Question

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

Answer 1

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)