Calculating the limit of a sequence using Stirling's approximation

Question

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.

Answer 1

You appear to have made some errors applying Stirling’s approximation. I get

$$\frac{(2n)!}{n!^2}\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{2\pi n\left(\frac{n}e\right)^{2n}}=\frac{2^{2n}}{\sqrt{\pi n}}=\frac{4^n}{\sqrt{\pi n}}\;.$$

Answer 2

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: $$ 2^{2n}\cdot \frac{\sqrt{\pi}}{\pi 4^n} = 2^{2n}\cdot \frac{\sqrt{\pi}}{\pi 2^{2n}} = \frac{\sqrt{\pi}2^{2n}}{\pi 2^{2n}} = \frac {\sqrt \pi}{\pi} = \pi^{1/2} \cdot \pi^{-1} = \pi^{-1/2} = \frac 1 {\sqrt{\pi}} $$