Evaluation of $L=\lim_{ n \to \infty } \Big(\frac{n^{n+1}\cdot e}{(n+1)^n}-n\Big)$ Find the value of following limit:
$$L=\lim_{ n \to \infty } \Big(\frac{n^{n+1}\cdot e}{(n+1)^n}-n\Big)$$
How to solve this indeterminate form? After taking do we have apply L-Hopital's rule or is there any better approach?

Hint. One may use some standard Taylor series expansions writing, as $n \to \infty$,
$$
\begin{align}
\frac{n^{n+1}.e}{(n+1)^n}&=n\cdot e\cdot e^{\large -n\log(1+1/n)}
\\\\&=n\cdot e\cdot e^{\large -n\left(\frac1n-\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)}
\\\\&=n\cdot e\cdot e^{\large -1+1/(2n)+O(1/n^2)}
\\\\&=n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)
\\\\&=n+\frac{1}{2}+O\left(\frac{1}{n} \right)
\end{align}
$$ giving, as $n \to \infty$,
$$
\frac{n^{n+1}.e}{(n+1)^n}-n=\frac{1}{2}+O\left(\frac{1}{n} \right) \to L=\frac12.
$$
