Does $\sum\limits_{n=1}^{\infty}\frac{1}{P_n\ln(P_n)}$ converge to the golden ratio?
With $P_n \approx n \ln(n)$, we should have $$\sum_{N}^\infty \dfrac{1}{P_n \ln(P_n)} \approx \int_N^\infty \dfrac{dx}{x \ln(x)^2} = \dfrac{1}{\ln N}$$ If the sum for $n$ up to $\pi(19999999) = 1270607$ is $1.57713$, we'd expect the remainder to be about $.071$, which would push the total to about $1.648$, too high for $\phi$.
1.63661632335... See http://oeis.org/A137245 and links therein (also http://en.wikipedia.org/wiki/Prime_zeta_function, scroll down to integral section)