The Riemann's 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*} and 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 teh Riemann's 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*} wich in Gauss' notation is (1)

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?

  • $\begingroup$ Better in what way? $\endgroup$ – Igor Rivin Dec 22 '11 at 20:05
  • $\begingroup$ This question could be improved: mathoverflow.net/howtoask#specific $\endgroup$ – Stopple Dec 22 '11 at 20:19
  • 3
    $\begingroup$ 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. $\endgroup$ – Noam D. Elkies Dec 23 '11 at 0:30
  • 3
    $\begingroup$ @Noam: it's somewhat standard in CF literature. I'm told it's originally Gauss's. $\endgroup$ – J. M. isn't a mathematician Dec 23 '11 at 1:17
  • $\begingroup$ I don't really understand Gauss's notation. Why is it not $\zeta(z)=\left(1+\bigk\cdots\right)^{-1}$? $\endgroup$ – 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.



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

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

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.