每日一题[1063]单调与有界的纠缠

已知数列 \(\{a_n\}\) 满足对任意正整数 \(n\in\mathbb N^{\ast}\),都有 \(a_n>0\) \(a_{n+1}+\dfrac{1}{a_n}<2\)

1)求证:\(a_{n+2}<a_{n+1}<2\)

2)求证:\(a_n>1\)


分析与解 (1)根据题意,有\[a_{n+1}<2-\dfrac{1}{a_n}<2,\]\[a_{n+2}-a_{n+1}<2-\dfrac{1}{a_{n+1}}-a_{n+1}=-\dfrac{(a_{n+1}-1)^2}{a_{n+1}}<0,\]于是 \(a_{n+2}<a_{n+1}\),原命题得证.

 (2)用反证法.
假设存在
\(a_k\leqslant 1\),则 \(a_{k+1}<1\),且易知\[a_n>\dfrac{1}{2-a_{n+1}}>\dfrac 12,\]于是 \(\dfrac 12<a_{k+1}<1\).当 \(n\geqslant k+1\) 时,有\[\dfrac{a_{n+1}}{a_n}<\dfrac{2-\dfrac{1}{a_n}}{a_n}=\dfrac{2a_n-1}{a_n^2}\leqslant \dfrac{2a_{k+1}-1}{a_{k+1}^2},\]因此可得\[a_{k+m}<\left(\dfrac{2a_{k+1}-1}{a_{k+1}^2}\right)^{m-1}\cdot a_{k+1},\]由于\[0<\dfrac{2a_{k+1}-1}{a_{k+1}^2}<1,\]于是必然存在正整数 \(m\),使得\[a_{k+m}<\dfrac 12,\]矛盾.

因此原命题得证.

 因为函数 \(y=\dfrac{2x-1}{x^2}=-\left(\dfrac 1x-1\right)^2+1\) \(\left(\dfrac 12,1\right)\) 上单调递增,而 \(n\geqslant k+1\) 时,\(a_n\leqslant a_{k+1}\),所以有\[\dfrac{2a_n-1}{a_n^2}\leqslant \dfrac{2a_{k+1}-1}{a_{k+1}^2}.\]

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