Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The convergence and absolute-convergence abscissae differ by at most one

Statement

For every Dirichlet series, its abscissae satisfy

σc≤σa≤σc+1

in the extended real line.

Facts & Assumptions

Given: A Dirichlet series D(s)=∑n≥1ann−s with abscissae σc and σa.

[L1]

The two abscissae are defined by right-half-plane convergence and absolute convergence (The convergence and absolute-convergence abscissae of a Dirichlet series).

[L2]

Convergence at one point gives convergence on the entire open half-plane to its right (Convergence at one point of a Dirichlet series forces local uniform convergence on the open half-plane to its right).

Proof

technique · direct
1.1L1givenalgebra

Absolute convergence implies ordinary convergence term by term, so every half-plane counted for σa is also counted for σc. Therefore σc≤σa.

1.2L2givenalgebra

Let s0 be any point of convergence and write σ0:=ℜs0. Then the terms ann−s0 tend to 0, so they are bounded: ∣an∣≤Cnσ0 for some C. Hence for every s with ℜs>σ0+1, ∣an∣n−ℜs≤Cn−1−δ for some δ>0, and the right-hand side is summable. Thus absolute convergence holds throughout ℜs>σ0+1.

2.1step 1.1step 1.2∎

Since step 1.2 applies at every point of convergence, taking infima in [L1] gives σa≤σc+1. Combined with step 1.1, this is the claimed gap bound.

Depends on

Used by

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Dependency tree · two levels

6 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