Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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A multiplicative Dirichlet series factors as an Euler product on its absolute half-plane

Statement

Let f be a multiplicative arithmetic function. If

n1f(n)nσ<,

then for every s with sσ,

n1f(n)ns=pk0f(pk)pks,

where the infinite product is the limit of the finite prime products.

Facts & Assumptions

Given: A multiplicative arithmetic function f and a complex number s with sσ.

[L1]

Multiplicative functions satisfy f(mn)=f(m)f(n) for coprime m,n (Multiplicative arithmetic functions).

[L3]

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

Proof

technique · direct
1.1

For a finite set P of primes, expand pPk0f(pk)pks. Using multiplicativity [L1] and unique factorization [L2], this is exactly n1all prime factors of n lie in Pf(n)ns.

L1L2L3givenalgebra
2.1

As P increases, these partial Euler products exhaust the original Dirichlet series. Because sσ and the series f(n)nσ converges, the omitted tail tends to 0 absolutely. Hence the finite prime products converge to n1f(n)ns.

step 1.1givenalgebra

Depends on

Used by

Dependency tree · two levels

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Sources