infinite limit with $n^2$ root of $n!$

Question

If $\displaystyle p=\lim_{n\rightarrow \infty}\frac{\sqrt[n^2]{1!\cdot 2!\cdot 3!\cdots n!}}{n^q}.$ Then finding value of ordered pair $(p,q)$, Where $p>0 ,q\neq 0$.

What I try:

$\displaystyle (1!\cdot 2!\cdot 3!\cdots n!)^{\frac{1}{n^2}}=e^{\ln(1!\cdot 2!\cdots n!)^{\frac{1}{n^2}}}$

$\displaystyle =e^{\frac{1}{n^2}\ln(1!\cdot 2!\cdot 3!\cdots n!)}=e^{\frac{1}{n^2}(n\ln(1)+(n-1)\ln(2)+(n-2)\ln(3)+\cdots +1\ln(n)}$

I have seems that it must be in Riemann sum , but could not understand how do i Convert it, please have a look , Thanks

Presumably, you want $p>0,$ or there are infinitely many examples $q$ with $p=0.$ — Thomas Andrews, Oct 04 '23 at 18:22
Computing $\lim_{n \to \infty} \left[\left(\prod_{i=1}^{n}i!\right)^{1\over n^{2}} (n^{x})\right] $ if exists for certain $x\in\mathbb R$ — user10354138, Oct 04 '23 at 18:28
You can use AM/GM to show $$1!2!3!\cdots n!\leq \left(\frac{n(n+1)(n+2)}6\right)^{n(n+1)/2}$$ So you definitely want $q\leq 3.$ — Thomas Andrews, Oct 04 '23 at 18:28

Claude Leibovici · Answer 1 · 2023-10-05T09:53:10.527

$$1!\cdot2!\cdots n! = \prod_{k=1}^n k! =G(n+2)$$ which is the Barnes G-function. Its logarithm can be expanded almost like for the factorial function

$$\log (G(n+2))=n^2 \log \left(\frac{\sqrt{n}}{e^{3/4}}\right)+n \log \left(\frac{\sqrt{2 \pi } n}{e}\right)+$$ $$\log \left(\frac{\sqrt[12]{e} \sqrt{2 \pi } n^{5/12}}{A}\right)+\frac{1}{12 n}+O\left(\frac{1}{n^2}\right)$$ where $A$ is Glaisher constant.

So, $$\log(p)=\frac{1}{4} ((2-4 q) \log (n)-3)+\frac 1 {2n}\log \left(\frac{2 \pi n^2}{e^2}\right)+O\left(\frac{1}{n^2}\right)$$

So, to have a finite limit, we need $q=\frac 12$ in order to cancel the $\log(n)$ and then $\log(p)=-\frac 34$ so $p=e^{-\frac 34}$.

Beside the limit, we then have the approximation $$p \sim e^{-\frac 34}\,\exp\Big(\frac 1{2n} \log \left(\frac{2 \pi n^2}{e^2}\right) \Big)$$ whose relative error is smaller than $0.10$% for $n \geq 49$ and smaller than $0.01$% for $n \geq 171$.

score 3 · Accepted Answer · answered Oct 04 '23 at 18:37

Rewrite the product as

$$1!\cdot2!\cdots n! = \prod_{k=1}^n k! \sim \prod_{k=1}^n\sqrt{2\pi k}\left(\frac{k}{e}\right)^k = \left(2\pi\right)^{\frac{n}{2}}\sqrt{n!}\exp\left(\sum_{k=1}^n k\log k - k\right)$$

by using Stirling's approximation. Then applying the root we get

$$\sqrt[n^2]{1!\cdot2!\cdots n!} \sim \left(2\pi\right)^{\frac{1}{2n}}\sqrt[n^2]{n!}\exp\left(\frac{1}{n}\sum_{k=1}^n \frac{k}{n}\log k - \frac{k}{n}\right)$$

$$\sim\left(2\pi\right)^{\frac{1}{2n}+\frac{1}{2n^2}}\left(\frac{n}{e}\right)^{\frac{1}{n}}\exp\left(\frac{1}{n}\sum_{k=1}^n \frac{k}{n}\log \frac{k}{n} - \frac{k}{n}\right)\exp\left(\frac{\log n}{n}\sum_{k=1}^n\frac{k}{n}\right)$$

again by Stirling's approximation. The terms outside the exponentials tend to $1$ in the limit. The first exponential is a straightforward Riemann sum

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \frac{k}{n}\log \frac{k}{n} - \frac{k}{n} = \int_0^1 x\log x - x \:dx = -\frac{3}{4}$$

The second exponential, however, diverges as it tends to the expression $n^R$, with $R$ being the Riemann sum

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{k}{n} = \int_0^1 x\:dx = \frac{1}{2}$$

This means choosing $\boxed{q = \frac{1}{2}}$ gives us the limit

$$\lim_{n\to\infty}\frac{\sqrt[n^2]{1!\cdot2!\cdots n!} }{\sqrt{n}} = 1\cdot 1 \cdot e^{-\frac{3}{4}}\cdot\left(\lim_{n\to\infty}n^{\frac{1}{n}\sum_{k=1}^n\frac{k}{n}-\frac{1}{2}}\right) = e^{-\frac{3}{4}}\cdot \lim_{n\to\infty}n^{\frac{1}{2n}} = \boxed{e^{-\frac{3}{4}}=p}$$

infinite limit with $n^2$ root of $n!$

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