Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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The Dirichlet series of the Möbius function is the reciprocal of the zeta Dirichlet series on Re s greater than 1

Statement

For ℜs>1,

∑n≥1μ(n)n−s=1ζ(s),

where ζ(s)=∑n≥1n−s.

Facts & Assumptions

Given: A complex number s with ℜs>1.

[L1]

The Möbius function satisfies ∑d∣nμ(d)={1,n=1,0,n>1, that is, μ∗1=ε (Classical Möbius inversion over positive divisors, The Dirichlet-convolution identity and the constant-one function, The number-theoretic Möbius function μ(n) from prime factorisation).

[L2]

Products of absolutely convergent Dirichlet series multiply by Dirichlet convolution (Multiplying absolutely convergent Dirichlet series gives Dirichlet convolution).

[L3]

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

Proof

technique · direct
1.1L1L3givenchoosealgebra

Write σ:=ℜs>1 and choose a rational q with 1<q<σ. Since μ(n)∈{−1,0,1} by [L1], one has ∣μ(n)n−s∣≤n−σ≤n−q,∣n−s∣=n−σ≤n−q. By [L3], both Dirichlet series therefore converge absolutely.

2.1L1L2step 1.1algebra

By [L2], (∑n≥1μ(n)n−s)(∑n≥1n−s)=∑n≥1(μ∗1)(n)n−s.

3.1L1step 2.1∎

Step 2.1 and the identity in [L1] make the right-hand side equal to 1, so the first factor is the reciprocal of ζ(s)=∑n≥1n−s.

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