设 $a_1=2$，$a_{n+1}=a_n^2-a_n+1$．证明：$\displaystyle 1-\dfrac 1{2018^{2018}}<\sum\limits_{n=1}^{2018}{\dfrac 1{a_n}}<1$．
解析 根据题意，有\[\dfrac{1}{a_n}=\dfrac{1}{a_n-1}-\dfrac{1}{a_{n+1}-1}\implies \sum_{n=1}^{2018}\dfrac{1}{a_n}=1-\dfrac{1}{a_{2019}-1}.\]而\[a_{n+1}-a_n=(a_n-1)^2>0\implies a_{2018}>a_1=2.\]又\[a_n=\dfrac{a_{n+1}-1}{a_n-1}\implies a_1a_2\cdots a_n=a_{n+1}-1,\]于是\[\dfrac{1}{a_{n+1}-1}=\dfrac{1}{a_1a_2\cdot a_n}\leqslant \left(\dfrac 1n\sum_{k=1}^n\dfrac{1}{a_k}\right)^n<\dfrac 1{n^n}\implies a_{n+1}-1>n^2,\]因此 $a_{2018}>2018^{2018}$，命题得证．