Formulation of the Wallis product for $\pi/2$ using the $\Gamma$ function I am trying to write the Wallis Product (WP) using the $\Gamma$ function, thereby hoping for either a new identity or a simple derivation of the formula, but I get a result I am not sure its correct.
This is the WP
$$ \frac{\pi}{2} = \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots $$
Now 
$$ \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot \dots\ (2n)(2n)}{1\cdot1\cdot3\cdot3\cdot5\cdot5\cdot\dots\ (2n-1)(2n-1)(2n+1)}\cdots $$
$$=\frac{1\cdot2\cdot1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot2\cdot3\cdot2\cdot \hspace{12pt} \dots\hspace{12pt}n\cdot2\hspace{12pt}\cdot\hspace{12pt} n\cdot2\hspace{24pt}}{\frac{1}{2}\cdot2\cdot\frac{1}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\dots\ (n-\frac{1}{2})\cdot2\cdot(n-\frac{1}{2})\cdot2\cdot(2n+1)}\cdots $$
$$=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot \hspace{12pt} \dots\hspace{12pt}n\hspace{12pt}\cdot\hspace{12pt} n\hspace{24pt}}{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot(n-\frac{1}{2})\cdot(2n+1)}\cdots $$
$$=\big(\frac{1\cdot2\cdot3\cdot\ \dots\ n\hspace{24pt}}{\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot\sqrt{2n+1}}\cdots\big)^2  $$
$$ = \lim_{n\rightarrow\infty}\;\;\Bigg(\frac{\Gamma{(n+1)}}{\frac{\Gamma{(n+\frac{1}{2})}}{\sqrt{\pi}}}\Bigg)^2\frac{1}{(2n+1)}$$
$$=\lim_{n\rightarrow\infty}\;\; \frac{\pi}{2n+1}\Big(\frac{\Gamma{(n+1)}}{\Gamma{(n+\frac{1}{2})}}\Big)^2$$
$$= \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2.$$
Question 1) Is this a correct fully equivalent reformulation of rhs of WP? 
When using WP we would get:
$$\frac{\pi}{2} = \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2,$$
$$\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2 = 1,$$ which implies
$$\frac{\Gamma(n)}{\Gamma(n-\frac{1}{2})} \overset{n\rightarrow \infty}{\longrightarrow} \sqrt{n}. \tag{1}$$ 
Alternatively if (1) is known otherwise (e.g. from the Sterling approximation (Question 2) Is it possible to derive (1) from the Sterling approximation or from any other elementary considerations?)) one could derive the WP easily from that. 
Question 3) Is this conclusion correct?
 A: Let's say $\rho(n) = \Gamma(n)/\Gamma(n-1/2).$ Then, yes, $\rho(n) \approx \sqrt n$ for large $n$. More precisely, $\rho(n)/\sqrt n \to 1$ as $n \to \infty$.
You should expect this. After all, $$\rho(n)\cdot \rho(n-1/2) = \frac{\Gamma(n)}{\Gamma(n-1/2)} \cdot \frac{\Gamma(n-1/2)}{\Gamma(n-1)} = \frac{\Gamma(n)}{\Gamma(n-1)} \approx n$$
for large $n$, so if $\rho(n) \approx \rho(n-1/2)$ for large $n$, then we must have $\rho(n) \approx \sqrt n$ for large n.
Indeed, this observation is one way to pick out the Gamma function as the "best" generalization of factorial to non-integer arguments. For any fixed $t$ between $0$ and $1$ this sort of heuristic argument suggests that you want $$\Gamma(n+t) \approx n^t \cdot \Gamma(n) $$ for large integer $n$. Combined with the functional equation $\Gamma(x+1) = x \cdot \Gamma(x)$, that means for large $n$ you want
$$(n-1+t)\cdot\dots\cdot(1+t) \cdot \Gamma(1+t) \approx n^t \cdot (n-1)\cdot \dots \cdot 2\cdot 1.$$
And it turns out that taking the definition
$$\Gamma(1+t) = \lim_{n\to \infty} n^t \frac{(n-1)\cdot \dots \cdot 2\cdot 1}{(n-1+t)\cdot\dots\cdot(2+t)\cdot(1+t)}$$
gives you the same Gamma function that we know and love.
This is closely related to the observation that the Gamma function is the unique log-convex extension of the factorial function which satisfies the right functional equation.
A: By Stirling,
$$\frac{\Gamma(n)}{\Gamma(n-\frac12)}\approx\frac{\sqrt{2\pi n}\left(\dfrac ne\right)^n}{\sqrt{2\pi (n-\frac12)}\left(\dfrac {n-\frac12}e\right)^{n-\frac12}}.$$
After simplifications,
$$\frac1{\left(1-\dfrac1{2n}\right)^n}\left(\frac {n-\frac12}e\right)^{1/2}\to e^{1/2}\sqrt {\frac ne}.$$
