已知数列${a_n}$满足$a_1=dfrac{1}{2},a_{n+1}=sinleft(dfrac{pi}{2}a_n ight),S_n$ 为${a_n}$的前$n$项和,求证:$S_n>n-dfrac{5}{2}$
证明:显然$a_nin(0,1)$故由约旦不等式:
$a_{n+1}=sinleft(dfrac{pi}{2}a_n
ight)gedfrac{2}{pi}cdot(dfrac{pi}{2}a_n)=a_n$, 即$a_n$单调递增,故$a_nge a_1=dfrac{1}{2}$,所以$a_nin[dfrac{1}{2},1)$
考虑到不动点$x_0=1,$
$dfrac{1-a_{n+1}}{1-a_n}=dfrac{1-sinleft(dfrac{pi}{2}a_n
ight)}{1-a_n}=dfrac{2sin^2left(dfrac{pi}{4}(1-a_n)
ight)}{1-a_n}ledfrac{2left(dfrac{pi}{4}(1-a_n)
ight)^2}{1-a_n}le dfrac{pi^2}{8}(1-a_n)ledfrac{pi^2}{16}$
故$1-a_nle(1-a_1)left(dfrac{pi^2}{16}
ight)^{n-1}$
所以$sumlimits_{k=1}^n(1-a_k)lesumlimits_{k=1}^n{(1-a_1)left(dfrac{pi^2}{16}
ight)^{n-1}}=dfrac{(1-a_1)(1-left(dfrac{pi^2}{16}
ight)^{n})}{1-dfrac{pi^2}{16}}<dfrac{5}{2}$
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$limlimits_{n o infty}{x_n}$
答案:$dfrac{pi}{2}$,提示不动点$x_0=dfrac{pi}{2}$