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.


1 Answer 1


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 $\lambda_n$.

  • $\begingroup$ 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? $\endgroup$
    – user469908
    Nov 28, 2021 at 11:34
  • $\begingroup$ 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/… $\endgroup$
    – user469908
    Nov 28, 2021 at 11:40
  • $\begingroup$ @Rama1729 yes the RH is not required, but the argument is slightly more complex. But apparently you knew the answer. $\endgroup$
    – juan
    Nov 28, 2021 at 11:57
  • $\begingroup$ Thank you. It would be of great help if you would edit your answer and prove it without assuming RH. Thanks again. $\endgroup$
    – user469908
    Nov 28, 2021 at 11:58
  • $\begingroup$ How can we associate $\rho$ with $\bar{\rho}$. Will this association not require the absolute convergence of the sum? $\endgroup$
    – user469908
    Nov 28, 2021 at 12:00

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