MT【159】单调有界有极限

已知数列$\{a_n\}$满足:$a_n>0,a_{n+1}+\dfrac{1}{a_n}<2,n\in N^*$.
求证:
(1)$a_{n+2}<a_{n+1}<2 (n\in N^*)$
(2)$a_n>1 (n\in N^*)$

第二题:分析:由题意$\{a_n\}$单调递减又有下界,故有极限,记$\lim\limits_{n\longrightarrow +\infty}a_n=x$
则由$a_{n+1}+\dfrac{1}{a_n}<2$两边取极限得$x+\dfrac{1}{x}\le2$,又由于$x+\dfrac{1}{x}\ge2$故$\lim\limits_{n\longrightarrow +\infty}a_n=1$
由单调递减得$a_n>1$

注:也可以用反证法，提示:关键递推式$\dfrac{1}{a_{n+1}-1}>1+\dfrac{1}{a_n-1}$