Not Counting Primes Afterall

 No, sum of the reciprocals of primes is not the count of primes,

$\pi(x)\ne log(x)$

but,

$\sum^n_{i=1}\cfrac{1}{p_n}\rightarrow log(n)$

more correctly, piecewise,

$\sum^n_{i}\cfrac{1}{p_n}\rightarrow log(x)|^{n}_{i}$

for $n$ and $i$ large, positive.  Which adds no information because, the sum of $\cfrac{1}{x}$ is smaller than the integration of $\cfrac{1}{x}$, ie $log(x)$.  And rightfully,

$p_n=n$

which means a log(x) factor is the result of possible intersections of $\cfrac{1}{x}$ with $\cfrac{x}{p_n+2}$.

Shifting $\cfrac{x}{p_n+2}$ against $\cfrac{1}{x}$ to find an integer intercept is,

$\cfrac{x_1}{p_n+2}=y=\cfrac{1}{x_2}$

$x_1x_2=p_n+2$

looking for a product of $x_1$ and $x_2$, that rules out $p_n+2$ as prime.

There is still no new information that makes prime $p_n$ different from any number $x$.