每日一题[1626]反向迭代

已知数列 $\{a_n\}$满足$a_{n+1}^2-a_{n+1}=a_n$（$n\in\mathbb N^{\ast}$）．若数列 $\{a_n\}$ 各项单调递增，则首项 $a_1$ 的取值范围是_______；当 $a_1=\dfrac 23$ 时，记 $b_n=\dfrac{(-1)^{n+1}}{a_n-1}$，若 $k<b_1+b_2+\cdots+b_{2019}<k+1$，则整数 $k=$ _______．

答案     $\left[-\dfrac 14,2\right)$；$-5$．

解析     根据题意，有

$$a_{n+1}=\dfrac{1+\sqrt{1+4a_n}}{2},$$ 对应的迭代函数函数 $f(x)=x^2-x$（$x\geqslant \dfrac 12$）的反函数 $f^{-1}(x)$，由迭代函数法可得首项 $a_1$ 的取值范围是 $\left[-\dfrac 14,2\right)$．

题中递推公式可变形为

$$\dfrac{1}{a_{n+1}-1}=\dfrac{1}{a_n}+\dfrac{1}{a_{n+1}},$$ 于是 $$\begin{split}\sum_{k=1}^{2019}b_k&=\dfrac{1}{a_1-1}-\dfrac{1}{a_2-1}+\dfrac{1}{a_3-1}-\dfrac{1}{a_4-1}+\cdots+\dfrac{1}{a_{2019}-1}\\ &=\dfrac{1}{a_1-1}-\dfrac1{a_1}-\dfrac{1}{a_2}+\dfrac{1}{a_2}+\dfrac1{a_3}-\dfrac1{a_3}-\dfrac1{a_4}+\dfrac1{a_4}+\dfrac{1}{a_5}-\cdots+\dfrac{1}{a_{2018}}+\dfrac{1}{a_{2019}}\\ &=\dfrac{1}{a_1-1}-\dfrac{1}{a_1}+\dfrac{1}{a_{2019}}\\ &=-\dfrac 92+\dfrac{1}{a_{2019}},\end{split}$$ 而 $$\dfrac{2-a_{n+1}}{2-a_n}=\dfrac{1}{1+a_{n+1}}<\dfrac{1}{1+\dfrac23}=\dfrac 35,$$ 于是 $$0<\dfrac{1}{a_{2019}}<\dfrac 12,$$ 从而 $k=-5$．