LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Zeta reciprocal zero sum bound

Statement

For $T \geq 2$ , the sum of $1 / ∣ ρ ∣$ over nontrivial zeros with $0 < ∣ Im ρ ∣ \leq T$ is $O (\log^{2} T)$ , with multiplicity. Adjoining any real nontrivial zeros preserves the estimate.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

A unit-interval bound for zeta zeros: The number of nontrivial zeta zeros, with multiplicity, whose ordinates lie in $[T, T + 1]$ is $O (\log (T + 2))$ for $T \geq 0$ .

[F2]

The Riemann zeta zero-counting function: For $T > 0$ , $N (T)$ is the number, with multiplicity, of nontrivial zeros $ρ = β + i γ$ of the meromorphic continuation of zeta satisfying $0 < γ \leq T$ . Thus a zero on the top boundary is included.

[F3]

The Riemann xi function has its genus-one Hadamard product over the nontrivial zeros of zeta: There exist constants $A, B \in C$ such that $ξ (s) = e^{A + B s} \prod_{ρ} E_{1} (s / ρ),$ where the product runs over the nontrivial zeros $ρ$ of $ζ$ , counted with multiplicity, and $E_{1} (w) = (1 - w) e^{w} .$ The product converges in the genus-one canonical sense.

Proof

1.1

Nontrivial zeros have no accumulation in a compact subset of the plane. The finitely many with $∣ Im ρ ∣ \leq 1$ have nonzero denominator: the product for xi has no zero at zero, and its factors have exactly the nontrivial zeros. Their reciprocal sum is a fixed finite constant.

2.1

For each integer $n \geq 1$ , zeros with $n \leq ∣ Im ρ ∣ \leq n + 1$ each contribute at most $1 / n$ . The count in either band is $O (\log (n + 2))$ ; the negative band follows from $ζ (\overline{s}) = \overline{ζ (s)}$ , first on its defining half-plane and then by continuation. Therefore the remaining sum is at most $C \sum_{n \leq T} \log (n + 2) / n = O (\log^{2} T)$ . Endpoint overlap only increases this upper bound.

F1F2step 1.1∎

Depends on

Used by

Zeta explicit formula zero free error balance Lemma

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

§7.2, proof of Theorem 7.7 after Theorem 7.6 (standard reference, not scraped)