TAOCP 1.2.5 Exercise 18
From Euler's limit formula for the factorial function (Equation 13), n!
Section 1.2.5: Permutations and Factorials
Exercise 18. [**] [M20] Assume that
$$ \frac{\pi}{2} = \frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{5}{6}\cdot\frac{7}{6}\cdots $$
(This is "Wallis's product," obtained by J. Wallis in 1655, and we will prove it in exercise 1.2.6-43.) Using the previous exercise, prove that $(\tfrac12)! = \sqrt{\pi}/2$.
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From Euler's limit formula for the factorial function (Equation 13),
$$ n! = \lim_{m\to\infty} \frac{m^n,m!}{(n+1)(n+2)\cdots(n+m)}. $$
Setting $n=\tfrac12$, we obtain
$$ \left(\frac12\right)! = \lim_{m\to\infty} \frac{\sqrt m,m!} {\left(\frac32\right)\left(\frac52\right)\cdots\left(m+\frac12\right)}. $$
The denominator is
$$ \left(\frac32\right)\left(\frac52\right)\cdots\left(m+\frac12\right) = \prod_{k=1}^{m}\frac{2k+1}{2} = \frac{3\cdot5\cdot7\cdots(2m+1)}{2^m}. $$
Hence
$$ \left(\frac12\right)! = \lim_{m\to\infty} \frac{\sqrt m,2^m m!} {3\cdot5\cdot7\cdots(2m+1)}. $$
Since
$$ 2^m m! = 2\cdot4\cdot6\cdots(2m), $$
this becomes
$$ \left(\frac12\right)! = \lim_{m\to\infty} \sqrt m, \frac{2\cdot4\cdot6\cdots(2m)} {3\cdot5\cdot7\cdots(2m+1)}. $$
Now consider the $m$-th partial Wallis product:
$$ W_m = \prod_{k=1}^{m} \frac{(2k)^2}{(2k-1)(2k+1)}. $$
By Wallis's formula,
$$ \lim_{m\to\infty} W_m=\frac{\pi}{2}. $$
Writing $W_m$ in factored form gives
$$ W_m = \frac{2\cdot4\cdot6\cdots(2m)} {1\cdot3\cdot5\cdots(2m-1)} \cdot \frac{2\cdot4\cdot6\cdots(2m)} {3\cdot5\cdot7\cdots(2m+1)}. $$
The first factor simplifies to
$$ \frac{2\cdot4\cdot6\cdots(2m)} {1\cdot3\cdot5\cdots(2m-1)} = \frac{2\cdot4\cdot6\cdots(2m)} {3\cdot5\cdot7\cdots(2m+1)} \cdot(2m+1), $$
because multiplying the denominator of the second factor by $2m+1$ produces the denominator of the first factor.
Therefore
$$ W_m = (2m+1) \left( \frac{2\cdot4\cdot6\cdots(2m)} {3\cdot5\cdot7\cdots(2m+1)} \right)^2. $$
Hence
$$ \frac{2\cdot4\cdot6\cdots(2m)} {3\cdot5\cdot7\cdots(2m+1)} = \sqrt{\frac{W_m}{2m+1}}. $$
Substituting into the expression for $(\tfrac12)!$,
$$ \left(\frac12\right)! = \lim_{m\to\infty} \sqrt m,\sqrt{\frac{W_m}{2m+1}} = \lim_{m\to\infty} \sqrt{\frac{m}{2m+1}};\sqrt{W_m}. $$
Taking limits,
$$ \left(\frac12\right)! = \sqrt{\lim_{m\to\infty}\frac{m}{2m+1}} ; \sqrt{\lim_{m\to\infty}W_m} = \sqrt{\frac12}, \sqrt{\frac{\pi}{2}} = \frac{\sqrt{\pi}}{2}. $$
Therefore
$$ \boxed{\left(\frac12\right)! = \frac{\sqrt{\pi}}{2}}. $$
$\square$