Theorem: Sum of Reciprocals of Primes is Divergent

Let $n \in \N: n \ge 1$.
There exists a (strictly) positive real number $C \in \R_{>0}$ such that:
:$(1): \quad \ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - C$
where $\Bbb P$ is the set of all prime numbers.
:$(2): \quad \ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = +\infty$

Proof:
{{AimForCont}} the contrary.
If the prime reciprocal series converges then there must exist some $k \in \N$ such that:
:$\ds \sum_{n \mathop = k + 1}^\infty \frac 1 {p_n} < \frac 1 2$
Fixing $p_k$ from the series above, we create a set of natural numbers called $M_x$.
The set, $M_x$ only contains natural numbers between $1$ and $x$, inclusive, that are divisible '''only''' by the primes in the set: $\set {p_1, p_2, \cdots, p_k}$
Let:
:$M_x = \set {n \in \N: 1 \le n \le x \text { and } n \text { is not divisible by any primes greater than } p_k}$.
Every natural number $n$ in $M_x$ can be written as the product of a square number $\paren {m^2}$ and a square-free number $\paren r$:
:$n = m^2r$
Since $n \le x$, then
:$m \le \sqrt x$
Furthermore, since there are only $k$ primes and in the prime decomposition of the square-free number $\paren r$, each of the k primes has an exponent of either $0$ or $1$, then there are at most $2^k$ distinct non-square prime compositions using primes less than or equal to $p_k$.
With these $2$ observations, we can put an upper bound on the number of elements of $M_x$:
:$\size {M_x} \le {2^k} \sqrt x$
Once again, fixing $p_k$ from the series above, we create a different set of natural numbers called $N_{i, x}$.
The set, $N_{i, x}$ only contains natural numbers between $1$ and $x$, inclusive, that are divisible by at least one prime $p_i$ (for a fixed $i$, where $i > k$), outside of the set: $\set {p_1, p_2, \cdots, p_k}$
Let:
:$N_{i, x} = \set {n \in \N: 1 \le n \le x \text{ and } n \text { is divisible by } p_i, \text { where } i > k }$
Therefore, $x$ less the cardinality of $M_x$ will be less than or equal to the sum of the cardinalities of $\size {N_{i, x} }$ which implies...
{{begin-eqn}}
{{eqn | l = x - \size {M_x}
      | o = \le
      | r = \sum_{i \mathop = k + 1}^\infty \size {N_{i, x} }
      | c = 
}}
{{eqn | o = \le
      | r = \sum_{i \mathop = k + 1}^\infty \floor {\dfrac x {p_i} }
      | c = 
}}
{{eqn | o = \le
      | r = \sum_{i \mathop = k + 1}^\infty \dfrac x {p_i}
      | c = Floor is between Number and One Less
}}
{{eqn | o = \le
      | r = x \sum_{i \mathop = k + 1}^\infty \dfrac 1 {p_i}
      | c = 
}}
{{eqn | o = \lt
      | r = \frac x 2
      | c = from our initial assumption
}}
{{eqn | ll = \leadsto
      | l = \size {M_x}
      | o = \gt
      | r = \frac x 2
      | c = 
}}
{{end-eqn}}
We now have a lower bound on the number of elements of $M_x$:
:$\size {M_x} \gt \dfrac x 2$
Finally notice that whenever:
:$x \ge 2^{2 k + 2}$
then it cannot be the case that both:
:$\dfrac x 2 < \size {M_x} \le {2^k} \sqrt x$
Since that implies:
:$2^{2 k + 1} < 2^{2 k + 1}$
This contradicts our assumption that the prime reciprocal series converges.
Therefore, from Proof by Contradiction, the prime reciprocal series diverges.
{{qed}}