已知数列 $\{a_n\}$ 满足 $a_0=\dfrac 12$，$a_n=a_{n-1}+\dfrac {1}{n^2}a_{n-1}^2$．
1、求证：$\dfrac {1}{a_{n-1}}-\dfrac {1}{a_{n}}<\dfrac {1}{n^2}$．
2、求证：$a_n<n$（$n \in \mathbb N^{\ast}$）．
3、求证：$\dfrac {1}{a_{n}}\leqslant \dfrac {5}{6}+\dfrac {1}{n+1}$（$n \in \mathbb N^{\ast}$）．
解析
1、显然 $\{a_n\}$ 单调递增，根据题意，有\[LHS=\dfrac{1}{a_{n-1}}-\dfrac{1}{a_{n-1}+\dfrac1{n^2}a_{n-1}^2}=\dfrac{\dfrac{1}{n^2}}{1+\dfrac{1}{n^2}a_{n-1}}<\dfrac{1}{n^2},\]命题得证．
2、根据第 $(1)$ 小题的结论，可得\[\sum_{k=1}^n\left(\dfrac1{a_{k-1}}-\dfrac{1}{a_k}\right)<\sum_{k=1}^n\dfrac{1}{k^2}\iff 2-\dfrac{1}{a_n}<\sum_{k=1}^n\dfrac{1}{k^2},\]因此只需要证明\[\sum_{k=1}^n\dfrac{1}{k^2}<2-\dfrac 1n,\]而\[\sum_{k=1}^n\dfrac{1}{k^2}<1+\sum_{k=2}^n\dfrac{1}{(k-1)k}=2-\dfrac 1n,\]因此命题得证．
3、根据第 $(1)$ 和第 $(2)$ 小题的结论，有\[\dfrac{1}{a_{n-1}}-\dfrac{1}{a_n}=\dfrac{\dfrac{1}{n^2}}{1+\dfrac{1}{n^2}a_{n-1}}>\dfrac{\dfrac{1}{n^2}}{1+\dfrac{n-1}{n^2}}=\dfrac{1}{n^2+n-1}>\dfrac 1n-\dfrac1{n+1},\]因此\[\sum_{k=2}^n\left(\dfrac{1}{a_{k-1}}-\dfrac1{a_k}\right)\geqslant \dfrac 12-\dfrac1{n+1}\implies \dfrac{1}{a_1}-\dfrac{1}{a_n}\geqslant \dfrac 12-\dfrac1{n+1},\]结合 $a_1=\dfrac 34$ 整理即得命题成立．