I can show that
$$\log(\zeta (s)) = \sum _{p\in\Bbb P} \frac{1}{p} + R(s)$$where $$R(s) = \sum _{m\geq 2} \sum_{p\in\Bbb P} \frac{1}{m} \frac{1}{p^{ms}}$$
where $\Bbb P$ is the set of all primes, and that the LHS diverges as $s\rightarrow 1^+$. What I'm having trouble with is is showing that $R(s)$ is bounded as $s\rightarrow 1^+$.
I can get to
$$R(s) \leq \sum_{n\geq 2} \frac{1}{n^s (n^s - 1)} $$
but I don't think this helps.
Consider that: $$\lim_{s\to 1^+}R(s)\le \lim_{s\to 1^+} \sum_{n=2}^\infty \frac{1}{n^s(n^s-1)}\le \sum_{n=2}^\infty \frac{1}{n(n-1)}=1 $$ And we know that: $$\lim_{s\to 1^+}\zeta(s)=\infty$$ By using this fact that the function $\ln(n)$ is increasing and $\lim_{n\to \infty} \ln(n)=\infty$, we can get: $$\lim_{s\to 1^+} \ln(\zeta(s))=\infty$$ From here and your equation: $$\ln(\zeta(s))=\sum_{p\in \mathbb P}\frac 1p +R(s)$$ When $s\to 1^+$, we have: $$\lim_{s\to 1^+}\ln(\zeta(s))=\sum_{p\in \mathbb P}\frac 1p +\lim_{s\to 1^+}R(s)$$ But the left side of above equation tend to infinity and in right side, $\lim_{s\to 1^+}R(s)\le 1$ . So, $$\sum_{p\in \mathbb P}\frac 1p $$ is divergence.