CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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Completely multiplicative Dirichlet series have geometric Euler factors

Statement

If $f$ is completely multiplicative and its Dirichlet series converges absolutely at $s$ , then

$\sum_{n \geq 1} f (n) n^{- s} = \prod_{p} \frac{1}{1 - f (p) p^{- s}} .$

Facts & Assumptions

Given: A completely multiplicative function $f$ and a point of absolute convergence.

[L1]

Completely multiplicative means $f (p^{k}) = f (p)^{k}$ for every prime power (Completely multiplicative arithmetic functions).

[L2]

Multiplicative Dirichlet series factor into Euler products (A multiplicative Dirichlet series factors as an Euler product on its absolute half-plane).

Proof

technique · direct

1.1L2givenalgebra

By [L2], $\sum_{n \geq 1} f (n) n^{- s} = \prod_{p} \sum_{k \geq 0} f (p^{k}) p^{- k s} .$

2.1L1step 1.1algebra∎

By [L1], each local factor is the geometric series $\sum_{k \geq 0} (f (p) p^{- s})^{k} = \frac{1}{1 - f (p) p^{- s}},$ whose denominator is nonzero because absolute convergence forces $∣ f (p) p^{- s} ∣ < 1$ . Substituting into step 1.1 gives the claimed Euler product.

Depends on

Used by

Dependency tree · two levels

6 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

Kiran S. Kedlaya, Notes on Analytic Number Theory, Definition 2.7 (standard reference, not scraped)