Sum of Reciprocals of Primes is Divergent/Proof 4

Proof
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:


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