每日一题[390]级数放缩

2016年浙江省宁波市高三期末联考压轴题：

对任意正整数$n$，设$a_n$是方程$x^2+\dfrac xn=1$的正根．

（1）求证：$a_{n+1}>a_n$；

（2）求证：$\dfrac{1}{2a_2}+\dfrac{1}{3a_3}+\cdots +\dfrac{1}{na_n}<1+\dfrac 12+\dfrac 13+\cdots +\dfrac 1n$．

分析    根据题意有$na_n^2+a_n=n$，对此式进行适当变形来获得解题的突破口．

解    （1）法一　显然有$0<a_n<1$．根据题意，有$$\begin{split} (n+1)a_{n+1}^2+a_{n+1}&=n+1,\\na_n^2+a_n&=n, \end{split} $$两式相减得$$n(a_{n+1}^2-a_n^2)+a_{n+1}^2+a_{n+1}-a_n=1,$$即$$(a_{n+1}-a_n)\cdot\left[n(a_{n+1}+a_n)+1\right]=1-a_{n+1}^2>0,$$因此$a_{n+1}>a_n$．

法二　(由meiyun提供）由题意知$a_n>0$，所以$$1=a_n^2+\dfrac {a_n}{n}>a_n^2+\dfrac {a_n}{n+1},$$从而有$$a_{n+1}^2+\dfrac {a_{n+1}}{n+1}=1>a_n^2+\dfrac {a_n}{n+1},$$整理得$$(a_{n+1}-a_n)\left(a_{n+1}+a_n+\dfrac {1}{n+1}\right )>0,$$从而有$a_{n+1}>a_n$．

（2）法一    利用界放缩

由于$$na_n^2+a_n=n,$$于是$$na_n=\dfrac{n}{a_n}-1>n-1,$$因此$$\dfrac{1}{na_n}<\dfrac{1}{n-1},$$于是$$\dfrac{1}{2a_2}+\dfrac{1}{3a_3}+\cdots +\dfrac{1}{na_n}<1+\dfrac 12+\cdots +\dfrac 1{n-1},$$原不等式得证．

法二    裂项放缩

由已知可得$na_n=\dfrac{\sqrt{4n^2+1}-1}2$，$n\in\mathcal N^*$．于是\[\begin{split} \dfrac{1}{na_n}-\dfrac 1n&=\dfrac{2}{\sqrt{4n^2+1}-1}-\dfrac 1n\\ &=\dfrac{\sqrt{4n^2+1}+1}{2n^2}-\dfrac 1n\\ &=\dfrac{\sqrt{4n^2+1}-2n+1}{2n^2}\\ &=\dfrac{\frac{1}{\sqrt{4n^2+1}+2n}+1}{2n^2}\\ &<\dfrac{1}{8n^3}+\dfrac{1}{2n^2} \\&<\dfrac{1}{8(n-1)n(n+1)}+\dfrac{1}{2\left(n-\frac 12\right)\left(n+\frac 12\right)}\\ &=\dfrac 1{16}\left[\dfrac{1}{(n-1)n}-\dfrac{1}{n(n+1)}\right]+\dfrac 12\left(\dfrac{1}{n-\frac 12}-\dfrac{1}{n+\frac 12}\right) ,\end{split} \]因此$$\left(\dfrac{1}{2a_2}+\dfrac{1}{3a_3}+\cdots +\dfrac{1}{na_n}\right)-\left(\dfrac 12+\dfrac 13+\cdots +\dfrac 1n\right)<\dfrac{1}{16}\cdot\dfrac 12+\dfrac 12\cdot \dfrac 23=\dfrac{35}{96}<1,$$于是原不等式得证．

注    对（2）中的不等式，右边的$\dfrac 1n$是多余的，甚至由法二可知，右边可以缩小$\dfrac{61}{96}$．