=1}^\infty$ notation standard? The only other place I've seen it is in other recent questions from A.Neves. Dec 23, 2011 at 0:30
• @Noam: it's somewhat standard in CF literature. I'm told it's originally Gauss's. Dec 23, 2011 at 1:17
• I don't really understand Gauss's notation. Why is it not $\zeta(z)=\left(1+\bigk\cdots\right)^{-1}$? Nov 11, 2013 at 11:13

(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...