Prove that Wallis' product and Euler's formula directly imply that $(-1/2)!=\sqrt{\pi}$

Question

(This occured to me recently, and I was pretty sure that it was true, so I was pleased that it really was. This has almost certainly been published many times before, but I didn't see it in either of the Wikipedia articles on the Wallis product and Euler's limit formula for factorial so I thought that I would propose it here.)

Euler's formula for general factorial is

$$z! =\lim_{n \to \infty} \frac{n!n^z}{\displaystyle\prod_{k=1}^n (z+k)} . $$

Wallis' product is

$$\frac{\pi}{2} =\prod_{k=1}^{\infty} \frac{4k^2}{4k^2-1} $$

Prove that Wallis' product and Euler's formula directly imply that $(-1/2)! =\sqrt{\pi} $.

I'll post my answer in a few days if no one else does.

Answer 1

Rewrite the limit as an infinite product $$ \begin{align} z! &=\lim_{n\to\infty}\frac{n!n^z}{\prod\limits_{k=1}^n(z+k)}\tag{1a}\\ &=\lim_{n\to\infty}\frac{n!(n+1)^z}{\prod\limits_{k=1}^n(z+k)}\tag{1b}\\ &=\lim_{n\to\infty}(n+1)^z\prod_{k=1}^n\frac{k}{z+k}\tag{1c}\\[6pt] &=\lim_{n\to\infty}\prod_{k=1}^n\frac{k}{z+k}\left(\frac{k+1}{k}\right)^z\tag{1d}\\[6pt] &=\prod_{k=1}^\infty\frac{k}{z+k}\left(\frac{k+1}{k}\right)^z\tag{1e} \end{align} $$ Explanation:
$\text{(1a):}$ Euler's Formula
$\text{(1b):}$ $\lim\limits_{n\to\infty}\frac{n+1}n=1$
$\text{(1c):}$ $n!=\prod\limits_{k=1}^nk$
$\text{(1d):}$ $n+1=\prod\limits_{k=1}^n\frac{k+1}k$
$\text{(1e):}$ definition of an infinite product

According to $(3)$ in this paper, $\text{(1b)}$ is Euler's Formula.

$\text{(1e)}$ is also proven in $(4)$ of this answer.

---

Evaluate $\boldsymbol{\left(-\frac12\right)!}$ $$ \begin{align} \left(-\frac12\right)! &=\prod_{k=1}^\infty\frac{k}{-\frac12+k}\left(\frac{k+1}{k}\right)^{-\frac12}\tag{2a}\\ &=\prod_{k=1}^\infty\frac{k}{-\frac12+k}\sqrt{\frac{k}{k+1}}\tag{2b}\\ &=\prod_{k=0}^\infty\frac{k+1}{\frac12+k}\sqrt{\frac{k+1}{k+2}}\tag{2c}\\ &=\sqrt2\prod_{k=1}^\infty\frac{k}{\frac12+k}\frac{k+1}k\sqrt{\frac{k+1}{k+2}}\tag{2d}\\ &=\left(\sqrt2\prod_{k=1}^\infty\frac{k^2}{k^2-\frac14}\frac{k+1}k\sqrt{\frac k{k+2}}\right)^{1/2}\tag{2e}\\ &=\left(2\prod_{k=1}^\infty\frac{k^2}{k^2-\frac14}\right)^{1/2}\tag{2f}\\[9pt] &=\sqrt\pi\tag{2g} \end{align} $$ Explanation:
$\text{(2a):}$ plug $z=-\frac12$ into $\text{(1e)}$
$\text{(2b):}$ apply exponent of $-\frac12$
$\text{(2c):}$ substitute $k=k+1$
$\text{(2d):}$ pull out the $k=0$ term
$\text{(2e):}$ take the geometric mean of $\text{(2b)}$ and $\text{(2d)}$
$\text{(2f):}$ $\prod\limits_{k=1}^\infty\frac{k+1}k\sqrt{\frac k{k+2}}=\lim\limits_{n\to\infty}(n+1)\sqrt{\frac2{(n+1)(n+2)}}=\sqrt2$
$\text{(2g):}$ Wallis' Product

Answer 2

Here is my answer.

There is my usual excruciating detail since, as always, I wrote nothing down on paper but directly entered this into MacDown, my MathJax editor of choice.

Euler's formula for general factorial is

$z! =\lim_{n \to \infty} \dfrac{n!n^z}{\prod_{k=1}^n (z+k)} $.

Wallis' product is

$\dfrac{\pi}{2} =\prod_{k=1}^{\infty} \dfrac{4k^2}{4k^2-1} $

I first tried to put Wallis' formula into a form that I hoped could match with Euler's formula for $(-1/2)!$.

$\begin{array}\\ \dfrac{\pi}{2} &=\prod_{k=1}^{\infty} \dfrac{4k^2}{4k^2-1}\\ &=\prod_{k=1}^{\infty} \dfrac{2k}{2k-1}\dfrac{2k}{2k+1}\\ &=\lim_{n \to \infty}\prod_{k=1}^{n} \dfrac{2k}{2k-1}\dfrac{2k}{2k+1}\\ &=\lim_{n \to \infty}\prod_{k=1}^{n} \dfrac{2k}{2k-1}\prod_{k=1}^{n}\dfrac{2k}{2k+1}\\ \end{array} $

I then plugged $-1/2$ into Euler's formula, manipulated it a bit, squared that, split and regrouped terms, did some index fiddling, and, shazam, it matched!

$\begin{array}\\ -\dfrac12! &=\lim_{n \to \infty} \dfrac{n!n^{-1/2}}{\prod_{k=1}^n (-1/2+k)}\\ &=\lim_{n \to \infty} \dfrac{n!n^{-1/2}2^n}{\prod_{k=1}^n (-1+2k)}\\ &=\lim_{n \to \infty} \dfrac{\prod_{k=1}^n (2k)n^{-1/2}}{\prod_{k=1}^n (-1+2k)}\\ \text{so}\\ (-\dfrac12!)^2 &=\lim_{n \to \infty} \dfrac{\prod_{k=1}^n (2k)^2}{n\prod_{k=1}^n (-1+2k)^2}\\ &=\lim_{n \to \infty} \dfrac{\prod_{k=1}^n (2k)}{n\prod_{k=1}^n (-1+2k)} \dfrac{\prod_{k=1}^n (2k)}{\prod_{k=1}^n (-1+2k)}\\ &=\lim_{n \to \infty} \dfrac1{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \prod_{k=1}^n\dfrac{ (2k)}{(2k-1)}\\ &=\lim_{n \to \infty} \dfrac1{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} 2\prod_{k=2}^n\dfrac{ (2k)}{(2k-1)}\\ &=\lim_{n \to \infty} \dfrac{2}{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \prod_{k=1}^{n-1}\dfrac{ (2(k+1))}{2k+1}\\ &=\lim_{n \to \infty} \dfrac{2}{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \dfrac{ \prod_{k=1}^{n-1}(2(k+1))}{\prod_{k=1}^{n-1}(2k+1)}\\ &=\lim_{n \to \infty} \dfrac{2(2n+1)}{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \dfrac{ \prod_{k=1}^{n-1}(2(k+1))}{\prod_{k=1}^{n}(2k+1)}\\ &=\lim_{n \to \infty} \dfrac{2(2n+1)}{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \dfrac{ \prod_{k=2}^{n}(2k)}{\prod_{k=1}^{n}(2k+1)}\\ &=\lim_{n \to \infty} \dfrac{(2n+1)}{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \dfrac{ \prod_{k=1}^{n}(2k)}{\prod_{k=1}^{n}(2k+1)}\\ &=\lim_{n \to \infty} \dfrac{(2n+1)}{n}\prod_{k=1}^n\dfrac{ (2k)}{(2k-1)} \prod_{k=1}^{n}\dfrac{(2k)}{(2k+1)}\\ &=2(\dfrac{\pi}{2})\\ &=\pi\\ \end{array} $