TITLE: Upper bound for $(1-1/x)^x$
QUESTION [5 upvotes]: 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 $x>0$?

REPLY [3 votes]: 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 $x\ge1$.