TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Stirling's formula for factorials

Statement

$n! \sim \sqrt{2 π n} (n / e)^{n}$ as the positive integer $n$ tends to infinity.

Equivalently,

$\frac{n!}{\sqrt{2 π n} (n / e)^{n}} ⟶ 1 (n \to \infty, n \geq 1) .$

Facts & Assumptions

Given: Positive integer $n$ tending to infinity.

[F1]

There is a constant $C > 0$ such that $n! \sim C \sqrt{n} (n / e)^{n}$ (Stirling's factorial asymptotic holds up to a positive constant).

[F2]

The constant $C$ in that asymptotic is $\sqrt{2 π}$ (Wallis's product determines the Stirling constant as $\sqrt{2 π}$ ).

Proof

technique · direct

1.1

Insert [F2] into [F1] and combine $\sqrt{2 π} \sqrt{n} = \sqrt{2 π n}$ .

F1F2algebra

2.1

By the definition of asymptotic equivalence, step 1.1 is exactly the displayed ratio limit. The restriction $n \geq 1$ keeps its denominator positive.

step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 7 §6 (standard reference, not scraped)