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
$y < c^{1/(n-1)}$.

Here's my elementary proof by induction that
$(1+x/n)^n < 1/(1-x)$.
Let $y = x/n$.
This becomes
$(1+y)^n < 1/(1-ny)$
when $0 < y < 1/n$.
For $n=1$,
this is
$1+y < 1/(1-y)$
or
$1-y^2 < 1$
which is true.
Suppose it it true for $n$,
so that
$(1+y)^n < \frac1{1-ny}$.
Then,
if $0 < y < \frac1{n+1}$,
$\begin{align}
(1-(n+1)y)(1+y)^{n+1}
&=(1-(n+1)y)(1+y)(1+y)^{n}\\
&<\frac{(1-(n+1)y)(1+y)}{1-ny}\\
&=\frac{1-ny-(n+1)y^2}{1-ny}\\
&=\frac{1-ny}{1-ny}\\
&=1
\end{align}
$
so
$(1+y)^{n+1} < \frac1{1-(n+1)y}$.
