complex analysis - Sum over all inverse zeta nontrivial zeros

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros)
$$\zeta(s)=\frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)e^{s/\rho}$$
can one derive the exact value of $\sum_{\rho} \frac{1}{\rho}$
to be
$$\sum_{\rho} \frac{1}{\rho} = -\log(2\sqrt{\pi})+1+\gamma/2$$
What implications does this have?

Answer

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $\;\psi(z):=\dfrac d{dz}\log\Gamma(z)\,$ with $\psi$ the digamma function) :
\begin{align}
\log\zeta(s)&=(\log(2\pi)-1-\gamma/2)s-\log\left(2(s-1)\Gamma(1+s/2)\right)+\sum_{\rho}\log\left(1-\frac{s}{\rho} \right)+\frac s{\rho}\\
\frac{\zeta'(s)}{\zeta(s)}&=\log(2\pi)-1-\gamma/2-\left(\frac 1{s-1}+\frac 12\psi\left(\frac s2+1\right)\right)+\sum_{\rho}\frac 1{s-\rho}+\frac 1{\rho}\\
\frac{\zeta'(s)}{\zeta(s)}+\frac 1{s-1}&=\log(2\pi)-1-\gamma/2-\frac 12\psi\left(\frac s2+1\right)+\sum_{\rho}\frac 1{s-\rho}+\frac 1{\rho}\\
\end{align}
taking the limit as $s\to 1$ and remembering that $\;\displaystyle \zeta(s)-\frac 1{s-1}=\gamma+O(s-1)\;$ and $\psi\left(\frac 32\right)=\psi\left(\frac 12\right)+2=-\gamma-\log(4)+2\;$ we get :
\begin{align}
\gamma&=\log(2\pi)-1-\gamma/2+\left(\gamma/2+\log(2)-1\right)+\sum_{\rho}\frac 1{1-\rho}+\frac 1{\rho}\\

\sum_{\rho}\frac 1{1-\rho}+\frac 1{\rho}&=-\log(4\pi)+2+\gamma
\end{align}
With the answer the double of your series (as it should since if $\rho$ is a root then $1-\rho$ will also be one).

I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.

For the sum of integer powers of the roots see Mathworld starting at $(5)$.

Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).