calculus - Evaluate $lim (n!)^{-1/n ln n}$

Problem :

Evaluate $$\lim_{n\to\infty} \left( \frac{1}{n!}\right)^\frac{1}{n \ln n}$$

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My Attempts:

Suppose
$$\begin{align} &L=\lim_{n\to\infty} \left( \frac{1}{n!}\right)^\frac{1}{n \ln n}\\ &\ln L=\lim_{n\to\infty}\frac{1}{n\ln n} \ln \left(\frac{1}{n!} \right) \\ & =-\lim_{n\to\infty}\frac{\ln n!}{n\ln n}\\ & =-\lim_{n\to\infty}\frac{\ln n + \ln(n-1) + \cdots+\ln1}{n\ln n} \\ & = 0 \\ &\Leftrightarrow L=1 \end{align}$$

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But the answer is not. Where am I wrong?
And how can I solve this without using stirling approximation?

Answer

You can use Riemann integral to handle the limit. In fact,
$$\begin{eqnarray} \frac{\ln(n!)}{n\ln n}&=&\frac{\sum_{k=1}^n\ln k}{n\ln n}=1+\frac1{\ln n} \sum_{k=1}^n\frac1n\ln(\frac{k}{n}). \end{eqnarray}$$
Since
$$\lim_{n\to\infty}\frac1{\ln n}=0,\lim_{n\to\infty}\sum_{k=1}^n\frac1n\ln(\frac{k}{n})=\int_0^1\ln x dx=-1$$
one has

$$\begin{eqnarray} \lim_{n\to\infty}\frac{\ln(n!)}{n\ln n}=\lim_{n\to\infty}\bigg[1+\frac1{\ln n} \sum_{k=1}^n\frac1n\ln(\frac{k}{n})\bigg]=1. \end{eqnarray}$$