Let $\rho$ denote a non-trivial zero of $\zeta(s)$ and $l_n(\rho)$ the $n$th coefficient in the Laurent expansion of $1/\zeta(s)$ about $\rho$. For example, if the pole is simple then we have $l_{-1}(\rho)=1/\zeta'(\rho)$.

My motivation is to understand the asymptotics of these sequences, the reasons for which I will briefly explain: Firstly, on intuitive grounds, these sequences encode the radii of convergence of the expansions at each $\rho$ and thus the vertical distribution of poles. However, I believe that a lot more can be said in connection with the horizontal distribution of zeros by considering the sequence $s:\mathbb{N}\rightarrow\mathbb{R}$, defined by $$s(n)=\sum_{\rho}\frac{l_{n}(1-\rho)}{\rho\zeta'(\rho)},$$ assuming all the poles are simple (one would have to modify $s(n)$ otherwise). The quantity $l$ in $$\frac{1}{l}=\limsup_{n\rightarrow\infty}\sqrt[n]{|s(n)|}$$(if it exists) is the radius of convergence of a certain power series, and measures twice the horizontal distance from the critical line to a zero of $\zeta(s)$ off the line, if any exist. Thus, establishing that $l>0$ establishes that there is a strip of non-zero width about the critical line containing no zeros off the line, and establishing that $l\geq 1$ is the Riemann Hypothesis. One might like to think of this as attacking the strip from the inside, rather than out.

I would be very interested to know if anything is known about $l_n(\rho)$, or indeed $s(n)$, and where I can find the details if so? If anyone would like me to explain how this series is derived, I will happily do so. Thanks!

  • $\begingroup$ Oops: I got the conditions on $l$ upside down - it has been corrected now. $\endgroup$ – Kevin Smith Oct 31 '11 at 9:45
  • $\begingroup$ I think perhaps a better way to look at this would be to ask if $\limsup_{n\rightarrow\infty} n^{-1}\log |s(n)|$ exists. $\endgroup$ – Kevin Smith Oct 31 '11 at 10:43
  • $\begingroup$ Then $\log|s(n)|=O(n)$ implies that there are no zeros arbitrarily close to the critical line, and $\log|s(n)|=o(n)$ implies that there are none off the line. I should add that I am aware of the Bohr-Landau Theorem, so I am considering how one might assess whether the zeros off the line are necessarily repelled further away. $\endgroup$ – Kevin Smith Oct 31 '11 at 10:51

The answer is that $l=0$, so the power series does not converge. This is not necessarily because there are poles arbitrarily close to the critical line yet not on it; it is because there are, somewhere on the critical line, poles that are arbitrarily close together. Indeed, Littlewood has shown that the sequence of differences of the imaginary parts $(\gamma_n)$ of the zeros of $\zeta(s)$, that is, the sequence $|\gamma_{n+1}-\gamma_{n}|\rightarrow 0$. Thus, since there are infinitely many zeros, there are infinitely many $\rho$ such that $$\limsup\sqrt[n]{|l_n(\rho)|}>C$$
for any positive constant $C$, so the sequence $s(n)$ is unbounded.

I should add that my statement that the radius of convergence of the power series measures only horizontal distance was clearly incorrect.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.