Deducing Identity from Stirling Formula

Question

Stirling's Formula states that $\Gamma(z+1) \sim \sqrt{2 \pi z} (\frac{z}{\mathbb{e}})^{z}$ as $z \rightarrow \infty$. I need to prove the following identity using Stirling's formula:

$$ (2n)! \sim \frac{2^{2n} (n!)^{2}}{\sqrt{\pi n}} $$ as $n \rightarrow \infty$.

In Stirling's formula I plugged in $z = 2n$ to get:

$$ \Gamma(2n+1) = 2n\Gamma(2n) \sim \sqrt{4n \pi} (\frac{2n}{\mathbb{e}})^{2n} $$ where the first equality follows from the functional equation for the gamma function. Simplifying a little bit more I achieve:

$$ (2n)! \sim \frac{\sqrt{\pi n} 2^{2n} n^{2n-1}}{\mathbb{e}^{2n}} $$

But, I don't know how to prove the identity from here. Can someone give me some hints on how to move on from this step?

Answer 1

You're on the right track:

$$(2n)! \sim \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n} as \ n \to \infty$$ $$(2n)! \sim 2\sqrt{\pi n}\left(\frac{2n}{e}\right)^{2n} =\frac{2\sqrt{\pi n}\sqrt{\pi n}}{\sqrt{\pi n}}\left(\frac{2n}{e}\right)^{2n} $$ $$(2n)! \sim \frac{2^{2n}}{\sqrt{n\pi}}2n\pi \left(\frac{n}{e}\right)^{2n} $$

Observe from definition that $(n!)^2 \sim 2n\pi \left(\dfrac{n}{e}\right)^{2n} $

I think you can finish off from here

Answer 2

Using Stirling formula, we get $$ \frac{(2n)!}{(n!)^2}\underset{n\rightarrow +\infty}{\sim}\frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}{2\pi n\left(\frac{n}{e}\right)^{2n}}=\frac{2^{2n}}{\sqrt{\pi n}} $$