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
$y < c^{1/(n-1)}$.
 A: \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 $n\geq 2$, $0<x<1$.
A: 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.
A: In response to "more fully...", but not necessarily "in addition...":
Note that $\lim_{n\rightarrow \infty} (1+(x/n))^n = e^x$ by definition and that, for any finite $n$, $(1+(x/n))^n \leq e^x$ for $0 \leq x \leq 1$ (can check this on graph, or use Taylor series).
Now consider the end points of the interval: $e^{0+} < 1/(1-(0+))$ and $e^1 < 1/(1-1)$. Since both $e^x$ and $1/(1-x)$ are strictly monotonic in $]0,1]$ this means that $e^x < 1/(1-x)$ throughout this interval (no crossings). A fortiori, because of the previous part, $(1+(x/n))^n < 1/(1-x)$.
A: 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
$(1+y)^{n+1} < \frac1{1-(n+1)y}$.
A: Because
$$ (1 + \frac{x}{n})^n < e^x $$
We can rewrite the expression as
$$ e^x < \frac{1}{1-x} $$
Replacing the terms with their infinite sumation
$$ \sum_{k=0}^\infty \frac{x^k}{k!} < \sum_{k=0}^\infty x^k $$
After trivial manipulations we get
$$ \sum_{k=2}^\infty \frac{x^k (1 - k!)}{k!} < 0 $$
The inequality is trivially proved
