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


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?

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Better in what way? – Igor Rivin Dec 22 '11 at 20:05
This question could be improved: mathoverflow.net/howtoask#specific – Stopple Dec 22 '11 at 20:19
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. – Noam D. Elkies Dec 23 '11 at 0:30
@Noam: it's somewhat standard in CF literature. I'm told it's originally Gauss's. – J. M. Dec 23 '11 at 1:17
I don't really understand Gauss's notation. Why is it not $\zeta(z)=\left(1+\bigk\cdots\right)^{-1}$? – Sylvain JULIEN Nov 11 '13 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...

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this $e^{-2arcoth((2k+1)z)}$ is not what I mean the correct expression is $e^{-2(arcoth(2k+1))z}$ – A.Neves Dec 23 '11 at 10:15
OK, I see${}{}$. – A.Neves Dec 23 '11 at 12:31