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.