求一道数列题已知数列an的首项a13,通项an与前n项和Sn满足2an=Sn*S(n-1),（1)求证1/Sn是等差数列,并求公差,(2)求数列an的通项公式,(3)数列an中是否存在自然数k,使得不等式ak大于a(k+1)对于任意大于k或者等于k的自然数都成立?若存在,求出最小的k,若不存在,说明理由

1.
n≥2时,
2an=2[Sn-S(n-1)]=2Sn-2S(n-1)
2Sn-2S(n-1)=SnS(n-1)
等式两边同除以2SnS(n-1)
1/S(n-1)- 1/Sn=1/2
1/Sn -1/S(n-1)=-1/2
1/S1=1/a1=1/3,数列{1/Sn}是以1/3为首项,-1/2为公差的等差数列,公差=-1/2
2.
1/Sn=(1/3)+(-1/2)(n-1)=(5-3n)/6
Sn=6/(5-3n)
n≥2时,an=Sn-S(n-1)=6/(5-3n)- 6/[5-3(n-1)]=6/(5-3n) -6/(8-3n)
n=1时,a1=6/(5-3)-6/(8-3)=9/5≠3
数列{an}的通项公式为
an=3 n=1
6/(5-3n)-6/(8-3n) n≥2
3.
假设存在满足题意的k
k=1时,
a2=6/(5-3×2)-6/(8-3×2)=-9
k≥2时,
ak>a(k+1)
6/(5-3k) -6/(8-3k)>6/[5-3(k+1)]-6/[8-3(k+1)]
1/(3k-8)+1/(3k-2)>2/(3k-5)
1/[(3k-2)(3k-8)]>1/(3k-5)²
不等式右边1/(3k-5)²恒>0,要不等式有解,(3k-2)(3k-8)>0 k>8/3,k为正整数,k≥3
(3k-2)(3k-8)0,不等式恒成立,即k≥3时,恒满足题意.
综上,得k的最小值为3.