TITLE: Evaluate the limit $\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right)$
QUESTION [5 upvotes]: Evaluate
$$\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right).$$
This sequence looks extremely horrible and it makes me crazy. How can we evaluate this?

REPLY [5 votes]: Stirling's asymptotic formula (without any series enhancement)
$$
n!\sim\frac{n^n}{e^n}\sqrt{2\pi n}\tag{1}
$$
put into Landau notation is
$$
n!=\frac{n^n}{e^n}\sqrt{2\pi n}\,(1+o(1))\tag{2}
$$
which gives
$$
\begin{align}
(n!)^{1/n}
&=\frac ne(2\pi n)^{\frac1{2n}}(1+o(1/n))\\
&=\frac ne\left(1+\frac{\log(2\pi n)}{2n}\right)(1+o(1/n))\\
\frac{(n!)^{1/n}}{n}
&=\frac1e\left(1+\frac{\log(2\pi n)}{2n}\right)(1+o(1/n))\\
\frac{(n!)^{1/n}}{n}-\frac1e
&=\frac1e\frac{\log(2\pi n)}{2n}+o(1/n)\\
\frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right)
&=\frac1{2e}+\frac{\log(2\pi)}{2e\log(n)}+o(1/\log(n))\tag{3}
\end{align}
$$
Taking the limit as $n\to\infty$, $(3)$ yields
$$
\lim_{n\to\infty}\frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right)
=\frac1{2e}\tag{4}
$$
which is the result sought.
However, $(3)$ gives more
$$
\lim_{n\to\infty}\log(n)\left(\frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right)-\frac1{2e}\right)=\frac{\log(2\pi)}{2e}\tag{5}
$$