Sum of Reciprocals of Primes is Divergent/Proof 4

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Theorem

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

Aiming for a contradiction, suppose the contrary.

If the prime reciprocal series converges then there must exist some $k \in \N$ such that:

$\ds \sum_{n \mathop = k + 1}^\infty \frac 1 {p_n} < \frac 1 2$

Let:

$M_x = \set {n \in \N: 1 \le n \le x \text { and } n \text { is not divisible by any primes greater than } p_k}$.

Notice that:

$\size {M_x} \le {2^k} \sqrt x$

because if:

$n = m^2r$

then

$m \le \sqrt x$

and there are at most $2^k$ distinct non-square prime compositions using primes less than $p_k$.

Now let:

$N_{i, x} = \set {n \in \N: 1 \le n \le x \text{ and } n \text { is divisible by } p_i}$

Notice:

$\ds \size {\set {1, \dots, x} \setminus M_x} \le \sum_{i \mathop = k + 1}^\infty \size {N_{i, x} } < \sum_{i \mathop = k + 1}^\infty \dfrac x {p_i} \implies \dfrac x 2 < \size {M_x}$

Finally notice that whenever:

$x \ge 2^{2 k + 2}$

then it cannot be the case that both:

$\dfrac x 2 < \size {M_x} \le {2^k} \sqrt x$


Historical Note

This proof is credited to Paul Erdős.