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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!

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  • $\begingroup$ 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
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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.

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