Riemann's zeta function is a function with many faces, I mean representations. I recently derived this one, bellow, as a continued fraction over prime numbers.

$$ \zeta(s)=1 +\cfrac{\frac{1}{2^{s}}}{1-\frac{1}{2^{s}} -\cfrac{\frac{2^{s}-1}{3^{s}}}{1+\frac{2^{s}-1}{3^{s}} -\cfrac{\frac{3^{s}-1}{5^{s}}}{1+\frac{3^{s}-1}{5^{s}} -\cfrac{\frac{5^{s}-1}{7^{s}}}{1+\frac{5^{s}-1}{7^{s}} -\cfrac{\frac{7^{s}-1}{11^{s}}}{1+\frac{7^{s}-1}{11^{s}} -\ddots}}}}} $$

... and I'd like to know if this is known in the literature and if so I'd appreciate to have references about it.


  • $\begingroup$ "Found" in the sense of derived it yourself, or found it in a written source somewhere? If the latter, you might include that source here. $\endgroup$ – Jack M Apr 8 '14 at 11:50
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    $\begingroup$ @Jack M, I derived it myself. $\endgroup$ – Neves Apr 8 '14 at 11:54
  • $\begingroup$ Please show your derivation. Thank you. $\endgroup$ – marty cohen Apr 11 '14 at 2:46
  • $\begingroup$ @marty, ok, I'll do that. $\endgroup$ – Neves Apr 11 '14 at 6:23
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    $\begingroup$ @FredKline, Thanks, but I already knew this one... and none of the others is "over primes". $\endgroup$ – Neves Apr 12 '14 at 6:14

The continued fraction representation above had its origins on another problem I was working on sometime ago.

It's based on a very simple way of looking at the Euler's product representation of $\frac{1}{\zeta(s)}$. Interestingly it applies to every infinite product.

And this is as follows

$$ \frac{1}{\zeta(s)}=\left(1-\frac{1}{2^s}\right)-\left(1-\frac{1}{2^s}\right)\frac{1}{3^s}-\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\cdots $$

From here its easy to derive the above continued fraction using Euler's continued fraction formula.

And thats it, It's nice and eventually a new thing.


Just to make it clear, note that $$ \begin{align*} \frac{1}{\zeta(s)}&=\left(1-\frac{1}{2^s}\right)\left[\left(1-\frac{1}{3^s}\right)-\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\ &=\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left[\left(1-\frac{1}{5^s}\right)-\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\ &\vdots\\ &=\prod_{p\in\mathbb{P}}\left(1-\frac{1}{p^{s}}\right) \end{align*} $$ where $\mathbb{P}$ is the set of the prime numbers.


To derive the continued faction just put $\frac{1}{\zeta(s)}$ in the form $$ \frac{1}{\zeta(s)}=1-\frac{1}{2^s}\left(1+\frac{2^s-1}{3^s}\left(1+\frac{3^s-1}{5^s}\left(1+\frac{5^s-1}{7^s}\left(1+\frac{7^s-1}{11^s}\left(1+\ddots\right ) \right ) \right ) \right ) \right) $$ and then just apply the Euler continued fraction formula. So we can write this as $$ \frac{1}{\zeta(s)}=1-\frac{1}{2^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}\frac{5^s-1}{7^s}-\cdots $$ Now, let $a_1=-\frac{1}{2^s};a_2=\frac{2^s-1}{3^s};a_3=\frac{3^s-1}{5^s};a_4=\frac{5^s-1}{7^s}\cdots$ and we'll get the Euler continued fraction formula.

  • $\begingroup$ Found this: arxiv.org/abs/1003.4015 $\endgroup$ – Fred Kline Apr 15 '14 at 3:50
  • $\begingroup$ Can you expand on the last step.? I didn't get it. $\endgroup$ – user230452 Mar 29 '16 at 23:01
  • $\begingroup$ @user230452, I added some clarification. $\endgroup$ – Neves Mar 30 '16 at 21:08

By using Mathematica to simplify the first 7 primes, we get: $$\frac{510510^s}{\left(2^s-1\right) \left(3^s-1\right) \left(5^s-1\right) \left(7^s-1\right) \left(11^s-1\right) \left(13^s-1\right) \left(17^s-1\right)},$$ which is equivalent to: $$\prod _{p\text{ prime}} \frac{p^s}{p^s-1} = \zeta(s).$$ Product does not converge when $s=1.$

Scroll down to Euler product formula (2nd paragraph).
When $s=1\text{, }\frac{1}{1-\frac{1}{p^s}}$ simplifies to $\frac{p}{p-1}.$ When $s>1,$ there is no simplification.

Neves's formula puts the exponents back onto the primes when it is simplified, $\frac{p^s}{p^s-1}.$

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    $\begingroup$ Thats $\zeta(s)$ in the Euler's product form... $\endgroup$ – Neves Apr 11 '14 at 6:18
  • $\begingroup$ @Neves, we crossed paths. My last edit explains. $\endgroup$ – Fred Kline Apr 11 '14 at 7:03

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