Riemann's zeta as a continued fraction over prime numbers. 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.
Thanks.

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.
EDIT
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.
EDIT
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.
