MT【131】$a_{n+1}\cdot a_n=\dfrac 1n$

已知数列$\{a_n\}$满足$a_1=1$，$a_{n+1}\cdot a_n=\dfrac 1n$($n\in\mathbb N^*$)．
(1) 求证：$\dfrac{a_{n+2}}{n}=\dfrac{a_n}{n+1}$；
(2) 求证：$2\left(\sqrt{n+1}-1\right)\leqslant \dfrac{1}{2a_3}+\dfrac{1}{3a_4}+\cdots+\dfrac{1}{(n+1)a_{n+2}}\leqslant n$．

解:(1) 根据题意，有 $$\begin{split} \dfrac{a_{n+2}}{n}=&\dfrac{\dfrac{1}{n+1}\cdot \dfrac{1}{a_{n+1}}}{n}\=&\dfrac{na_n}{n(n+1)}=\dfrac{a_n}{n+1}.\end{split}$$
(2) 根据第$(1)$小题的结论，有$\dfrac{1}{2a_3}+\dfrac{1}{3a_4}+\cdots+\dfrac{1}{(n+1)a_{n+2}}=a_2+a_3+\cdots+a_{n+1}.$
右边不等式　根据第$(1)$小题的结论，有$\dfrac{a_{n+2}}{a_n}=\dfrac{n}{n+1}<1,$于是数列的奇子列和偶子列均单调递减，结合$a_1=a_2=1$，可得$a_n\leqslant 1,n\in\mathbb N^*$，于是右边不等式得证．
左边不等式　由于$\begin{split}\dfrac{1}{a_n\cdot a_{n+1}}&=n,\dfrac{1}{a_{n+1}\cdot a_{n+2}}&=n+1,\end{split}$于是$\dfrac{1}{a_{n+1}}\left(\dfrac{1}{a_{n+2}}-\dfrac{1}{a_n}\right)=1,$从而$a_{n+1}=\dfrac{1}{a_{n+2}}-\dfrac{1}{a_n}.$ 因此 $$\begin{split} a_2+a_3+\cdots+a_{n+1}=&\dfrac{1}{a_{n+2}}+\dfrac{1}{a_{n+1}}-\dfrac{1}{a_1}-\dfrac{1}{a_2}\geqslant &\dfrac{2}{\sqrt{a_{n+1}a_{n+2}}}-2\=&2\left(\sqrt{n+1}-1\right),\end{split}$$ 于是左边不等式得证．
综上所述，原命题得证．
评:这类题目最后往往要用基本不等式把$a_{n+1}\cdot a_n=\dfrac 1n$这个条件用进去.下面给一道类似的练习:
设$a_1=1,a_n\cdot a_{n+1}=n,n\in N^+$,求证:$\sum\limits_{k=1}^{n}{\dfrac{1}{a_k}}\ge2\sqrt{n}-1$