Question

`\rm \lim_( n \to \infty ) \frac{ \sqrt[ {n}^(2) ]{1! * 2! * 3! * 4! * \dots * n!} }{ √(n) } \\`​

Answer 1

Trying this again. Take a logarithm and expand it:

`\ln\left(\frac{\sqrt[n^2]{1! * 2! * \cdots * n!}}{\sqrt n}\right) = (\ln(1!*2!*\cdots* n!))/(n^2) - \frac{\ln(n)}2`

`=(\ln(1) + (\ln(1)+\ln(2)) + \cdots + (\ln(1)+\ln(2)+\cdots+\ln(n)))/(n^2) - \frac{\ln(n)}2`

`=\displaystyle \sum_(k=1)^n ((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(\frac 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(\frac 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 - \frac1n*\frac{n(n+1)}2 - \frac nn = \frac{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(((n+1)\ln(n))/(2n) - \frac{\ln(n)}2\right) = 0`

Therefore

`\displaystyle \lim_(n\to\infty) \frac{\sqrt[n^2]{1!*2!*\cdots* n!}}{\sqrt n} = \boxed{e^(-3/4)}`