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?

  • 4
    $\begingroup$ 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
  • 6
    $\begingroup$ @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


(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.



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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.