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I have been trying to figure out the following limit:

$$\lim_{n\to\infty} \sqrt[n(n+1)]{\prod_{r=0}^n {n\choose r}}=\mathscr L$$

It’s quite natural to take the natural log of both sides: $$\ln \mathscr L = \lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}}{n(n+1)} \\ =\lim_{n\to\infty} \frac{(n+1)\ln(n!) - 2\sum_{r=0}^n \ln(r!)}{n(n+1)}\\ =\lim_{n\to\infty} \frac{(n+1)\ln(n!) - 2\ln(\color{green}{n! \times (n-1)! \times (n-2)! \dots \times 2!\times 1!})}{n(n+1)}$$ I believe the green term is called the superfactorial of $n$, but I barely know anything about it. The last thing I tried was to write $$n! \times (n-1)! \times (n-2)! \dots 2!\times 1! =n^1 \times (n-1)^2 \times (n-2)^3 \dots \times 2^{n-1} \times 1^n = \prod_{r=1}^n r^{n+1-r}$$ but that doesn’t help. Any ideas on how to evaluate this limit?

Vishu
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  • I don't know if this will help, but: $$\left(\prod_{r=0}^n \frac{n^r}{r^r}\right)^\frac{1}{n(n+1)} \le \left(\prod_{r=0}^n {n \choose r}\right)^\frac{1}{n(n+1)} \le \left(\prod_{r=0}^n \frac{n^r}{r!}\right)^\frac{1}{n(n+1)} \ n^{1/2}\left(\prod_{r=0}^n \frac{1}{r^r}\right)^\frac{1}{n(n+1)} \le \left(\prod_{r=0}^n {n \choose r}\right)^\frac{1}{n(n+1)} \le n^{1/2}\left(\prod_{r=0}^n \frac{1}{r!}\right)^\frac{1}{n(n+1)}$$ – VIVID Dec 23 '20 at 15:45

3 Answers3

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In fact, by the Stolz thorem, one has \begin{eqnarray} \ln \mathscr L &=& \lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}}{n(n+1)}\\ &=&\lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}-\sum_{r=0}^{n-1} \ln {n-1\choose r}}{n(n+1)-(n-1)n}\\ &=&\lim_{n\to\infty} \frac{\sum_{r=0}^{n-1} \ln {n\choose r}-\sum_{r=0}^{n-1} \ln {n-1\choose r}}{2n}\\ &=&\lim_{n\to\infty} \frac{\sum_{r=0}^{n-1} \ln \frac{{n\choose r}}{{n-1\choose r}}}{2n}\\ &=&\lim_{n\to\infty} \frac{\sum_{r=0}^{n-1} \ln \frac{n}{n-r}}{2n}\\ &=&\lim_{n\to\infty} \frac{\ln \frac{n^n}{n!}}{2n}\\ &=&\lim_{n\to\infty} \frac{n\ln n-\ln n!}{2n}\\ &=&\lim_{n\to\infty} \frac{n\ln n-(n\ln n-n+O(\ln n))}{2n}\\ &=&\frac12. \end{eqnarray} Here the Stirling approximation $$ \ln n!=n\ln n-n+O(\ln n) $$ is used. So $$ \mathscr L=\sqrt{e}. $$

xpaul
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We have

$$\sum_{k=1}^n k \ln(n+1-k) = \sum_{t = n+1-k, 1\le t \le n} (n+1-t) \ln t = S_1 - S_2$$ where $$ S_1 = \sum_{t=1}^n (n+1) \ln t = (n+1) \ln (n!) $$ may be evaluated from Stirling's formula and $$S_2 = \sum_{t=1}^n t \ln t.$$ Hence it's sufficient to evaluate $S_2$. We have $$\frac{S_2}{n} = \sum_{t=1}^n \frac{t}n (\ln \frac{t}n + \ln n) = \sum_{t=1}^n \frac{t}n \ln \frac{t}n + \frac{\ln n}{n} \sum_{t=1}^n t = n S_3 + \frac{\ln n}{n} \frac{n(n+1)}{2},$$ where $$S_3 = \frac1n \sum_{t=1}^n \frac{t}n \ln \frac{t}n \to \int_{0}^1 x \ln x dx = - \frac14.$$

Botnakov N.
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The series expanision of the hyperfactorial is $$\log (H(n))=n^2 \left(\frac{\log (n)}{2}-\frac{1}{4}\right)+\frac{1}{2} n \log (n)+\log \left(A \sqrt[12]{n}\right)+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right)$$ where $A$ is Glaisher constant.

Use Stirling also.

Claude Leibovici
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