Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}$

Notice $\displaystyle \sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n}$ and $\displaystyle \binom{2n}{n} \ge \binom{2n}{k}$ for all $0 \le k \le 2n$, we have

$$ \frac{2^{2n}}{2n+1}\le \binom{2n}{n} \le 2^{2n} \quad\implies\quad \log 2 - \frac{\log{(2n+1)}}{2n} \le \frac{1}{2n} \log \binom{2n}{n} \le \log 2$$ Since $\quad\displaystyle \lim_{n\to\infty} \frac{\log(2n+1)}{2n} = 0\quad$, we get $\quad\displaystyle \lim_{n\to\infty} \frac{1}{2n}\log\binom{2n}{n} = \log 2$.

Beside the elegant demonstration given by achille hui, I think that the simplest manner to solve this problem is to use Stirling approximation.

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At the first order, Stirling's approximation is $n! = \sqrt{2 \pi n} (n/e)^n$. It is very good. Have a look at http://en.wikipedia.org/wiki/Stirling%27s_approximation. They have a very good page for that. For sure, you need to express the binomial coefficient as the ratio of factorials. Try that and you will be amazed to see how simple becomes your problem.