记 a_n:=\left(1+\frac{1}{n}\right)^{n+1},~~~n=1,2,\cdots, 则
\begin{align*} \frac{1}{a_n}&=\frac{1}{\left(1+\dfrac{1}{n}\right)^{n+1}}=\left(1-\frac{1}{n+1}\right)^{n+1}\\ &=\left(1-\frac{1}{n+1}\right)\cdot\left(1-\frac{1}{n+1}\right)\cdots\left(1-\frac{1}{n+1}\right)\cdot 1\\ &\leq \left[\dfrac{\left(1-\dfrac{1}{n+1}\right)+\left(1-\dfrac{1}{n+1}\right)+\cdots+\left(1-\dfrac{1}{n+1}\right)+1}{n+2}\right]^{n+2}\\ &= \left[\dfrac{(n+1)\left(1-\dfrac{1}{n+1}\right)+1}{n+2}\right]^{n+2}\\ &=\frac{1}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}\\ &=\frac{1}{a_{n+1}}, \end{align*}
这就表明 a_n \geq a_{n+1}.
