Evaluate $\lim (n!)^{-1/n \ln n}$ Problem :
Evaluate $$\lim_{n\to\infty} \left( \frac{1}{n!}\right)^\frac{1}{n \ln n}$$

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}

But the answer is not. Where am I wrong?
And how can I solve this without using stirling approximation?
 A: 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}
A: When you have  two strictly increasing sequences such that $a_n,b_n\rightarrow\infty$, then
$$ \lim_{n\rightarrow \infty} \frac{a_n}{b_n} = \lim_{n\rightarrow \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
assuming the second limit exists. It's called Stolz–Cesàro theorem, but you can also think about it as a discrete version of l'Hospital's rule.
From that for $a_n = \ln(n!)$, $b_n = n\ln n$ you have
\begin{align} -\ln L = \lim_{n\rightarrow \infty} \frac{\ln(n!)}{n\ln n} &= \lim_{n\rightarrow \infty} \frac{\ln((n+1)!)-\ln(n!)}{(n+1)\ln(n+1)-n\ln n} =\\
&= \lim_{n\rightarrow \infty} \frac{\ln(n+1)}{n\ln(1+\frac{1}{n})+\ln(n+1)} =\\
&= \lim_{n\rightarrow \infty} \frac{1}{\frac{n}{\ln(n+1)}\ln(1+\frac{1}{n})+1} =\\
&=  \frac{1}{\lim_{n\rightarrow \infty}\Big(\frac{n}{\ln(n+1)}\ln(1+\frac{1}{n})\Big)+1}\end{align}
Since $$ \lim_{n\rightarrow \infty} n\ln(1+\frac{1}{n}) = 1 $$
then $$ \lim_{n\rightarrow\infty}\frac{n}{\ln(n+1)}\ln(1+\frac{1}{n}) = 0$$
and $$ -\ln L = 1$$
$$ L = e^{-1}$$
A: You are wrong in the part $\lim_{n\to\infty}\frac{\ln n + \ln(n-1) + \cdots+\ln1}{n\ln n} = 0$. It's easy to see this limit is greater then $1 - \frac{1}{m} - \varepsilon$ for any $m$ and $\varepsilon > 0$:
$\frac{\ln n + \ln(n - 1) + \ldots + \ln 1}{n \ln n} > \frac{n\cdot\frac{m - 1}{m} \ln \frac{n}{m}}{n \ln n} = \frac{m - 1}{m}\cdot(1 - \frac{\ln m}{\ln n}) > 1 - \frac{1}{m} - \frac{\ln m}{\ln n}$.
So the sum will eventually be arbitrary close to $1$. And as it is always less than $1$, the limit is $1$.
(actually, the same line can give you Stirling's formula up to $\sqrt{2 \pi n}$ term)
A: Indeed, @mathworker21 mentions the use of the sterling approximation. Informally, going back to your last line 
\begin{align*}
-\lim\limits_{n \rightarrow \infty}\frac{\ln(n!)}{n \ln(n)}
\end{align*}
by the stirling approximation we have
\begin{align*}
\ln(L) &= -\lim\limits_{n \rightarrow \infty}\frac{n \ln(n) - n}{n \ln(n)}\\\\
&= -\lim\limits_{n \rightarrow \infty}1 - \frac{1}{\ln(n)}\\\\
&= -1.\\
\end{align*}
\begin{align*}
\end{align*}
\begin{align*}
\therefore L = e^{-1} = \frac{1}{e}
\end{align*}
