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Wallis's product determines the Stirling constant as $\sqrt{2 π}$

Statement

The constant $C$ in the preceding asymptotic is $\sqrt{2 π}$ .

Facts & Assumptions

Given: The positive constant $C$ from the preceding lemma.

[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 Wallis consequence is $(\binom{2 n}{n}) \sim 4^{n} / \sqrt{π n}$ (The central binomial coefficient is asymptotic to 4^n divided by the square root of pi n).

[F3]

For $n \geq 1$ , $(\binom{2 n}{n}) (n!)^{2} = (2 n)!$ , so $(\binom{2 n}{n}) = (2 n)! / (n!)^{2}$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

Proof

technique · direct

1.1

Insert the two asymptotics of [F1] into the quotient [F3]. Cancellation gives $(\binom{2 n}{n}) \sim (\sqrt{2} / C) 4^{n} / \sqrt{n}$ .

F1F3algebra

2.1

Comparing the positive leading coefficient in step 1.1 with [F2] gives $\sqrt{2} / C = 1 / \sqrt{π}$ . Since $C > 0$ , $C = \sqrt{2 π}$ .

step 1.1F2algebra∎

Depends on

Used by

Stirling's formula for factorials Theorem

Dependency tree · two levels

53 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

University of Toronto MAT237Y1, The Gamma Function and the Beta Function, §2.3(g) (standard reference, not scraped)
K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 7 §6 (standard reference, not scraped)

Wallis's product determines the Stirling constant as 2π2π​