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.
Best Answer
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 $$ \frac{{2 \cdot 2}}{{1 \cdot 3}} \cdots \frac{{(2n)(2n)}}{{(2n - 1)(2n + 1)}} \to \frac{\pi }{2},\quad \frac{{2n + 1}}{n} \to 2. $$