摘要:求证:2<(1+)n<3(n≥2.n∈N*).
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(本小题满分13分)
已知正项数列{an}的首项a1=,函数f(x)=,g(x)=.
(1)若正项数列{an}满足an+1=f(an)(n∈N*),证明:{}是等差数列,并求数列{an}的通项公式;
(2)若正项数列{an}满足an+1≤f(an)(n∈N*),数列{bn}满足bn=,证明:b1+b2+…+bn<1;
(3)若正项数列{an}满足an+1=g(an),求证:|an+1-an|≤·()n-1
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