Are Li's numbers conditionally or absolutely convergent? Li's numbers, $\lambda_n$ are defined by 
$\lambda_n = \sum_{\rho} \Big( 1- \Big(1-1/\rho\Big)^n\Big) $
where $n$ is real and$\rho$ runs over all the nontrivial zeros of the Riemann zeta function, and are always paired together with $1-\rho$.
My question is: are the $\lambda_n$'s absolutely or conditionally convergent ? The relevant Wikipedia article says they are conditionally convergent, while an article of Mark Coffey ''On certain sums over the nontrivial zeros, (page 2, line 1)'' states that Li's numbers are absolutely  convergent for $n>1$ ? Also, in Bombieri and Lagarias' article, ''Complements to Li's criterion and the Riemann Hypothesis, (page 2, lemma 1)'', it is mentioned that $\Re(\lambda_n)$ converges absolutely for every integer.
Is there a contradiction between the Wikipedia article and the articles of Coffey, Bombieri and Lagarias(especially that of Coffey) ?
 A: *

*The density of zeros  is $\displaystyle N(T) = \sum_{0 < Im(\rho) < T} 1 \ \ \sim \  \frac{T  \ln T}{2\pi}$ 
(the $\text{arg} \left[\frac{1}{2}s (s-1) \pi^{-s/2} \Gamma(s/2)\right]_{s = 1/2+IT}$ term in the contour integral)
hence $Im(\rho_k) \sim 2 \pi \frac{k}{\ln k}$ and $\rho_k \sim 2 i \pi \frac{k}{\ln k}$.

*As $x\to 0$ : $(1-x)^n = 1-nx+ \mathcal{O}(x^2)$ so that
$$\sum_{k=1}^\infty \left(1-\left(1-\frac{1}{\rho_k}\right)^n\right) = \sum_{k=1}^\infty  \frac{n}{\rho_k}+\mathcal{O}\left(\frac{1}{|\rho_k|^2}\right) =  \frac{ n}{2i\pi}\sum_{k=1}^\infty  \frac{\ln k}{k}+\mathcal{O}\left(\frac{1}{k^2}\right)$$
$\implies$ it doesn't converge

*but $(1-x)^n+(1-\overline{x})^n = 2-2n Re(x)+ \mathcal{O}(x^2)$, so that
$$\sum_{k=1}^\infty \left|1-\left(1-\frac{1}{\rho_k}\right)^n+1-\overline{\left(1-\frac{1}{\rho_k}\right)^n}\right| = \sum_{k=1}^\infty \left| 2n Re\left(\frac{1}{\rho_k}\right)\right|+\mathcal{O}\left(\frac{1}{|\rho_k|^2}\right)$$ $$= 
\sum_{k=1}^\infty  \left| 2n Re\left(\frac{\sigma_k-i2\pi\frac{k}{\ln k}}{\sigma_k^2+4\pi^2\frac{k^2}{\ln^2 k}}\right)\right|+\mathcal{O}\left(\frac{1}{k^2}\right) =
\frac{n}{4\pi^2} \sum_{k=1}^\infty  \frac{\ln^2 k}{k^2}+\mathcal{O}(1)$$
this time it converges absolutely, grouping $\rho$ with its complex conjugate $\overline{\rho}$
(that's why Bombieri requires that $\sum_\rho \frac{1+\left|\operatorname{Re}(\rho)\right|}{(1+|\rho|)^2} $ converges for applying the Li's criterion)
