已知数列$\{a_n\}$满足$a_1=\dfrac{1}{2},a_{n+1}=\sin\left(\dfrac{\pi}{2}a_n\right),S_n$ 为$\{a_n\}$的前$n$项和,求证:$S_n>n-\dfrac{5}{2}$
证明:显然$a_n\in(0,1)$故由约旦不等式:
$a_{n+1}=\sin\left(\dfrac{\pi}{2}a_n\right)\ge\dfrac{2}{\pi}\cdot(\dfrac{\pi}{2}a_n)=a_n$, 即$a_n$单调递增,故$a_n\ge a_1=\dfrac{1}{2}$,所以$a_n\in[\dfrac{1}{2},1)$
考虑到不动点$x_0=1,$
$\dfrac{1-a_{n+1}}{1-a_n}=\dfrac{1-\sin\left(\dfrac{\pi}{2}a_n\right)}{1-a_n}=\dfrac{2sin^2\left(\dfrac{\pi}{4}(1-a_n)\right)}{1-a_n}\le\dfrac{2\left(\dfrac{\pi}{4}(1-a_n)\right)^2}{1-a_n}\le \dfrac{\pi^2}{8}(1-a_n)\le\dfrac{\pi^2}{16}$
故$1-a_n\le(1-a_1)\left(\dfrac{\pi^2}{16}\right)^{n-1}$
所以$\sum\limits_{k=1}^n(1-a_k)\le\sum\limits_{k=1}^n{(1-a_1)\left(\dfrac{\pi^2}{16}\right)^{n-1}}=\dfrac{(1-a_1)(1-\left(\dfrac{\pi^2}{16}\right)^{n})}{1-\dfrac{\pi^2}{16}}<\dfrac{5}{2}$
即证$S_n>n-\dfrac{5}{2}$
练习:已知$x_1=\dfrac{3}{4}\pi,2x_{n+1}+\cos x_n-\pi=0$求$\lim\limits_{n\to \infty}{x_n}$
答案:$\dfrac{\pi}{2}$,提示不动点$x_0=\dfrac{\pi}{2}$