Asymptotic estimate of sum of reciprocals of $\log(p_k)$ Fix $n$. The sum $S(n)=\sum_{p_k\leq n} \frac{1}{\log p_k}$ taken over primes less than or equal to n, clearly diverges as $n$ goes to infinity, being lower bounded by the sum of the reciprocals of primes which itself diverges. Are good upper lower bounds to $S(n)$ known as $n$ increases? 
Experimentally it looks to be not too far from $n/\log(n)$.

My notes indicate this was from a note by E. Hertz at the Drexel Math forum. Any mistakes would be my own.
As usual $f(n)\sim g(n)\implies \lim_{n \to \infty }\frac{f(n)}{g(n)}= 1. $ Let
$$\sum_2^n\frac{1}{\log k}\sim\int_2^n\frac{dx}{\log x} = f(n)$$
Let $$g(n) = \frac{n}{\log n}.$$
Then f ' (n) $ = \frac{1}{\log n}$ and g' (n)$= \frac{\log n - 1 }{(\log n)^2}\sim \frac{1} {\log n}$ and by l'Hopital's rule $f \sim g.$
Let $p(n)$ be the n$^{th}$ prime.
By the PNT $p(n)=e(n)\cdot n\log n$ where $e(n)$ goes to unity and
$$\log p(n)= \log e(n)+ \log n + \log(\log(n)).$$
$$\sum_2^{n}\frac{1}{\log p(k)}\sim \int_2^n \frac{dx}{\log(e(x))+\log x+\log(\log x))} = I(n).$$
Since the integral is evaluated at $n \sim \frac{1}{\log n}$ and the integrand is the derivative of the integral $f(n) $ we have that $I(n)\sim f(n)\sim g(n).$
This is probably a pretty old result but at least it confirms your suspicion. I have a hunch this question was already answered in this forum but I couldn't find it. 
