Sum of Reciprocals of Primes is Divergent/Proof 2

Theorem

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


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

Proof of Limit
Observe the following simplification:

Fix $c \in \R$. It suffices to show there exists $N \in \N$, such that:


 * $(1):\quad n \ge N \implies \ln \left({\dfrac {6 \ln n} {\pi^2} }\right) > c$

Proceed as follows:

Let $N \in \N$ such that $N > \exp \left({\dfrac {\pi^2 \exp c} 6}\right)$.

By Logarithm is Strictly Increasing and Strictly Concave it follows that $N$ satisfies condition $(1)$.

Hence the result.