Is $\lim_{n \to \infty} n((1-\frac{1}{n})^n - \frac{1}{e}) = 0$?

Question

Intuitively I think this statement is true, but I am unable to proof it. Can someone help me? If possible, I would like bound $\vert(1-\frac{1}{n})^n - \frac{1}{\mathrm{e}}\vert$ (or even $\vert{(1+\frac{1}{n})^n - \mathrm{e}}\vert$) because I think this is a good think to know in general.

Answer 1

By the Binomial Theorem, we have $$ \begin{align} \left(1-\frac1{2n}\right)^{2n}-\left(1-\frac1n\right)^n &=\left(1-\frac1n+\frac1{4n^2}\right)^n-\left(1-\frac1n\right)^n\tag{1a}\\ &=n\left(1-\frac1n\right)^{n-1}\frac1{4n^2}+O\left(\frac1{n^2}\right)\tag{1b}\\ &=\frac1{4ne}+O\left(\frac1{n^2}\right)\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: expand $\left(1-\frac1{2n}\right)^2$
$\text{(1b)}$: use the first term and remainder from the Binomial Theorem
$\text{(1c)}$: as shown in this answer, $\left(1-\frac1n\right)^{n-1}$ is decreasing and $\left(1-\frac1n\right)^n$ is increasing,
$\phantom{\text{(1c):}}$ so $\left(1-\frac1n\right)^{n-1}=\frac1e+O\!\left(\frac1n\right)$

Repeating $(1)$ for $2^kn$ and summing gives a telescoping series that yields $$ \frac1e-\left(1-\frac1n\right)^n=\frac1{2ne}+O\left(\frac1{n^2}\right)\tag2 $$ Applying $(2)$ to the limit yields $$ \lim_{n\to\infty}n\left(\left(1-\frac1n\right)^n-\frac1e\right)=-\frac1{2e}\tag3 $$

Answer 2

$$\left(1-\frac{1}{n}\right)^n\sim \frac{1}{e}-\frac{1}{2 e n};\;n\to\infty$$ $$\lim_{n \to \infty}n\left(\left(1-\frac{1}{n}\right)^n - \frac{1}{e}\right) = \lim_{n \to \infty}n\left(\left(\frac{1}{e}-\frac{1}{2 e n}-\frac{1}{e}\right)\right)=$$ $$=-\frac{1}{2e}$$