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已知函数f(x)=(x+3)/(x+1) (x不等于-1),设数列{an}满足a1=1,an+1(n+1是下标)=f(an)
数列{bn}满足bn=绝对值(an-√3),sn=b1+b2+……+bn,求证:Sn
人气:140 ℃ 时间:2019-08-19 05:46:52
解答
a1=1,a=f(an)=1+2/(an+1),
∴an>=1.
√3-a=√3-(an+3)/(an+1)=(√3-1)(an-√3)/(an+1),
∴b=bn*(√3-1)/(an+1)<=bn*(√3-1)/2,
b1=√3-1,bn>0,
∴Sn<=b1/[1-(√3-1)/2]=2(√3-1)/(3-√3)=2√3/3.
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