Question

已知数列｛an｝前n项的和为Sn，满足an=Sn•Sn-1 (n≥2,n∈N),a1=2/9

,(1)求证｛1/Sn｝为等差数列；（2）求通项an及lim an/Sn
请详述过程。

Answer 1

(1)an=Sn-Sn-1=Sn*Sn-1
∴有1/Sn-1-1/Sn=1,即1/Sn-1/Sn-1=-1
1/S1=1/a1=9/2
∴{1/Sn}是以9/2为首项,-1为公差的等差数列
(2)1/Sn=9/2-(n-1)=11/2-n,∴Sn=2/(11-2n)
∴Sn-1=2/[11-2(n-1)]=2/(13-2n)
an=Sn*Sn-1=4/(11-2n)(13-2n)
当n=1时,4/(11-2)(13-2)=4/99≠2/9
∴an=2/9(n=1),an=4/(11-2n)(13-2n)(n≥2)
an/Sn=Sn-1=2/(13-2n),limn→∞,an/Sn=limn→∞,0/(-2)=0