Is the sum of the reciprocals of imaginary parts of the zeros of the zeta function divergent? 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).
 A: By the Riemann-von Mangoldt formula, we know that 
$$
\#\{\rho: 0< \rho \le T\} \sim c\, T\log T,
$$
as $T \to \infty$, for some $c>0$.
It follows by summation formula that
\begin{align}
\sum_{0<\rho\le T}\frac{1}{\rho}&\asymp \frac{1}{T}\cdot T\log T-\sum_{t\le T}t\log t\left(\frac{1}{t+1}-\frac{1}{t}\right)\\
&\asymp \log T+\sum_{t\le T}\frac{\log t}{t}\\
&\asymp \log T+\log^2 T \to \infty
\end{align}
$\text{ }$
$\text{ }$
(Or, if you remember that $\sum_{n\le x}1/n\sim \log x$, then it is sufficient to note that, here, you have "more elements".)
A: 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.
