Wallis formula is $$\prod ^{\infty }_{n=1}\dfrac{\left( 2n\right) ^{2}}{\left( 2n-1\right) \left( 2n+1\right) }=\dfrac{\pi }{2}$$
$$\lim _{n\rightarrow \infty }\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !} \tag{1}$$ I want to prove Stirling's formula$(\lim _{n\rightarrow \infty }\dfrac{n!}{\sqrt{n}}\left( \dfrac{e}{n}\right) ^{n}=\sqrt{2\pi } )$ using Wallis formula and $(1)$.
I know method not using $(1)$ but when I use $(1)$, I don't know how to do it.
Let $$c=\lim_{n\to\infty}\frac{n!}{\sqrt{2}n^{n+1/2}e^{-n}}$$
Rewrite Wallis formula as the following: $$\lim_{n\to\infty}\frac{2^{4n}(n!)^4}{((2n)!)^2(2n+1)}=\frac{\pi}{2}$$ Note that $$\frac{2^{4n}(n!)^4}{((2n)!)^2(2n+1)}\sim\frac{2^{4n}(c\sqrt{2}n^{n+1/2}e^{-n})^4}{(c\sqrt{2}(2n)^{2n+1/2}e^{-2n})^2(2n+1)}$$ Take the limit on both sides of the relation, and apply L'Hospital, $$\frac{\pi}{2}=\lim_{n\to\infty}\frac{n^2}{n(2n+1)}=\frac{c^2}{2}$$ which gives $c=\sqrt{\pi}$.
My notation is a little bit different from you, but the general method should work.
