Find the ordered pair $(\alpha,\beta)$ with non-infinite $\beta \ne 0$ such that $$\lim_{n\to\infty} \frac{\sqrt[n^2]{1!2!\cdots n!}}{n^\alpha} = \beta$$

My approach:

$$\ln (1!2!\cdots n!) = (n)\ln 1 + (n-1)\ln 2 + \cdots + (2)\ln (n-1) + \ln(n) \\ \begin{align} = n\ln\left(\frac{1}{n}\right) + (n-1)\ln\left(\frac{2}{n}\right) + \cdots + \ln\left(\frac{n}{n}\right) + \frac{(n)(n+1)}{2} \ln (n)\end{align}$$

Then $$\ln(\sqrt[n^2]{1!2!\cdots n!}) = \frac{1}{n^2} \ln(1!2!\cdots n!) = \frac{n+1}{n} \cdot \ln n + \frac{1}{n} \left[\sum_{m=1}^n \left(\frac{n+1-m}{n} \cdot \ln \frac{m}{n}\right)\right]$$

And that's about as far as I got. Any ideas about proceeding with this method or perhaps even with a different method?



  • $\begingroup$ Sorry, I don't know how to do that :\ Would you mind working it out? $\endgroup$ Jan 29, 2015 at 7:00
  • $\begingroup$ @AisforAmbition see my answer for help $\endgroup$
    – RE60K
    Jan 29, 2015 at 7:01
  • $\begingroup$ @r9m can you please just work out step-by-step the limit of a reimann sum? I'm having trouble seeing how to convert the summation function into the form of $f(\frac{m}{n})$. What I have so far is $f(\frac{m}{n}) = (1+\frac{1}{n} - \frac{m}{n})\ln \frac{m}{n} \implies f(x) = (1+\frac{x}{m} - x)\ln x.$ How do I get rid of the $\frac{x}{m}$?? $\endgroup$ Jan 29, 2015 at 7:19
  • $\begingroup$ $1/n(1+1/n-k/n)\ln(k/n)=1/n\ln(k/n)+1/n^2\ln(k/n)-k/n\ln(k/n)$ where first term becomes $\int\ln xdx$ , second is $\frac1n\int\ln xdx$ and third is $-\int x\ln xdx$ $\endgroup$
    – RE60K
    Jan 29, 2015 at 7:24
  • $\begingroup$ But that's what I'm having trouble with. If $x = \frac{k}{n},$ then shouldn't $\frac{1}{n} = \frac{x}{k},$ not $x?$ Instead, following your calculations, $\frac{1}{n} = x,$ but $\frac{k}{n}$ also equals $x$. Can you explain how that's possible? $\endgroup$ Jan 29, 2015 at 7:27

1 Answer 1


As you showed: $$\ln\left(\sqrt[n^2]{\prod_{k=1}^nk!}\right)=\frac{n^2}2\ln\left(\prod_{k=1}^nk!\right)=\frac1{n^2}\left(\sum_{k=1}^n(n+1-k)\ln(k/n) +\frac{n(n+1)}2\ln n\right)\\S=\sum_{k=1}^n\frac1n\left(1+\frac1n-\frac kn\right)\ln(k/n) +\frac{(1+1/n)}2\ln n$$ See if you can use as suggested by r9m, the relation: $$\lim_{n\to\infty}\sum_{k=a}^b\frac1nf(k/n)=\int_{\lim_{n\to\infty}a/n}^{\lim_{n\to\infty}b/n}\quad f(x){\rm d}x$$ When $n\to\infty$: $$S_{\infty}\sim\int_0^1\ln x{\rm d}x+0-\int_0^1x\ln x{\rm d}x+\frac12\ln n+0=\frac12\ln n-\frac34$$ Note that $\lim_{n\to\infty}\frac{\ln n}n=0$ and $\int(1-x)\ln x=\frac14 x (x-2 (x-2) \ln x-4)+\text{constant}$

So: $$\beta\sim\frac{e^{-3/4}\sqrt n}{n^{\alpha}}$$ So $\alpha=1/2$ and $\beta=e^{-3/4}$. See related question here

Interesting note

The Barnes G-function "G" is: $$G(n+2)=\prod_{k=1}^n(k!)$$ and our limit is: $$\lim_{n\to\infty}\frac{[G(n+2)]^{1/n^2}}{\sqrt n}=e^{-3/4}$$


Wolfram Check

  • $\begingroup$ @r9m here $\lim_{n\to\infty}1/n=0,\lim_{n\to\infty}n/n=1$ $\endgroup$
    – RE60K
    Jan 29, 2015 at 7:21
  • 1
    $\begingroup$ @r9m do you really know about R-sums? $\endgroup$
    – RE60K
    Jan 29, 2015 at 7:25

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