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:
 * $\displaystyle \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$

Let $\map S {r, j}$ be the sum of all of the terms from $\map S r$ 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:


 * $\map P r = \set {p_{k + 1}, p_{k + 2}, \ldots, p_{\map m r} }$

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


 * $\displaystyle \paren {\sum_{n \mathop = k \mathop + 1}^{\map m r} \frac 1 {p_n} }^j < \paren {\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }^j < \paren {\frac 1 2}^j$

and also by Sum of Infinite Geometric Progression:


 * $\displaystyle \map S r = \sum_{j \mathop = 1}^r \map S {r, j} < \sum_{j \mathop = 1}^r \paren {\frac 1 2}^j < 1$

for every $r$.

Finally notice that:


 * $\displaystyle \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n \mathop = 1}^r \frac 1 n$

which implies that $\map S r$ diverges towards $+\infty$ by Harmonic Series is Divergent, a contradiction.