<s>[INST] Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$ More fully,
if $n\ge 2$ is an integer
and
$0 < x < 1$,
prove that 
$(1+\frac{x}{n})^n < \frac1{1-x}$.
In addition,
if $c > 1$ and
$0 < x \le \frac{c-1}{c}$,
prove that
$(1+\frac{x}{n})^n < 1+cx$.
Proofs by elementary means
(no calculus or limits)
are particularly sought.
As an example of the utility of this result,
set $x = \frac12$.
Then
this shows that
$2 > (1+\frac1{2n})^n$
or
$2^{1/n} > 1+\frac1{2n}$
.
This is an example
of what I call a
contra-Bernoulli inequality
(CBI)
which gives an upper bound to
$(1+y)^n$
as opposed to Bernoulli's inequality,
which gives a lower bound to
$(1+y)^n$
of $1+ny$.
Note that
any CBI of the form
$(1+y)^n < 1+c y$
for $n \ge 2$
requires that $y$ is bounded,
since
$(1+y)^n > 1+y^n$
so
$1+cy > 1+y^n$
or
$cy > y^n$
or
$y < c^{1/(n-1)}$.
 [/INST] \begin{align}
\left ( 1+\frac{x}{n} \right )^n 
&=1+n\cdot \frac{x}{n}+\frac{n(n-1)}{2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}\\
&<1+x+x^2+\cdots\\
&=\frac{1}{1-x}
\end{align}
for $n\geq 2$, $0<x<1$.
</s>