题目 已知数列 $\{a_n\}$ 满足 $a_1=1$，$a_{n+1}\cdot a_n=\dfrac 1n$（$n\in\mathbb N^{\ast}$）． 问题1 求证：$\dfrac{a_{n+2}}{n}=\dfrac{a_n}{n+1}$； 答案1 略 解析1 根据题意，有\[\dfrac{a_{n+2}}{n}=\dfrac{\dfrac{1}{n+1}\cdot \dfrac{1}{a_{n+1}}}{n}=\dfrac{na_n}{n(n+1)}=\dfrac{a_n}{n+1}.\] 备注1 问题2 求证：$2\left(\sqrt{n+1}-1\right)\leqslant \dfrac{1}{2a_3}+\dfrac{1}{3a_4}+\cdots+\dfrac{1}{(n+1)a_{n+2}}\leqslant n$． 答案2 略 解析2 根据第 $(1)$ 小题的结论，有\[\dfrac{1}{2a_3}+\dfrac{1}{3a_4}+\cdots+\dfrac{1}{(n+1)a_{n+2}}=a_2+a_3+\cdots+a_{n+1}.\]<sol>右边不等式</sol>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 $a_n\leqslant 1,n\in\mathbb N^{\ast}$，于是右边不等式得证．<br/><sol>左边不等式</sol>由于\[\begin{aligned}\dfrac{1}{a_n\cdot a_{n+1}}&=n,\\ \dfrac{1}{a_{n+1}\cdot a_{n+2}}&=n+1,\end{aligned}\]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\[a_2+a_3+\cdots+a_{n+1}=\dfrac{1}{a_{n+2}}+\dfrac{1}{a_{n+1}}-\dfrac{1}{a_1}-\dfrac{1}{a_2}\geqslant \dfrac{2}{\sqrt{a_{n+1}a_{n+2}}}-2=2\left(\sqrt{n+1}-1\right),\]于是左边不等式得证．<br/>综上所述，原命题得证． 备注2 每日一题[1025]数列递推与求和．