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$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$

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.

user91500
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Guy
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  • You could use Stirling's approximation. – Pedro Mar 06 '14 at 06:53
  • It's not legitimate because $n \log n$ is just the dominant term in an asymptotic expansion, and as the $n\log n$-sized terms cancel out in subsequent calculations (leaving only $O(n)$ terms) the conclusion is unwarranted without looking at the lower-order terms. – Erick Wong Mar 06 '14 at 06:54
  • @ErickWong I suspected as much. What are my alternatives here? – Guy Mar 06 '14 at 06:55
  • @PedroTamaroff I am not familiar with Stirling's approximation. Care to post it as an answer then? – Guy Mar 06 '14 at 06:55
  • @Sabyasachi As Pedro suggested, Stirling gives you the more accurate estimate $\log(n!) = n \log n - n + O(\log n)$, which is plenty of precision for your purposes (you only need $o(n)$ rather than $O(\log n)$, as it transpires). – Erick Wong Mar 06 '14 at 06:58
  • try to show that $\frac{4^n}{n+1}\le {2n \choose n} \le 4^n$, take log and use squeeze theorem ! – r9m Mar 06 '14 at 06:59
  • See Stirling's approximation and the central binomial coefficient. – Lucian Mar 06 '14 at 07:01
  • @Lucian thanks. – Guy Mar 06 '14 at 07:07
  • @Hawk I think it would be better to mark as a duplicate of the question your question is marked as a duplicate of. Better to make the links straightforward. – Guy Apr 04 '14 at 06:50
  • @Sabyasachi The reasons I linked my question are: 1)I could not see the link to the original question due to some server problems. 2)There are more answers on my question than the original one. – Hawk Apr 04 '14 at 06:52
  • @Hawk ah fair enough. – Guy Apr 04 '14 at 07:19

3 Answers3

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One easy way is to observe in the binomial expansion of $(1+1)^{2n} = \sum\limits_{k=0}^{2n} \binom{2n}{k}$, the term $\binom{2n}{n}$ is the largest one among all the $2n+1$ terms of the form $\binom{2n}{k}$. This gives us a bound

$$\frac{2^{2n}}{2n+1} \le \binom{2n}{n} \le 2^{2n} \quad\implies\quad \log 2 - \frac{\log(2n+1)}{2n} \le \frac{1}{2n}\log\binom{2n}{n} \le \log 2$$ which implies the desired limit is $\log 2$.

achille hui
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Alternative computation:

$$\begin{align*} a_n &= \frac{1}{2n} \log \binom{2n}{n} = \frac{1}{2n} \log \frac{(2n)!}{(n!)^2} \\ &= \frac{1}{2n} \sum_{k=1}^n \log \frac{n+k}{k} = \frac{1}{2n} \sum_{k=1}^n \log \Bigl( 1 + \frac{1}{k/n} \Bigr). \end{align*}$$ Therefore, as $n \to \infty$, we get a Riemann sum: $$\begin{align*} \lim_{n \to \infty} a_n &= \frac{1}{2} \int_{x=0}^1 \log\Bigl(1 + \frac{1}{x}\Bigr) \, dx = \frac{1}{2} \int_{x=0}^1 \log (x+1) - \log x \, dx \\ &= \log 2. \end{align*}$$

In fact, we can easily extend this solution to evaluate limits of the form $$a_n(z) = \frac{1}{zn} \log \binom{zn}{n}$$ for integers $z > 1$: $$\lim_{n \to \infty} a_n(z) = \log \frac{z}{(z-1)^{1-1/z}}.$$ (The case $z = 1$ is trivial.) What about limits of sequences of the form $$a_n(w,z) = \frac{1}{\binom{z}{w}n} \log \binom{zn}{wn}, \quad z > w \ge 1?$$

heropup
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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).

Claude Leibovici
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