Upper bound for $(1-1/x)^x$

Question

I remember the bound $$\left(1-\frac1x\right)^x\leq e^{-1}$$ but I can't recall under which condition it holds, or how to prove it. Does it hold for all $x>0$?

Answer 1

Starting from $e^x \geq 1+x$ for all $x \in \mathbb{R}$:

For all $x \in \mathbb{R}$ $$ e^{-x} \geq 1-x. $$ For all $x \neq 0$ $$ e^{-1/x} \geq 1-\frac{1}{x}. $$ And, since $t \mapsto t^x$ is increasing on $[0,\infty)$, for $x \geq 1$ $$ e^{-1} \geq \left(1-\frac{1}{x}\right)^x. $$

Answer 2

Bernoulli's Inequality says that $$ 1+nt\le\left(1+t\right)^n\tag{1} $$ for all $t\ge-1$ and $n\ge1$. Thus, if $y\ge x\ge1$, $$ \begin{align} 1-\frac1x &=1-\frac yx\frac1y\\ &\le\left(1-\frac1y\right)^{y/x}\\ \left(1-\frac1x\right)^x &\le\left(1-\frac1y\right)^y\tag{2} \end{align} $$ We can send $y\to\infty$ to get $$ \left(1-\frac1x\right)^x \le\frac1e\tag{3} $$ for $x\ge1$.