Calculating the limit of a sequence using Stirling's approximation 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.
 A: You appear to have made some errors applying Stirling’s approximation. I get
$$\frac{(2n)!}{n!^2}\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{2\pi n\left(\frac{n}e\right)^{2n}}=\frac{2^{2n}}{\sqrt{\pi n}}=\frac{4^n}{\sqrt{\pi n}}\;.$$
A: Take:
$$
lim_{n \to \infty} \frac{(2n)!\sqrt{n}}{n!^24^n} = lim_{n \to \infty} \frac{(2n)!}{n!^2} \cdot \frac{\sqrt{n}}{4^n}
$$
Recall Stirling's Approximation:
$$n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$$
Whereby:
$$(2n)! \sim \sqrt{2\pi 2n}\left(\frac {2n} e\right)^{2n} = \sqrt{4\pi n}\left(\frac {2n} e\right)^{2n} = \sqrt{4}\sqrt{\pi n}\left(\frac {2n} e\right)^{2n} = 2\sqrt{\pi n}\left(\frac {2n} e\right)^{2n}$$
And:
$$n!^2 \sim \left(\sqrt{2\pi n}\left(\frac n e\right)^n\right)^2 = \left(\sqrt{2\pi n}\right)^2 \cdot \left(\left(\frac n e\right)^n\right)^2
=2\pi n \cdot \left(\frac n e\right)^{2n}
$$
So:
$$
lim_{n \to \infty} \frac{(2n)!}{n!^2} \cdot \frac{\sqrt{n}}{4^n}
 = 
\frac{2\sqrt{\pi n}\left(\frac {2n} e\right)^{2n}}
{2\pi n \cdot \left(\frac n e\right)^{2n}}
\cdot 
\frac{\sqrt{n}}{4^n}
=
\frac{\sqrt{\pi n}\left(\frac {2n} e\right)^{2n}}
{\pi n \cdot \left(\frac n e\right)^{2n}}
\cdot 
\frac{\sqrt{n}}{4^n}
=\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi n}\sqrt{n}}{\pi n 4^n}
$$
Continuing:
$$
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi n}\sqrt{n}}{\pi n 4^n}
= \frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}\sqrt{n}\sqrt{n}}{\pi n 4^n}
 = 
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}n}{\pi n 4^n}
 = 
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
$$
Continuing:
$$
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
= \left( \frac{\frac{2n}e}{\frac n e} \right)^{2n}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
=\left( \frac{2n}e\cdot\frac e n \right)^{2n}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
= 2^{2n}\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
$$
Now we note that $4=2^2$, so $4^n= (2^2)^n = 2^{2n}$, whereby:
$$
2^{2n}\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
= 
2^{2n}\cdot
\frac{\sqrt{\pi}}{\pi 2^{2n}}
= \frac{\sqrt{\pi}2^{2n}}{\pi 2^{2n}}
= \frac {\sqrt \pi}{\pi}
= \pi^{1/2} \cdot \pi^{-1} = \pi^{-1/2} = \frac 1 {\sqrt{\pi}}
$$
