I am trying to find or get a numerical approximation of $$ \sum_{\rho \text{ non-trivial zeros of } \zeta} \frac{1}{\rho} $$

In The Riemann Hypothesis: Arithmetic and Geometry Lagarias gives the identity:

$$\hat{\zeta}(s) := \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2})\zeta(s)$$

$$ \frac{\hat{\zeta}^\prime(s)} {\hat{\zeta}(s)} = \frac{d}{ds} [ \log \hat{\zeta}(s) ] = -\frac{1}{s} - \frac{1}{s-1} + {\sum_{\rho \text{ zeros of } \zeta }}^\prime \frac{1}{s-\rho} \qquad(1)$$

where the prime indicates the zeros must be summed in pairs $\rho,1-\rho$

Q1 Does the last sentence mean that the sum is over the non-trivial zeros?

Maple gives:
$$\lim_{s \to 0} {\sum_{\rho \text{ zeros of } \zeta }}^\prime \frac{1}{s-\rho} = -\gamma + \frac{1}{2} \log\left(\pi\right) + \log\left(2\right) - 1$$
~~
If the above result is correct, is it true that:~~

~~$$ \sum_{\rho \text{ non-trivial zeros of } \zeta} \frac{1}{\rho} = \gamma - \frac{1}{2} \log\left(\pi\right) - \log\left(2\right) + 1 $$
~~
**EDIT** As Micah Milinovich kindly answerd the above is wrong.

Trying to save the quiestion, is it true that: $$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)} = \gamma - \frac{1}{2} \log\left(\pi\right) - \log\left(2\right) + 1 $$

Assuming RH $1-\rho = \bar{\rho}$ and the LHS is $\sum_{\rho} \frac{1}{|\rho|^2}$

According to RH Equivalence 5.3. $$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2} = 2 + \gamma - \log 4\pi$$.

And the constants still don't match.