Question

数列与恒成立问题

Answer 1

(1)由题意得,

Sn-S(n-1)+3SnS(n-1)=0

等式两边同时除以SnS(n-1),得

1/S(n-1)-1/Sn+3=0

即{1/Sn}是等差数列,公差为3

(2)1/S1=3

所以1/Sn=3n

Sn=1/(3n)

n>1时,an=Sn-S(n-1)=1/(3n)-1/(3(n-1))=-1/[3n(n-1)]

n=1不满足此式.

所以an=1/3(n=1);=-1/[3n(n-1)](n>1)

(3)易得bn=n

对于Tn,有T(n+1)-Tn=[1/(1+n+1)+1/(2+n+1)+...+1/(n+n+1)+1/(n+1+n+1)]-[1/(1+n)+1/(2+n)=...+1/(n+n)]=-1/(n+1)+1/(2n+1)+1/(2n+2)=1/(2n+1)-1/(2n+2)>0

所以Tn单调递增

故T1为最小值T1=1/(b1+1)=1/2

k的取值范围为(-∞,3/2]