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I would like to know if $$\sum_{k \geq 1} \frac{1}{\gamma_k}$$ is convergent or not. ($\gamma_k$ is the imaginary part of the k-th non trivial zero of $\zeta$).

In fact, I think this series is actually divergent, because of the "Explicit Formula" : $$\psi(x) = x-\sum_\rho \frac{x^\rho}{\rho} - \ln(2 \pi) - \ln(1-x^{-2})/2$$Where $\psi$ is the Chebyshev function. So we have $$\sum_\rho \frac{x^\rho}{\rho} = x - \psi(x) - \ln(2\pi) - \frac{1}{2} \ln (1-x^{-2})$$But at $x=1$, the function $\ln(1-x^{-2})$ diverges. So the series $\sum_\rho 1/\rho$ must too. However, this "proof" may not be one because the "$\rho$" are $1/2+i \gamma_k$ (iff the RH is true) and not just $\gamma_k$.

Now, what I would like to know, is the value (if it converges) of $$\sum_{k \geq 1} \frac{1}{\gamma_k^s}$$For any $s \in \mathbb{C}$. I'm aware that this series propably diverges for $\Re(s) \leq 1$ but perhaps there is a way to extend this function analytically as in the case of Riemann's zeta function. Even if I can have an asymptotic approximation.

K. Kapa
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2 Answers2

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Let $N(T)$ be the number of non-trivial zeros of $\zeta$ such that $0<\gamma\leqslant T$ ($\gamma$ being their imaginary part). It is known that $\displaystyle N(T)\underset{T\rightarrow +\infty}{\sim}\frac{T}{2\pi}\log T$, thus if $\gamma_n$ is the imaginary part of the $n$-th non-trivial zero (I rank them such that $0<\gamma_1\leqslant\ldots\leqslant\gamma_n$ and $\gamma_{-n}=-\gamma_n$ for $n\geqslant 1$). Since $N(\gamma_n)=n$, we have $\displaystyle n\underset{n\rightarrow +\infty}{\sim}\frac{\gamma_n}{2\pi}\log\gamma_n$, taking the logarithm gives $\log n\underset{n\rightarrow +\infty}{\sim}\log\gamma_n$ and thus $\gamma_n\underset{n\rightarrow +\infty}{\sim}\frac{2\pi n}{\log n}$. We can then conclude that $\sum_{n\geqslant 1}\frac{1}{\gamma_n^s}$ converges if and only if ${\rm Re}(s)>1$, in particular the series $\sum_{n\geqslant 1}\frac{1}{\gamma_n}$ diverges. As for the precise value of $\sum_{n\geqslant 1}\frac{1}{\gamma_n^s}$ when ${\rm Re}(s)>1$, I don't have the answer (and I'm not sure if we know anything about it).

Tuvasbien
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If you have a look at this question, you will see an approximation $$\Im\left(\rho _{n}\right)\sim \frac{2 \pi n}{W\left(\frac{n}{e}\right)}$$ where $W(.)$ is Lambert function. So $$\sum_{n=1}^\infty \frac 1 {\big[\Im\left(\rho _{n}\right)\big]^s}\sim\sum_{n=1}^\infty\left( \frac{W\left(\frac{n}{e}\right)}{2 \pi n}\right)^s\sim \sum_{n=1}^\infty \left(\frac{\log (n)}{n}\right)^s$$ which converges only if $s>1$.

Claude Leibovici
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