Are Li's numbers $\lambda_n$ absolutely convergent for $n>1$?

Li's numbers $$\{\lambda_n\}$$ are defined as $$\lambda_n=\frac{1}{(n-1)!}\frac{d^n}{ds^n} [s^{n-1}\log\xi(s)]_{s=1}$$ for all positive integers $$n$$. Also $$\lambda_n$$ is given as a sum over the non trivial zeros of $$\zeta(s)$$ by $$\lambda_n=\sum_{\rho}\left[1-\left(1-\frac{1}{\rho}\right)^n\right]$$ where $$\rho$$ are the non trivial zeros of the Riemann zeta function.

My question is: Are the $$\lambda_n$$'s absolutely convergent for $$n>1$$?

An article of Mark W. Coffey "On certian sums over the non trivial zeta zeros" the year being (2010) says in line 1 p.2 that fot $$n>1$$, $$\lambda_n$$ is absolutely convergent, while for $$\lambda_1$$ the sum should be taken over complex conjugate pairs of zeros of increasing imaginary part.

An answer or a reference is desired.

Assuming the Riemann hypothesis $$\rho=\frac12+i\gamma$$, then $$1-\frac{1}{\rho}=-\frac{\frac12-i\gamma}{\frac12+i\gamma}=e^{2i\theta},\qquad \theta=\arctan\frac{1}{2\gamma}.$$ $$\sum_\rho\Bigl[1-\Bigl(1-\frac{1}{\rho}\Bigr)^n\Bigr]=\sum_\gamma\bigl(1-e^{2in\theta}\bigr).$$ Here $$n$$ is fixed and for $$\gamma\to+\infty$$ $$1-e^{2in\theta}\sim 2n\theta\sim \frac{n}{\gamma}$$ Since $$\gamma_k\sim 2\pi k/\log k$$ the series is equivalent to $$n\sum_k\frac{\log k}{2\pi k}$$ therefore is divergent.

When $$\rho$$ is associated with $$\overline{\rho}$$ we obtain $$2\sum_{\gamma>0} \Re(1-e^{2in\theta})=4\sum_{\gamma>0}\sin^2(n\theta)\sim 4n^2\sum_k\frac{(\log k)^2}{4\pi^2k^2},$$ that is absolutely convergent.

There are several papers dealing with the asymptotic value of $$\lambda_n$$.

• Thank you. So you assumed the Riemann Hypothesis. What if we do not assume the Riemann Hypothesis. Will combining $\rho$ with $\bar{\rho}$ will we have absolute convergence?
– user469908
Nov 28, 2021 at 11:34
• In this post on Math. SE, the answer is given irrespective of the truth of the Riemann Hypothesis. Please see here- math.stackexchange.com/questions/1921175/…
– user469908
Nov 28, 2021 at 11:40
• @Rama1729 yes the RH is not required, but the argument is slightly more complex. But apparently you knew the answer.
– juan
Nov 28, 2021 at 11:57
• Thank you. It would be of great help if you would edit your answer and prove it without assuming RH. Thanks again.
– user469908
Nov 28, 2021 at 11:58
• How can we associate $\rho$ with $\bar{\rho}$. Will this association not require the absolute convergence of the sum?
– user469908
Nov 28, 2021 at 12:00