Help with Limit: $\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}$

Question

Help with Limit: $$\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}$$

EDIT: Posted a similar question just yesterday but bad formating from my part dropped a term, and lead me to strange places. Honestly, I am just plain stuck, I have been hitting my head against it for $2$ days straight. I know the solution should be $2^{-1/2}$ but...

Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy.

Answer 1

Hint: Use Stirling's approximation $$n!\approx n^n e^{-n}\sqrt{2\pi n}$$ for large $n$.

Answer 2

Hint. Use Stirling approximation: $$n!\sim\sqrt{2\pi n} \cdot\frac{n^n}{e^n}.$$ which implies that $$(2n)!\sim2\sqrt{\pi n} \cdot\frac{(2n)^{2n}}{e^{2n}}.$$

Answer 3

Using Stirling's approximation, we have that \begin{align} \frac{(2n)!e^nn^n}{n!(2n)^{2n}} \approx \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n} \frac{e^nn^n}{(2n)^{2n}} \frac{1}{\sqrt{2\pi n}}\left( \frac{e}{n}\right)^n = \sqrt{2}. \end{align}