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 \(x\).
