Does $\sum\limits_{n=1}^{\infty}\frac{1}{P_n\ln(P_n)}$ converge to the golden ratio?

Question

The sum $\displaystyle\sum\limits_{n=2}^{\infty}\frac{1}{n\ln(n)}$ does not converge.

But the sum $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{P_n\ln(P_n)}$ where $P_n$ denotes the $n$th prime number appears to be.

Is that correct, and if so, how can we calculate the value of convergence?

Is it possible that this sum converges to the golden ratio ($\dfrac{1+\sqrt{5}}{2}$)?

Answer 1

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 $\phi$.

Answer 2

1.63661632335... See http://oeis.org/A137245 and links therein (also http://en.wikipedia.org/wiki/Prime_zeta_function, scroll down to integral section)