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 Res>1

Statement

For every complex number s with Res>1, the series

n=1ns,ns:=exp(slogn),

converges absolutely. Moreover, if K{sC:Res>1} is compact, then the same series converges uniformly on K.

Facts & Assumptions

Given: A compact set K{sC:Res>1}.

[L1]

The complex exponential is expz=m0zm/m! (The complex exponential by its power series).

[L2]

The real logarithm is defined on (0,), so logn is defined for every integer n1 (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 n1np converges (For rational p>0, 1/kp converges iff p>1).

Proof

technique · direct
1.1

Because K is compact and lies in the open half-plane Res>1, there is a real number σ>1 with Resσ for every sK. For n1 and sK, [L1] and [L2] give ns=exp(slogn), so ns=exp(Reslogn)exp(σlogn)=nσ.

givenL1L2choosealgebra
2.1

Taking σ>1 rational if necessary, [L4] makes n1nσ convergent. Step 1.1 and [L3] therefore give uniform convergence of ns on K. Since ns is bounded by the same summable majorant, the series also converges absolutely at each point of K, hence at each s with Res>1.

step 1.1L3L4algebra

Depends on

Used by

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

Dependency tree · two levels

28 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