Is this sum of reciprocals of zeta zeros correct? 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.

 A: The series $\sum_\rho \rho^{-1}$ over the non-trivial zeros  is not absolutely convergent, this is proved in Davenport p. 80.  But as Davenport says and proves in 
page 81-82  the series converges conditionally provided one groups together
the terms from $\rho$ and its conjugate $\overline{\rho}$. And the value of 
the sum can be given, independently of RH, as the constant $-B$ where
$B = -\frac12 \gamma-1+\frac12\log4\pi$.  (This value was known to Riemann, as Siegel 
says in his paper about the Riemann Nachlass).
A: To answer your modified question, according to Mathematica:
$$ \lim_{s\to 0} \left(\frac{\hat{\zeta}'}{\hat{\zeta}}(s)+\frac{1}{s}\right) = -\frac{\gamma}{2} + \tfrac{1}{2}\log(4\pi).$$
This implies that
$${\sum_\rho}'\frac{1}{\rho} = \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 1 +\frac{\gamma}{2} - \tfrac{1}{2}\log(4\pi). $$
Therefore
$$\sum_{ \rho } \frac{1}{\rho(1-\rho)} = 2 \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 2 +\gamma - \log(4\pi).$$
A: These identities are not mysterious. They are simply the fact that the Riemann Zeta function has a Weierstrass product like any other meromorphic function of finite exponential order. Note here that $f'/f$ is called logarithmic derivative for a reason;)
Then it follows immediately
1) Yes, the zeros of the completed Riemann zeta function are exactly the nontrivial ones.
2) If RH hold, they come in pairs $\rho = 1/2 \pm \mathrm{i}t$ for $t>0$.
A suggestion for computing the sum: 
Let $\Omega_T$ be the boundary of $ -T \leq Im s \leq T$ and $-1/2 < Re s < 3/2$. For an approximation consider the integral
$$\frac{1}{2 \pi i} \int\limits_{\Omega_T} s^{-1} \frac{\zeta'(s)}{\zeta(s)} d \; s = \sum\limits_{-T \leq Im \rho \leq T} \frac{1}{\rho} + $$
some contribution coming from the poles, which are slightly delicate for $s=0$, since you encounter a double pole. I am pretty sure that you have missed that, and that this is why your computation fails.
Wikipedia tells you that 
$$ \zeta(s) = \frac{2^{s-1}}{s-1}-2^s \int_0^{\infty}\frac{\sin(s\arctan t)}{(1+t^2)^\frac{s}{2}(\mathrm{e}^{\pi\,t}+1)}\,\mathrm{d}t,$$
is pretty convenient for computing $\zeta$ numerically.
A: Edit: I endorse Juan's answer to the original question. The sum $\displaystyle{\sum_{\rho} \tfrac{1}{|\rho|}}$, running over the non-trivial zeros $\rho$ of $\zeta(s)$, is known to diverge, so at best $\displaystyle{\sum_{\rho} \tfrac{1}{\rho}}$ is conditionally convergent so you cannot re-arrange the terms.
In your second to last displayed equation, you removed the assumption that the sum runs over pairs of zeros $\rho$ and $1-\rho$. So it seems that Lagarias' result can be used to evaluate the sum
$$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)}.$$
As you observed, assuming the Riemann Hypothesis $1-\rho =\overline{\rho}$ for any non-trivial zero $\rho$ of $\zeta(s)$. This implies that
$$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2}.$$
