TITLE: Evaluating $\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$
QUESTION [6 upvotes]: $$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$
Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write, 
$$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$
$$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$
$$\log\left(2n\right) - \log\left(n\right) = \log(2)$$
Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.

REPLY [2 votes]: As suggested in several comments, the simplest form of Stirling approximation for $n!$ is $$\sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$
(http://en.wikipedia.org/wiki/Stirling%27s_approximation) 
Take the logarithms and develop first
$$\log\left(2n!\right) - 2\log\left(n!\right)$$ which results to be $(2 n+1) \log (2)$. The remaining is obvious.  
If I may, remember Stirling approximation; it is extremlely useful almost any time we need to play with factorials. If required, higher orders can be used (as given in Wikipedia).