Show that $$\frac{\zeta'(z)}{\zeta(z)}=-\sum_{n=2}^{\infty}\frac{f(z)}{n^z}$$ for $\Re z\gt 1$ Where
$f(z)= \ln p$ if $n=p^m$ for some prime $p$ and some $m\in \Bbb N^+$
Or $f(z)=0$ otherwise.
I think that I need to use product formula $$\frac{1}{\zeta(z)}=\prod_p (1-1/p^z)$$ for $\Re z>1$
Please show me the solution more explicitly. Thank you.
Let $\;P:=$ the set of prime numbers in $\;\Bbb N =\;\{2,3,5,7,...\}\;$ , then
$$\zeta(s):=\prod_{p\in P}\frac1{1-p^{-s}}\implies\log\zeta(s)=-\sum_{p\in P}\log\left(1-\frac1{p^s}\right)\implies$$
$$\left(\log\zeta(s)\right)'=\frac{\zeta'(s)}{\zeta(s)}=-\sum_{p\in P}\frac{(\log p)p^{-s}}{1-p^{-s}}=-\sum_{p\in P}\log p\left(\frac1{1-p^{-s}}-1\right)=$$
$$=\sum\log p\sum_{n=1}^\infty p^{-ns}=\sum_{n=1}^\infty \Lambda (n)n^{-s}$$
The last equality following from here , with $\;\Lambda(n)=$ the Mangoldt Function.