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 s>1,

n1μ(n)ns=1ζ(s),

where ζ(s)=n1ns.

Facts & Assumptions

Given: A complex number s with s>1.

[L1]

The Möbius function satisfies dnμ(d)={1,n=1,0,n>1, that is, μ1=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 n1nq converges (For rational p>0, 1/kp converges iff p>1).

Proof

technique · direct
1.1

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

L1L3givenchoosealgebra
2.1

By [L2], (n1μ(n)ns)(n1ns)=n1(μ1)(n)ns.

L1L2step 1.1algebra
3.1

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)=n1ns.

L1step 2.1

Depends on

Used by

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