LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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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.
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Stirling's factorial asymptotic holds up to a positive constant

Statement

There is a constant $C > 0$ such that $n! \sim C \sqrt{n} (n / e)^{n}$ . Here $n$ tends to infinity through the positive integers.

Facts & Assumptions

Given: Positive integers $r, n$ and the logarithm on positive reals.

[F1]

For $- 1 < x \leq 1$ , $\log (1 + x) = \sum_{j = 1}^{\infty} (- 1)^{j + 1} x^{j} / j$ (The power series for log(1+x), including the Abel endpoint).

[F2]

The positive series $\sum_{k \geq 1} k^{- p}$ converges exactly when $p > 1$ (The p-series for a real exponent p converges exactly when p is greater than one).

[F3]

For $x > 0$ , $\log^{'} (x) = 1 / x$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

Proof

technique · direct

1.1

Put $e_{r} : = \log r - \int_{r - 1 / 2}^{r + 1 / 2} \log t d t$ . After $t = r + u$ , expand $\log (1 + u / r)$ by [F1]. Integration over the symmetric interval cancels the odd powers, and the remaining absolutely convergent even series gives $∣ e_{r} ∣ \leq K / r^{2}$ for one constant $K$ and all $r \geq 1$ .

F1algebra

2.1

By step 1.1 and [F2] with $p = 2$ , the series $\sum_{r \geq 1} e_{r}$ converges absolutely.

step 1.1F2

3.1

Summing the definition of $e_{r}$ from $1$ to $n$ telescopes the integrals to $\int_{1 / 2}^{n + 1 / 2} \log t d t$ . Fact [F3] gives the primitive $t \log t - t$ , and comparison of $n + 1 / 2$ with $n$ shows that $\log (n!) - ((n + 1 / 2) \log n - n)$ converges to a real constant $c$ .

step 2.1F3algebra

4.1

Exponentiating step 3.1 and putting $C = e^{c} > 0$ gives $n! / (\sqrt{n} (n / e)^{n}) \to C$ , which is the stated asymptotic.

step 3.1algebra∎

Depends on

Used by

Dependency tree · two levels

57 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 6 §1(v) (standard reference, not scraped)