I have the following limit:
$$
\lim_{n \to \infty} \frac{(2n)!\sqrt{n}}{n!^24^n}
$$
Now, in order to get somewhere further in calculating this limit, I used Stirling's approximation that led me to the following:
$$
\lim_{n \to \infty} \left[\frac{(2n)!}{(2n)^{2n}e^{-2n}\sqrt{4\pi n}} \times \frac{(2n)^{2n}ne^{-2n}\sqrt{4\pi}}{n!^24^n} \right]= \lim_{n \to \infty} \frac{(2n)^{2n}ne^{-2n}\sqrt{4\pi}}{n!^24^n}
$$
But am having trouble getting anywhere further on simplifying this. What is a direction I can take from this point on? Or should I have done a different approach perhaps? Any insights/hints are much welcome.
Best Answer
You appear to have made some errors applying Stirling’s approximation. I get
$$\frac{(2n)!}{n!^2}\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{2\pi n\left(\frac{n}e\right)^{2n}}=\frac{2^{2n}}{\sqrt{\pi n}}=\frac{4^n}{\sqrt{\pi n}}\;.$$