how to prove that $(1 + \frac{1}{n})^{n+1}$ is decreasing? Please, help me to prove that $$x_n=\left(1+\frac{1}{n}\right)^{n+1}$$decreases.
I know I must to prove that that $$\frac{x_n}{x_{n+1}}> 1$$
What to do next?
 A: Write out 
$$
   \frac{x_{n+1}}{x_n} = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( \frac{1+\frac{2}{n}}{\left(1+\frac{1}{n} \right)^2} \right)^n = 
\frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{\frac{1}{n^2}}{\left(1+\frac{1}{n} \right)^2} \right)^n = 
\frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{1}{\left(n+1 \right)^2} \right)^n = \left(1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3} \right) \left( 1 - \frac{1}{\left(n+1 \right)^2} \right)^n
$$
we now have to use $(1-x)^n \leqslant \frac{1}{1+n x}$ for $0<x<1$ and $n \geqslant 1$:
$$
  \frac{x_{n+1}}{x_n} \leqslant \frac{1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3}}{ 1+\frac{n}{(n+1)^2} } < 1
$$
A: Let $f(x) = (1+\frac{1}{x})^{x+1}$. In the following, we show that $f(x)$ is decreasing for $x>0$. We have $f(x) = \exp[(x+1)\log(1+\frac{1}{x})]$ and it is thus sufficient to show that $g(x) = (x+1)\log(1+\frac{1}{x})$ is decreasing. Indeed, $$g'(x) = \log\left(1+\frac{1}{x}\right) - \frac{1}{x} < \frac{1}{x} - \frac{1}{x} = 0,$$
as $\log(1+y)<y,\,\forall y> 0$. 
A: This can be shown using AM $\ge$ GM or using Bernoulli's ineqaulity.
1) Using AM $\ge$ GM
Take one copy of $1$, and $n$ copies of $1 - \frac{1}{n}$, ($n \gt 1$) to get
$$ \frac{1 + n(1 - \frac{1}{n})}{n+1} \gt \left(1-\frac{1}{n}\right)^{n/(n+1)}$$
which simplies to
$$ \left(\frac{n}{n+1}\right)^{n+1} \gt \left(\frac{n-1}{n}\right)^n$$
Taking the reciprocals, yields
$$ \left(1 + \frac{1}{n}\right)^{n+1} \lt \left(1 + \frac{1}{n-1}\right)^n $$
2) Using (generalized) Bernoulli's inequality
$$(1 + x)^r \lt 1 + rx $$
 for the case when $x \gt -1$, $x \neq 0$, and $0 \lt r \lt 1 $
Here we set $x = -\frac{1}{n}$, $r = \frac{n}{n+1}$ and obtain the result easily.

For a similar problem and multiple proofs of that (including AM,GM and bernoulli's inequality), see: Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $
