TITLE: Is the sum of the reciprocals of imaginary parts of the zeros of the zeta function divergent?
QUESTION [5 upvotes]: Is the sum $\sum_{\rho} \frac {1}{\rho}$ divergent? Or does it converge to any special value? Could you provide a proof to this? Here, $\rho$ is the imaginary part of the zeros of the Riemann zeta function.
Edit 1:
I also want to know if there is a way to approximate how many $\rho$'s there are less or equal to x (under the condition that the RH is true).

REPLY [2 votes]: The Riemann-von Mangoldt function gives that the number $N(T)$ of zeros $z$ of the zeta function with $0 < \operatorname{Im} z < T$ is asymptotically distributed like
$$N(T) = \frac{T}{2 \pi} \log \frac{T}{2 \pi} - \frac{T}{2 \pi} + \frac{7}{8} + \frac{1}{\pi} \arg \zeta\left(\frac{1}{2} + i T\right) + O\left(\frac{1}{T}\right),$$
and the penultimate term is $O(\log T)$.
As $T \to \infty$, the term $\frac{T}{2 \pi} \log \frac{T}{2 \pi}$ dominates, which implies the sum of reciprocals of the imaginary parts of the zeros with $0 < \operatorname{Im} z < T$ is $$\geq \frac{1}{T} \left(\frac{T}{2 \pi} \log \frac{T}{2 \pi} + O(T)\right) = \frac{1}{2 \pi} \log T + O(1) ,$$ and so diverges. With a little more work one could probably find a much better asymptotic expression for that sum of reciprocals.