Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04
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The Dirichlet series for zeta converges absolutely and locally uniformly on the half-plane Re⁡s>1

Statement

For every complex number s with Re⁡s>1, the series

∑n=1∞n−s,n−s:=exp⁡(−slog⁡n),

converges absolutely. Moreover, if K⊆{s∈C:Re⁡s>1} is compact, then the same series converges uniformly on K.

Facts & Assumptions

Given: A compact set K⊆{s∈C:Re⁡s>1}.

[L1]

The complex exponential is exp⁡z=∑m≥0zm/m! (The complex exponential by its power series).

[L2]

The real logarithm is defined on (0,∞), so log⁡n is defined for every integer n≥1 (The natural logarithm as the inverse of the exponential function).

[L3]

If ∣fn(s)∣≤Mn on a set and ∑Mn converges, then ∑fn converges uniformly there (Weierstrass M-test for complex-valued function series).

[L4]

For rational p>1, the series ∑n≥1n−p converges (For rational p>0, ∑1/kp converges iff p>1).

Proof

technique · direct
1.1givenL1L2choosealgebra

Because K is compact and lies in the open half-plane Re⁡s>1, there is a real number σ>1 with Re⁡s≥σ for every s∈K. For n≥1 and s∈K, [L1] and [L2] give n−s=exp⁡(−slog⁡n), so ∣n−s∣=exp⁡(−Re⁡slog⁡n)≤exp⁡(−σlog⁡n)=n−σ.

2.1step 1.1L3L4algebra∎

Taking σ>1 rational if necessary, [L4] makes ∑n≥1n−σ convergent. Step 1.1 and [L3] therefore give uniform convergence of ∑n−s on K. Since ∣n−s∣ is bounded by the same summable majorant, the series also converges absolutely at each point of K, hence at each s with Re⁡s>1.

Depends on

Used by

Cited to discharge well-definedness by The Riemann zeta function on the half-plane Res>1.

Dependency tree · two levels

29 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