每日一题[1806]左右放缩

已知数列 $\{a_n\}$ 满足 $a_1=\dfrac 13$，$a_{n+1}=a_n+\dfrac{a_n^2}{n^2}$（$n\in \mathbb N^{\ast}$）．

1、证明：对一切 $n\in \mathbb N^{\ast}$，有 $a_n<a_{n+1}<1$．

2、证明：对一切 $n\in \mathbb N^{\ast}$，有 $a_n>\dfrac 12 -\dfrac 1{4n}$．

解析    

1、显然 $a_n>0$，根据题意，有 $$a_{n+1}-a_n=\dfrac{a_n^2}{n^2}>0,$$ 因此 $a_n<a_{n+1}$（$n\in\mathbb N^{\ast}$）．进而 $$\dfrac{1}{a_k}-\dfrac{1}{a_{k+1}}=\dfrac{a_{k+1}-a_k}{a_ka_{k+1}}=\dfrac{a_k}{a_{k+1}}\cdot \dfrac{1}{k^2}=\dfrac{1}{k^2},$$ 因此 $$\dfrac1{a_1}-\dfrac{1}{a_{n+1}}<\sum_{k=1}^n\dfrac{1}{k^2}<1+\sum_{k=2}^n\left(\dfrac 1{k-1}-\dfrac1k\right)=2,$$ 从而 $a_{n+1}<1$（$n\in\mathbb N^{\ast}$）．

2、根据第 $(1)$ 小题的结果，有 $$\dfrac1{a_k}-\dfrac{1}{a_{k+1}}=\dfrac{a_k}{a_{k+1}}\cdot \dfrac{1}{k^2}=\dfrac{1}{1+\dfrac{a_k}{k^2}}\cdot \dfrac{1}{k^2}>\dfrac{1}{k^2+1},$$ 于是 $$\dfrac{1}{a_1}-\dfrac{1}{a_n}\geqslant \sum_{k=1}^{n-1}\dfrac{1}{k^2+1}\geqslant \sum_{k=1}^{n-1}\dfrac1{k^2+k}=\sum_{k=1}^{n-1}\left(\dfrac 1k-\dfrac1{k+1}\right)=1-\dfrac1n,$$ 因此 $$a_n\geqslant \dfrac{n}{2n+1}>\dfrac {2n-1}{4n}=\dfrac 12-\dfrac1{4n}.$$