$\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



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.

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.

  • $\begingroup$ Thanks Claude Leibovici , would you like to explain me how can we get using Atriling approximation. $\endgroup$ – juantheron Oct 27 '13 at 10:07
  • $\begingroup$ I forgot to mention a simpler formula for large values of n. It is Lon[n!] = n Log[n] - n $\endgroup$ – Claude Leibovici Oct 27 '13 at 13:31
  • 1
    $\begingroup$ @Neal. Thanks for editing my post. $\endgroup$ – Claude Leibovici Oct 27 '13 at 13:32

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