Evaluate the limit $\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right)$ 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?
 A: 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
$$
\lim_{n\to\infty}\log(n)\left(\frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right)-\frac1{2e}\right)=\frac{\log(2\pi)}{2e}\tag{5}
$$
A: Use Stirling's approximation:
$$n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}=\left(\frac{n}{e}\right)^{n}e^{\frac12 \ln 2\pi n}$$
It transforms your limit into
$$\lim_{n\rightarrow\infty}\frac{n}{\ln n}\frac{e^{\frac{\ln n}{2n}}-1}{e}=\lim_{x\rightarrow 0}\frac{e^{\frac{x}{2}}-1}{ex}=\frac{1}{2e}.$$
When obtaining the first expression, we neglect $\frac{\ln 2\pi}{2n}$ in the exponential (it is $o\left(\frac{\ln n}{2n}\right)$), and then we make the change of variables $x=\frac{\ln n}{n}$.
A: There is a gap in @O.L.'s solution as I explained in a comment below that solution. I don't have time now to provide a complete solution but the OP should be aware of this. I see that the wiki page provides a formula with an error term, but the $O(1/n)$ occurs inside an argument and is not stated as a theorem, nor is there a source for this. It is probably true but note also that we need the error estimate for $\frac{\sqrt[n]{n!}}{n}$  rather than for $n!$ itself.
