n≥2时,
SnS(n-1)+(1/2)an=0
2SnS(n-1)+Sn-S(n-1)=0
等式两边同除以SnS(n-1)
2+1/S(n-1)-1/Sn=0
1/Sn -1/S(n-1)=2,为定值
1/S1=1/a1=1/1=1,数列{1/Sn}是以1为首项,2为公差的等差数列
1/Sn=1+2(n-1)=2n-1
Sn=1/(2n-1)
n=1时,S1=1/(2-1)=1,同样满足通项公式
数列{Sn}的通项公式为Sn=1/(2n-1)
n≥2时,an=Sn-S(n-1)=1/(2n-1)-1/[2(n-1)-1]=1/(2n-1)-1/(2n-3)
n=1时,a1=1/(2-1)-1/(2-3)=1+1=2≠1
数列{an}的通项公式为
an=1 n=1
1/(2n-1)-1/(2n-3) n≥2
1/(1-Sn)=1/[1- 1/(2n-1)]=(2n-1)/(2n-2)
n=1时,1/(1-S2)=3/2 √(1+1)=√2√(k+1),则当n=k+1时,
[1/(1-S2)]·[1/(1-S3)]·...·[1/(1-S(k+2))]
>√(k+1)·[2(k+2)-1]/[2(k+2)-2]
=√(k+1)·(2k+3)/[2(k+1)]
=(2k+3)/[2√(k+1)]
(2k+3)²-4(k+1)(k+2)
=4k²+12k+9-4(k²+3k+2)
=4k²+12k+9-4k²-12k-8
=1>0
(2k+3)²/[4(k+1)]>k+2
(2k+3)/[2√(k+1)]>√(k+2)
[1/(1-S2)]·[1/(1-S3)]·...·[1/(1-S(k+2))]>(2k+3)/[2√(k+1)]>√(k+2)=√[(k+1)+1]
不等式同样成立,k为任意正整数,因此对于任意正整数n,不等式恒成立.
[1/(1-S2)]·[1/(1-S3)]·...·[1/(1-S(n+1))]>√(n+1)
