Sum of Reciprocals of Primes is Divergent/Proof 3

Proof
the contrary.

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


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

Let:
 * $\displaystyle Q = \prod_{i \mathop = 1}^k {p_i}$

and:
 * $\displaystyle S \left({r}\right) = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$

Let $S \left({r, j}\right)$ be the sum of all of the terms from $S \left({r}\right)$ for which $1 + i Q$ has exactly $j$ prime factors.

Notice that $1 + i Q$ is coprime with every prime factor in $Q$.

Thus every prime factor of $1 + i Q$ where $i = 1, \ldots ,r$ falls into some finite sequence of consecutive primes:


 * $P \left({r}\right) = \left\{ {p_{k + 1}, p_{k + 2}, \ldots, p_{m \left({r}\right)} }\right\}$

Notice again that each term of $S \left({r, j}\right)$ occurs at least once in the expansion of:


 * $\displaystyle \left({\sum_{n \mathop = k \mathop + 1}^{m \left({r}\right)} \frac 1 {p_n} }\right)^j < \left({\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }\right)^j < \left({\frac 1 2}\right)^j$

and also by Sum of Infinite Geometric Progression:


 * $\displaystyle S \left({r}\right) = \sum_{j \mathop = 1}^r S \left({r, j}\right) < \sum_{j \mathop = 1}^r \left({\frac 1 2}\right)^j < 1$

for every $r$.

Finally notice that:


 * $\displaystyle S \left({r}\right) = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n\mathop = 1}^r \frac 1 n$

which implies that $S \left({r}\right)$ diverges towards $+\infty$ by Harmonic Series is Divergent, a contradiction.