数列竞赛题!在线等!数列{an},a1=2/3,a(n+1)=an^2+a(n-1)^2+.+a1^2(n∈N+),若对
数列竞赛题!在线等!
数列{an},a1=2/3,a(n+1)=an^2+a(n-1)^2+.+a1^2(n∈N+),若对于任意n属于N+,1/（a1+1）+2/（a2+1）+.+1/(an +1)＜M恒成立,求M最小值
数列{an},a1=2/3,a(n+1)=an^2+a(n-1)^2+.+a1^2(n∈N+),若对于任意n属于N+,1/（a1+1）+2/（a2+1）+.+1/(an +1)＜M恒成立,求M最小值
赞 白熊1977 幼苗
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题目抄错了吧,应该是1/(a2+1)吧.
n≥2时,
a(n+1)=an²+a(n-1)²+...+a1² (1)
an=a(n-1)²+a(n-2)²+...+a1² (2)
(1)-(2)
a(n+1)-an=an²
a(n+1)=an²+an=an(an +1)
1/a(n+1)=1/[an(an +1)]=1/an -1/(an +1)
1/(an +1)=1/an- 1/a(n+1)
1/(a1+1)+1/(a2+1)+...+1/(an +1)
=1/a1-1/a2+1/a2-1/a3+...+1/an -1/a(n+1)
=1/a1 -1/a(n+1)
a(n+1)/an=an(an +1)/an=an +1>1,数列是递增数列.
n->+∞时,a(n+1)->+∞ 1/a(n+1)->0
1/a1 -1/a(n+1)->1/a1=1/(2/3)=3/2
1/a1 -1/a(n+1)
n≥2时,
a(n+1)=an²+a(n-1)²+...+a1² (1)
an=a(n-1)²+a(n-2)²+...+a1² (2)
(1)-(2)
a(n+1)-an=an²
a(n+1)=an²+an=an(an +1)
1/a(n+1)=1/[an(an +1)]=1/an -1/(an +1)
1/(an +1)=1/an- 1/a(n+1)
1/(a1+1)+1/(a2+1)+...+1/(an +1)
=1/a1-1/a2+1/a2-1/a3+...+1/an -1/a(n+1)
=1/a1 -1/a(n+1)
a(n+1)/an=an(an +1)/an=an +1>1,数列是递增数列.
n->+∞时,a(n+1)->+∞ 1/a(n+1)->0
1/a1 -1/a(n+1)->1/a1=1/(2/3)=3/2
1/a1 -1/a(n+1)
1年前
8
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