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

$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} =$

$\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdot............\cdot (n+n)}{(1)\cdot (2)\cdot (3).........\cdot (n)}\right)$

$\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+..............+\ln \left(\frac{n+n}{n}\right)\right\}$

$\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$

Now Using Reinman Sum

$\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx = \ln (2)$

My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test

If yes then please explain here

Thanks

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 $\quad\displaystyle \lim_{n\to\infty} \frac{1}{2n}\log\binom{2n}{n} = \log 2$.
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.