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

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 $n$.