Prove that $ \lim_{n\to\infty}\frac{\left(n!\right)^{2}2^{2n}}{\left(2n\right)!\sqrt{n}}=\sqrt{\pi} $
What i want to do is to reach the form $ \sqrt{2\left(\prod_{k=1}^{n}\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)\right)} $
and then I want to use Wallis product, which will bring me to the result I seek. But I cant see how to reach this form. Any ideas would help.
We have \begin{align*} \frac{{n!^2 2^{2n} }}{{(2n)!\sqrt n }} & = \frac{{2^n n!2^n n!}}{{(2n)!\sqrt n }} = \frac{{(2 \cdot 4 \cdots (2n))(2 \cdot 4 \cdots (2n))}}{{1 \cdot 2 \cdots (2n)\sqrt n }} \\ & = \frac{{2 \cdot 4 \cdots (2n)}}{{1 \cdot 3 \cdots (2n - 1)\sqrt n }} = \sqrt {\frac{{2 \cdot 2}}{{1 \cdot 3}} \cdots \frac{{(2n)(2n)}}{{(2n - 1)(2n + 1)}}\frac{{2n + 1}}{n}} . \end{align*} Now note that $$ \frac{{2 \cdot 2}}{{1 \cdot 3}} \cdots \frac{{(2n)(2n)}}{{(2n - 1)(2n + 1)}} \to \frac{\pi }{2},\quad \frac{{2n + 1}}{n} \to 2. $$
By considering the odd and even numbers separately, it's easy to see that $$(2n)!=2^n(n!)\cdot(1\cdot3\cdot5\cdots2n-1)$$ so that $$\frac{2^{2n}(n!)^2}{(2n)!}=\frac{2^nn!}{1\cdot3\cdot5\cdots2n-1}=\prod_{k=1}^n\frac{2k}{2k-1}$$
Now compare $\prod_{k=1}^n\frac{2k}{2k-1}$ to $\prod_{k=1}^n\frac{2k}{2k+1}$ and you'll find the answer.
Note that splitting between even and odd $k$ in two different ways yields $\frac{n!^4}{(2n)!^2}=(2n+1)\prod_{k=1}^n{\frac{k^2}{(2k-1)(2k)}}\prod_{k=1}^n{\frac{k^2}{(2k)(2k+1)}}=(2n+1)2^{-4n-1}W_n^2$, when $W_n$ is your product.
Notice that : $$ \frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}=\prod_{k=1}^{n}{\frac{2k-1}{2k}}=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}{\sin^{2n}{x}\,\mathrm{d}x}=\frac{2}{\pi}W_{2n} $$
Where $ W_{n} $ is Wallis integral. And a well-known and easy to prove property of Wallis integral is that : $$ W_{n}\underset{n\to +\infty}{\sim}{\sqrt{\frac{\pi}{2n}}} $$
Thus : $$ \frac{\left(2n\right)!\sqrt{n}}{2^{2n}\left(n!\right)^{2}}=\frac{\sqrt{2}}{\pi}\sqrt{2n}W_{2n}\underset{n\to +\infty}{\longrightarrow}\frac{\sqrt{2}}{\pi}\times\sqrt{\frac{\pi}{2}}=\frac{1}{\sqrt{\pi}} $$
Hence : $$ \frac{2^{2n}\left(n!\right)^{2}}{\left(2n\right)!\sqrt{n}}\underset{n\to +\infty}{\longrightarrow}\sqrt{\pi} $$
Notice: $$\frac{(n!)^22^{2n}}{(2n)!\sqrt{n}}=\frac{1}{\binom{2n}n\frac1{4^n}\sqrt{n}}$$
According to this post, via Wallis:
$$\begin{aligned}\frac1{\sqrt{\pi\left(n+\frac12\right)}}&\leqslant\binom{2n}n{1\over 4^n}\leqslant {1\over\sqrt{\pi n}}\\\\\iff\sqrt{\pi\left( n+\frac12\right)}&\geqslant\frac1{\binom{2n}n\frac1{4^n}}\geqslant\sqrt{\pi n}\Big/:\sqrt{n}\end{aligned}$$ $$$$ $$\lim_{n\to\infty}\sqrt{\pi+\frac{\pi}{2n}}\geqslant\lim_{n\to\infty}\frac1{\binom{2n}{n}\frac1{4^n}\sqrt{n}}\geqslant\lim_{n\to\infty}\sqrt{\pi}$$
