I try to compute the following limit:

$$\lim_{n\to\infty}\left(\prod_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$$

I'm interested in finding some reasonable ways of solving the limit. I don't find any easy approach. Any hint/suggestion is very welcome.


All the binomial coefficients except the last one are at least $n$, so the $n$th root is at least $n^{\frac{n-1}{n}}$, so the limit is infinity.

  • $\begingroup$ @QiaochuYuan Oops, thanks. $\endgroup$ – Phira Jun 9 '12 at 17:13

Start with the identity $$\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}.$$ Since $k^k/k!\to \infty$, the same is true of its geometric mean: $$\left(\prod_{k=1}^n {k^k\over k!}\right)^{1/n}\to\infty.$$

  • $\begingroup$ an interesting and nice approach. Thanks! $\endgroup$ – user 1357113 Jun 9 '12 at 19:53

As noted by Phira and Byron Schmuland, the product diverges. I find an asymptotic expression for the product for large $n$, Eqn. (3) below.

I have found some inspiration for this solution in @sos440's answer here.

With a little work, one can show that $$\begin{equation*} \log \left(\prod_{k=1}^n{n\choose k}\right)^{1/n} = -\frac{n+1}{n}\log n! + (n+1)\log n + 2 \sum_{j=1}^n \frac{j}{n}\log\frac{j}{n}. \tag{1} \end{equation*}$$ For a derivation of (1), see below. Using Stirling's approximation, and the fact that $\sum_{j=1}^n \frac{j}{n}\log\frac{j}{n} \approx n\int_0^1 x \log x = -n/4$ (the error here is $O(\log(n)/n)$), we get $$\begin{equation*} \log \left(\prod_{k=1}^n{n\choose k}\right)^{1/n} \sim \frac{n}{2}+1 - \frac{1}{2} \log 2\pi n.\tag{2} \end{equation*}$$ Therefore, $$\begin{equation*} \left(\prod_{k=1}^n{n\choose k}\right)^{1/n} \sim \frac{e^{n/2+1}}{\sqrt{2\pi n}}. \tag{3} \end{equation*}$$ Clearly the product diverges. For $n=10$, $100$, and $1000$ the left and right side of (3) agree to $12\%$, $2.0\%$, and $0.28\%$, respectively.

From (3) we get the result $\lim_{n\to\infty} \left(\prod_{k=1}^n{n\choose k}\right)^{1/n^2} = \sqrt{e}$ for free. (Use $\lim_{n\to\infty} x^{1/n} = 1$ for $0<x<1$.)

Derivation of Eqn. (1)

Note that $$\begin{eqnarray*} \log \left(\prod_{k=1}^n{n\choose k}\right)^{1/n} &=& \log \left(\prod_{k=0}^n{n\choose k}\right)^{1/n} \\ &=& \frac{1}{n} \sum_{k=0}^n \log {n\choose k} \\ &=& \frac{1}{n} \sum_{k=0}^n \left(\log n! - \log k! - \log (n-k)! \right)\\ &=& \frac{1}{n} \left((n+1)\log n! - 2 \sum_{k=0}^n\log k! \right). \end{eqnarray*}$$ But $$\begin{eqnarray*} \sum_{k=0}^n\log k! &=& \sum_{k=1}^n\log k! \\ &=& \sum_{k=1}^n \sum_{j=1}^k \log j \\ &=& \sum_{j=1}^n \sum_{k=j}^n \log j \\ &=& \sum_{j=1}^n (n+1-j) \log j \\ &=& (n+1)\sum_{j=1}^n \log j - \sum_{j=1}^n j (\log j -\log n + \log n) \\ &=& (n+1)\log n! - \frac{n(n+1)}{2} \log n - \sum_{j=1}^n j\log\frac{j}{n}. \end{eqnarray*}$$ Eqn. (1) follows immediately.

Some key identities: $$\begin{eqnarray*} \log n! &=& \sum_{k=1}^n \log k \\ \sum_{k=j}^n 1 &=& n+1-j \\ \sum_{k=1}^n k &=& \frac{n(n+1)}{2} \\ \sum_{k=1}^n \sum_{j=1}^k a_{j k} &=& \sum_{j=1}^n \sum_{k=j}^n a_{j k} \end{eqnarray*}$$

  • $\begingroup$ Great answer. I was going to suggest maybe accelerating the $1/n$ would make it converge: interesting result! $\endgroup$ – Pedro Tamaroff Jun 10 '12 at 0:58
  • $\begingroup$ @PeterTamaroff: Thanks, Peter. As you probably know, if we hit the product with another power of $1/n$ we get this problem. Cheers! $\endgroup$ – user26872 Jun 10 '12 at 1:05
  • $\begingroup$ @oenamen: excellent work! Thanks! $\endgroup$ – user 1357113 Jun 10 '12 at 7:12
  • $\begingroup$ @Chris: Thank you for the interesting question! $\endgroup$ – user26872 Jun 10 '12 at 16:00
  • $\begingroup$ @oenamen: your answer was far more interesting than my question, a real lesson to study! Thanks! $\endgroup$ – user 1357113 Jun 10 '12 at 16:01

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