<s>[INST] 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?
 [/INST] 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 $n \geqslant 1$:
$$
  \frac{x_{n+1}}{x_n} \leqslant \frac{1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3}}{ 1+\frac{n}{(n+1)^2} } < 1
$$
</s>