(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.
Best Answer
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