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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 $\displaystyle e^{b_{n}}=\exp(b_{n})$

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

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    @Eric: I should have written unknown operations $ω_{1}$ and $ω_{2}$.2011-05-29

2 Answers 2

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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 $n$.