TITLE: Strange inequality about $e^{-x}$
QUESTION [5 upvotes]: Consider $\displaystyle e^{-x}-\left(1-\frac{x}{n} \right)^{n}$ (x>0).
Can we say that this difference less than $\displaystyle \frac{x^{2}e^{-x}}{n}$? 
I've tried to use Taylor formula and estimating $\displaystyle \frac{1}{m!} - \frac{1}{n^{m}}\binom{n}{m}$. But don't have any result. Any idea?

REPLY [4 votes]: For another approach to the inequality, since $1 - y \leqslant e^{-y}$ and $1+y \leqslant e^{y} $ for $0 \leqslant y < 1$ we have for $x < n$,
$$1 - x/n \leqslant e^{-x/n} \implies (1 - x/n)^n \leqslant e^{-x}\\ 1 + x/n \leqslant e^{x/n} \implies (1 + x/n)^n \leqslant e^x$$
and
$$e^{-x} - (1 - x/n)^n = e^{-x}[ 1 - e^{x}(1- x/n)^n] \leqslant e^{-x}[ 1 - (1 +x/n)^n(1 - x/n)^n] = e^{-x}[ 1 - (1 -x^2/n^2)^n] \\ \leqslant\frac{e^{-x}x^2}{n} .$$
The last step is obtained with Bernoulli's inequality.