Is this a relation between the Riemann zeta function and the Prime zeta function?

Question

I am partly repeating myself here again. Is this a correct relation between the Riemann zeta function and the Prime zeta function?

$$ \zeta (s) = \sum\limits_{n=1}^{\infty}\frac{1}{n^{s}}$$

$\displaystyle \log(\zeta (s)) = \sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$ where $\displaystyle a_{1} = 0\;$, $\displaystyle a_{n} = \frac{\Lambda(n)}{\log(n)}$ and $\Lambda(n)$ is the Mangoldt function defined by $\displaystyle \Lambda(n) = \log(p)$ if $n = p^{k}$ for $p$ a prime.

$\displaystyle ω_{1}(\log(\zeta (s))) = \sum\limits_{n=1}^{\infty}\frac{\log(a_{n})}{n^{s}},$ where $ω_{1}(\;)$ is an operation and $\displaystyle \log(a_n) = 0$ if $a_n = 0$ $$ω_{1}(\log(\zeta (s)))\zeta (s) = \sum\limits_{n=1}^{\infty}\frac{b_{n}}{n^{s}}$$

$\displaystyle ω_{2}(ω_{1}(\log(\zeta (s)))\zeta (s)) = \sum\limits_{n=1}^{\infty}\frac{e^{b_{n}}}{n^{s}}\;$ where $ω_{2}(\;)$ is another operation and $\displaystyle e^{b_{n}}=\exp(b_{n})$

$$\sum\limits_{p\;\text{prime}} \frac1{p^s} = \log(ω_2(ω_1(\log(\zeta (s)))\zeta (s)))$$

Answer 1

In the case $s=2$ we have $$\sum_{p\in\mathcal{P}}\frac{1}{p^2}\color{red}{\leq} \frac{1}{2}\sum_{p\in\mathcal{P}}\log\,\left(\frac{1+\frac{1}{p^2}}{1-\frac{1}{p^2}}\right)=\frac{1}{2}\log\frac{\zeta(2)^2}{\zeta(4)}=\log\sqrt{\frac{5}{2}}=0.458145365937\ldots $$ and that is quite a good approximation since $\frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots $ in a neighbourhood of the origin. In general, by Moebius inversion formula

$$ \sum_{p\in\mathcal{P}}\frac{1}{p^s}=\sum_{n\geq 1}\frac{\mu(n)}{n}\log\zeta(ns)\tag{1} $$ that is a series with a decent convergence speed, due to $\log\zeta(ns)\approx 2^{-ns}$ for any large $n$.

Answer 2

A variant of the relationship can be put like this in Mathematica:

s = 7;
nn = 1000;
N[Log[Sum[
   1/(Exp[FullSimplify[
        Sum[If[n/k == 1, Log[1], 
          If[Mod[n, k] == 0, 
           Log[Denominator[
             FullSimplify[MangoldtLambda[n/k]/Log[n/k]]]], 0]], {k, 1,
           n}]]]*n^s), {n, 1, nn}]], 12]
N[PrimeZetaP[s], 12]

which outputs two equal numbers:

0.00828383285613

0.00828383285613

The fake or the tweek is the use of Denominator[ which together with MangoldtLambda[n]/Log[n] should be possible to replace by something in terms of the PrimeZeta function and Riemann zeta function, which is a bit circular.

The same relationship as a product over the divisors instead of a sum:

s = 7;
nn = 1000;
N[Log[Sum[
   1/(FullSimplify[
       Product[If[n/k == 1, 1, 
         If[Mod[n, k] == 0, 
          Denominator[FullSimplify[MangoldtLambda[n/k]/Log[n/k]]], 
          1]], {k, 1, n}]]*n^s), {n, 1, nn}]], 12]
N[PrimeZetaP[s], 12]

0.00828383285613

0.00828383285613

The Dirichlet generating function is:

(*start*)
s = 7;
nn = 1000;
N[1 + Sum[
   Denominator[FullSimplify[MangoldtLambda[n]/Log[n]]]/n^s, {n, 2, 
    nn}], 12]
N[Zeta[s] + Sum[n*PrimeZetaP[(n + 1)*s], {n, 1, 20}], 12]
(*end*)

$$1+\sum _{n=2}^{\infty } \frac{\text{Denominator}\left[\text{FullSimplify}\left[\frac{\Lambda (n)}{\log (n)}\right]\right]}{n^s}=\zeta (s)+\sum _{n=1}^{\infty } n P((n+1) s)$$