Sum of Reciprocals of Primes is Divergent

Theorem
For $n \ge 1$:


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


 * $\displaystyle \lim_{n \to \infty} \left({\ln \left({\ln n}\right) - \frac 1 2 }\right) = + \infty$

Proof of Lower Bound
Assume all sums and product over $p$ are over the set of prime numbers.

Let $n \ge 1$.

Then:

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


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

Proceed as follows:

Now, obviously, any $N$ with $N > \exp \left({\exp \left({c + \dfrac 1 2}\right)}\right)$ satisfies condition $(2)$ by Logarithm is Strictly Increasing and Strictly Concave.