Sum of Reciprocals of Primes is Divergent/Proof 4

Proof
The following proof is due to Erdos

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


 * $ \quad \displaystyle \sum_{n=k+1}^{\infty} {\frac{1}{p_n} < \frac{1}{2}}$

Let


 * $ \quad \displaystyle M_x = \{n \in \N$ | $1 \leq n \leq x$ and $n$ is not divisible by any primes greater than $p_k\}$.

Notice that


 * $ \quad \displaystyle |M_x| \leq {2^k}\sqrt{x}$

because if


 * $ \quad \displaystyle n = m^2r$

then


 * $ \quad \displaystyle m \leq \sqrt{x}$

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


 * $ \quad \displaystyle N_{i,x} = \{n \in \N$ | $1 \leq n \leq x$ and $n$ is divisible by $p_i\}$.

Notice


 * $ \quad \displaystyle |\{1, \dots, x\}$\$M_x| \leq \sum_{i=k+1}^{\infty} {|N_{i,x}|} < \sum_{i=k+1}^{\infty} {\frac{x}{p_i}} \implies \frac{x}{2} < |M_x|$.

Finally notice that whenever


 * $ \quad \displaystyle x \geq 2^{2k+2}$

then it cannot be the case that both


 * $ \quad \displaystyle \frac{x}{2} < |M_x| \leq {2^k}\sqrt{x}$.