Question

$\rm \lim_{ n \to \infty } \frac{ \sqrt[ {n}^{2} ]{1! \times 2! \times 3! \times 4! \times \dots \times n!} }{ \sqrt{n} }$​

Answer 1

Trying this again. Take a logarithm and expand it:

$\ln\left(\dfrac{\sqrt[n^2]{1! \times 2! \times \cdots \times n!}}{\sqrt n}\right) = \dfrac{\ln(1!\times2!\times\cdots\times n!)}{n^2} - \dfrac{\ln(n)}2$

$=\dfrac{\ln(1) + (\ln(1)+\ln(2)) + \cdots + (\ln(1)+\ln(2)+\cdots+\ln(n))}{n^2} - \dfrac{\ln(n)}2$

$=\displaystyle \sum_{k=1}^n \frac{(n-k+1) \ln(k)}{n^2} - \frac{\ln(n)}2$

$=\displaystyle \frac1n \sum_{k=1}^n \left(1 - \frac{k-1}n\right) \ln(k) - \frac{\ln(n)}2$

With some rewriting, this is equivalent to

$=\displaystyle \frac1n \sum_{k=1}^n \left(1 - \frac{k-1}n\right) (\ln(k) - \ln(n) + \ln(n)) - \frac{\ln(n)}2$

$=\displaystyle \frac1n \sum_{k=1}^n \left(1 - \frac{k-1}n\right) \ln\left(\dfrac kn\right) + \frac{\ln(n)}n \sum_{k=1}^n \left(1 - \frac{k-1}n\right) - \frac{\ln(n)}2$

As n → ∞, the first sum converges to a definite integral,

$\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \left(1 - \frac{k-1}n\right) \ln\left(\dfrac kn\right) = \int_0^1 (1-x) \ln(x) \, dx = -\frac34$

while the second sum is

$\displaystyle \sum_{k=1}^n \left(1 - \frac{k-1}n\right) = n - \dfrac1n\times\dfrac{n(n+1)}2 - \dfrac nn = \dfrac{n+1}2$

so that the other two terms converge to zero,

$\displaystyle \lim_{n\to\infty} \left(\frac{\ln(n)}n \sum_{k=1}^n \left(1 - \frac{k-1}n\right) - \frac{\ln(n)}2\right) = \lim_{n\to\infty} \left(\dfrac{(n+1)\ln(n)}{2n} - \frac{\ln(n)}2\right) = 0$

Therefore

$\displaystyle \lim_{n\to\infty} \frac{\sqrt[n^2]{1!\times2!\times\cdots\times n!}}{\sqrt n} = \boxed{e^{-3/4}}$