Sum of Reciprocals of Primes is Divergent/Proof 2

Theorem
Let $n \in \N: n \ge 1$.

Then:
 * $(1): \quad \displaystyle \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac {\pi^2} 6}\right)$

where $\Bbb P$ is the set of all prime numbers.


 * $(2): \quad \displaystyle \lim_{n \mathop \to \infty} \left({\ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac {\pi^2} 6}\right)}\right) = + \infty$

Proof
By Sum of Reciprocals of Primes is Divergent: Lemma:
 * $\displaystyle \lim_{n \mathop \to \infty} \left({\ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac {\pi^2} 6}\right)}\right) = + \infty$

It remains to be proved that:


 * $\displaystyle \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac {\pi^2} 6}\right)$