Define $Z$ as a sort of zeta function defined over the powers of the non-trivial zeros $\rho$ of the Riemann zeta function (a meta-zeta function). So we have, $$Z(s) = \sum_{\rho} \frac{1}{\rho^s}$$ How can I prove that this sum converges absolutely for $\Re(s) >1$?
I am sure this function has been studied before, but I can't find any references on its convergence. I know that $Z(s)$ converges if the $S$ (below) converges. I was able to show that $S$ converges, but is this a valid proof, or the conventional proof? Thanks \begin{align} S &= \sum_{\gamma_n > 0} \frac{1}{\gamma_n^{\sigma}} = \int_{\gamma_1}^{\infty} \frac{1}{T^\sigma} \, dN(T) \\ &= \left[\frac{N(T)}{T^{\sigma}} \right]_{\gamma_1}^{\infty} - \int_{\gamma_1}^{\infty} N(T) \left( \frac{d}{dT} T^{- \sigma} \right) \, dT \\ &= \eta + \sigma \int_{\gamma_1}^{\infty} \frac{N(T)}{T^{\sigma+1}} \, dT \end{align}
where $\eta$ is some constant. Thus, the sum converges if the integral converges. Indeed, for the integral we may use the Riemann–von Mangoldt formula
$$\int_{\gamma_1}^{\infty} \frac{N(T)}{T^{\sigma + 1}} \, dT \sim \int_{\Omega}^{\infty} \frac{\frac{T}{2\pi} \log T}{T^{\sigma+1}} \, dT = \xi \int_{\Omega}^{\infty} \frac{\log T}{T^{\sigma}} \, dT$$
which converges if $\sigma > 1$ and diverges if $\sigma \le 1$.
