Sum of Reciprocals of Primes is Divergent/Proof 3

Proof
The following proof is due to James A. Clarkson

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 Q = \prod_{i=1}^{k} {p_i}, \quad S(r) = \sum_{i=1}^{r} {\frac{1}{1+iQ}}$

and let $S(r,j)$ be the sum of all of the terms from $S(r)$ for which $1+iQ$ has exactly $j$ prime factors.

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

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


 * $ \quad \displaystyle P(r) = \{p_{k+1}, p_{k+2},\dots,p_{m(r)}\}$.

Notice again that each term of $S(r,j)$ occurs at least once in the expansion of


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

< {\left[\frac{1}{2}\right]}^j$

and also that


 * $ \quad \displaystyle S(r) = \sum_{j=1}^{r} {S(r,j)} < \sum_{j=1}^{r} {\left[\frac{1}{2}\right]}^j < 1$ for every $r$ by Sum of Infinite Geometric Progression

Finally notice that


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

which implies that $S(r)$ diverges towards $+\infty$ by Sum of Reciprocals is Divergent, a contradiction.