Question

已知数列{an}满足a0=1/2,an=an-1+(1/n^2)an-1^2,求证:(n+1)/(n+2)

Answer 1

a(1) = a(0) + [a(0)]^2 = 3/4 < 1,

2/3 = 8/12 < 9/12 = 3/4 = a(1),

2/3 < a(1) < 1.

n=1时,2/3 < a(1) < 1,命题成立.

设n=k时,有 (k+1)/(k+2) < a(k) < k成立,

则,n=k+1时,a(k+1) = a(k) + [a(k)]^2/(k+1)^2 < k + k^2/(k+1)^2 < k + 1,

a(k+1) = a(k) + [a(k)]^2/(k+1)^2 > (k+1)/(k+2) + 1/(k+2)^2

(k+1)/(k+2) + 1/(k+2)^2 - (k+2)/(k+3) = [(k+2)^2-1 - (k+2)^2]/[(k+2)(k+3)] + 1/(k+2)^2

= 1/(k+2)^2 - 1/[(k+2)(k+3)]

= 1/(k+2)[1/(k+2) - 1/(k+3)]

> 0.

所以,a(k+1) > (k+1)/(k+2) + 1/(k+2)^2 > (k+2)/(k+3).

综合,有,

(k+2)/(k+3) < a(k+1) < k+1.

n=k+1时,命题成立.

由归纳法知,n>=1时,总有

(n+1)/(n+2) < a(n) < n

成立.