closed form for a series over the Riemann zeta zeros

You may start with the Hadamard product : $$ \zeta(s)=\frac{e^{s\left(\log(2\pi)-1-\frac{\gamma}2\right)}}{2(s-1)\Gamma\left(1+\frac s2\right)}\prod_{\rho} \left(1-\frac s{\rho}\right) e^{s/{\rho}} $$ or use the Riemann Xi function $\xi(s)=\frac 12 s(s-1)\pi^{-s/2}\Gamma\left(\frac s2\right)\zeta(s)$ to rewrite this as a $$\xi(s)=\xi(0)\prod_{\rho} 1-\frac s{\rho}$$

so that $\displaystyle \dfrac{\xi'(s)}{\xi(s)} = \sum^\prime_{\rho} \left(\ln\left(1-\frac s{\rho}\right)\right)' = \sum^\prime_{\rho} \frac 1{s-\rho}$
at least if I did'nt make a stupid mistake... (the ' over the sum meaning your summation convention)

You also may find the following derivation interesting: $$\xi(s)=\xi(0)e^{B s}\prod_{\rho} \left(1-\frac s{\rho}\right) e^{s/{\rho}}$$ with $B=\log 2+\frac 12 \log\pi-1-\frac{\gamma}2$ and $\rho$ the non-trivial zeros.

so that $\displaystyle \dfrac{\xi'(s)}{\xi(s)} = B+\sum_{\rho} \left(\ln\left(1-\frac s{\rho}\right)e^{s/{\rho}}\right)' = B+ \sum^\prime_{\rho} \frac 1{s-\rho} + \frac 1{\rho}$

There is no contradiction thanks to $\displaystyle B=-\sum^\prime_{\rho} \frac 1{\rho}=-\sum_{\Im(\rho)>0} \frac 1{\rho(1-\rho)}$

For more see Edwards' book,
Coffey's 'Toward verification of the Riemann hypothesis: Application of the Li criterion'
the MO article 'Is this sum of reciprocals of zeta zeros correct?'
and Lagarias' paper 'The Riemann Hypothesis: Arithmetic and Geometry'