(Ⅰ)设公差为d,且d≠0,
∵S 3=a 4+6,且a 1,a 4,a 13成等比数列
∴3a 1+3d=a 1+3d+6,(a 1+3d) 2=a 1(a 1+12d)
∴a 1=3,d=2
∴a n=3+2(n-1)=2n+1;
(Ⅱ)S n=
n(3+2n+1)
2 =n(n+2),∴
1
S n =
1
n(n+2) =
1
2 (
1
n -
1
n+2 )
∴数列{
1
S n }的前n项和为
1
2 (1-
1
3 +
1
2 -
1
4 +
1
3 -
1
5 +…+
1
n -
1
n+2 ) =
1
2 (1+
1
2 -
1
n+1 -
1
n+2 )
=
3 n 2 +5n
4(n+1)(n+2) .