calculus - Calculate $lim_{n to infty} ln frac{n!^{frac{1}{n}}}{n}$

How can I calculate the following limit? I was thinking of applying Cesaro's theorem, but I'm getting nowhere. What should I do?

$$\lim_{n \to \infty} \ln \frac{n!^{\frac{1}{n}}}{n}$$

Answer

First let's write $$\ln \frac{n!^{\frac{1}{n}}}{n} = \frac{1}{n} (\ln n! - n\ln n) = \frac{a_n}{b_n}$$ where $a_n = \ln n! - n\ln n$ and $b_n = n$.

Then

$$\begin{align*} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} & = \quad \frac{\ln (n+1)! - (n+1) \ln(n+1) - (\ln n! - n\ln n)}{1} \\ & = \quad \ln(n+1) - (n+1)\ln(n+1) + n \ln n \\ & = \quad -\ln(1 + 1/n)^n \\ & \longrightarrow -1 \ \text{ as } n \to \infty \end{align*}$$

Hence by the Cesàro theorem (a.k.a. Stolz-Cesàro theorem)

$$\lim_{n\to\infty} \ln \frac{n!^{\frac{1}{n}}}{n} = \lim_{n\to\infty} \frac{a_n}{b_n}= \lim_{n\to\infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = -1$$