How to show that $\phi(s)=\sum_p \frac{\log p}{p^s}$ converges absolutely and locally uniformly for $R(s)>1$?

Question

I know that $\zeta(s)=\sum_n \frac{1}{n^s}$ converges absolutely and locally uniformly for $R(s)>1$.

This implies that $\zeta'(s)=\sum_n \frac{\log n}{n^s}$ converges locally uniformly for $R(s)>1$.

However, how do we use that to show that $\phi(s)=\sum_p \frac{\log p}{p^s}$ converges locally uniformly.

i.e. I believe we need to use some manner of comparison test but that would require absolute convergence of $\zeta'(s)=\sum_n \frac{\log n}{n^s}$. If so how does one prove that?

Answer 1

Use $\log n\in o(n^{(s-1)/2})$ in a comparison test against $n^{-(s+1)/2}$.

Answer 2

There are many ways to go. One way is to prove that both the Riemann $\zeta$-function and the prime zeta-function are the analytic continuation of holomorphic series over $\text{Re}(s)>1$:

$$\zeta(s)=\sum_{n\geq 1}\frac{1}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}=\int_{0}^{+\infty}\frac{x^{s-1}}{\Gamma(s)}\cdot\frac{dx}{e^x-1} $$ $$P(s)=\sum_{p}\frac{1}{p^s}\stackrel{\text{Moebius}}{=}\sum_{n\geq 1}\frac{\mu(n)}{n}\log\zeta(ns) $$

$$ \sum_{p}\frac{\log p}{p^s}=-\frac{d}{ds}P(s) = \sum_{n\geq 1}\mu(n)\frac{\zeta'(ns)}{\zeta(ns)}=-\sum_{n\geq 1}\sum_{m\geq 2}\frac{\mu(n)\Lambda(m)}{m^{ns}}. $$