TITLE: $\lim_{n \to \infty} \frac{1}{2n}\log\binom{2n}{n}$
QUESTION [5 upvotes]: Please help me to solve this problem. I can find almost no clue regarding the log part. I tried to break the $\binom{2n}{n}$ part, but in vague...will the breaking help me in anyway?

REPLY [2 votes]: There are $2^{2n}=4^n$ subsets of a set of size $2n$, each of which has between $0$ and $2n$ elements.  Of these, ${2n}\choose{n}$ have exactly $n$ elements, and ${{2n}\choose{n}} > {{2n}\choose{k}}$ for any $k\neq n$.  Therefore,
$$
\frac{4^n}{2n+1} < {{2n}\choose{n}} \le 4^n,
$$
so
$$
n\log 4-\log(2n+1) < \log{{2n}\choose{n}} \le n \log 4,
$$
and
$$
\log 2-\frac{\log(2n+1)}{2n} < \frac{1}{2n}\log{{2n}\choose{n}} \le \log 2.
$$
We conclude that
$$
\lim_{n\rightarrow\infty}\frac{1}{2n}\log{{2n}\choose{n}} = \log 2.
$$