<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] For the first part, we only need the binomial theorem:
$$\left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k} = \sum_{k=0}^n \frac{\prod_{j=1}^k(n+1-j)}{k!n^k}x^k \leqslant \sum_{k=0}^n \frac{x^k}{k!} \leqslant \sum_{k=0}^n x^k < \frac{1}{1-x}.$$
For the second, we observe that
$$\frac{1}{1-x} \leqslant 1 + cx$$
for $0 < x < \frac{c-1}{c}$, since
$$(1-x)(1+cx) -1 = (c-1)x - cx^2$$
has zeros in $x = 0$ and $x = \frac{c-1}{c}$, and is positive between the zeros, since the coefficient of $x^2$ is negative.
</s>