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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?

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    $\begingroup$ 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. $\endgroup$ Dec 23, 2011 at 0:30
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    $\begingroup$ @Noam: it's somewhat standard in CF literature. I'm told it's originally Gauss's. $\endgroup$ Dec 23, 2011 at 1:17
  • $\begingroup$ I don't really understand Gauss's notation. Why is it not $\zeta(z)=\left(1+\bigk\cdots\right)^{-1}$? $\endgroup$ Nov 11, 2013 at 11:13

1 Answer 1

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(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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