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$Γ (1 / 2) = \sqrt{π}$ by Wallis's product

Statement

$Γ (1 / 2) = \sqrt{π}$ .

Facts & Assumptions

Given: Positive integers $n$ tending to infinity.

[F1]

For $x > 0$ and $0 \leq s \leq 1$ , $x^{1 - s} \leq Γ (x + 1) / Γ (x + s) \leq (x + 1)^{1 - s}$ (Gautschi's inequality for the real Gamma function).

[F2]

If $a_{n} = (\binom{2 n}{n}) / 4^{n}$ , then $\sqrt{π n} a_{n} \to 1$ (The central binomial coefficient is asymptotic to 4^n divided by the square root of pi n).

[F3]

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

[F4]

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

Proof

technique · direct

1.1F4algebra

Iterating [F4] gives $Γ (n + 1 / 2) = Γ (1 / 2) \prod_{k = 0}^{n - 1} (k + 1 / 2) = ((2 n)! / (4^{n} n!)) Γ (1 / 2)$ , with the empty product valid at $n = 0$ .

1.2F1

Apply [F1] with $x = n$ and $s = 1 / 2$ . After inversion, $\sqrt{n / (n + 1)} \leq \sqrt{n} Γ (n + 1 / 2) / n! \leq 1$ , so this middle sequence tends to $1$ .

2.1step 1.1step 1.2F2F3algebra

By step 1.1 and [F3], the middle sequence is $Γ (1 / 2) \sqrt{n} (\binom{2 n}{n}) / 4^{n}$ . Fact [F2] makes its limit $Γ (1 / 2) / \sqrt{π}$ , while step 1.2 makes the same limit $1$ .

3.1step 2.1∎

Positivity of Gamma and uniqueness of limits therefore give $Γ (1 / 2) = \sqrt{π}$ .

Remarks

This proof uses Gautschi and Wallis. The Gaussian-integral proof $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral is logically independent of it.

Depends on

Dependency tree · 2 levels

Sources

University of Toronto MAT237Y1, The Gamma Function and the Beta Function, §2.3

Γ(1/2)=πΓ(1/2)=π​ by Wallis's product