(Ⅰ)∵(a n-1)(a n+3)=4S n,当n≥2时,(a n-1-1)(a n-1+3)=4S n-1,
两式相减,得
a 2n -
a 2n-1 +2 a n -2 a n-1 =4 a n ,即(a n+a n-1)(a n-a n-1-2)=0,又a n>0,∴a n-a n-1=2.
当n=1时,(a 1-1)(a 1+3)=4a 1,∴(a 1+1)(a 1-3)=0,又a 1>0,∴a 1=3.
所以,数是以3为首项,2为公差的等差数列.
(Ⅱ)由(Ⅰ),a 1=3,d=2,∴a n=2n+1.
设 b n =
4
a n 2 -1 ,n∈N *;∵a n=2n+1,∴ a n 2 -1=4n(n+1)))
∴ b n =
4
4n(n+1) =
1
n(n+1) =
1
n -
1
n+1
∴T n=b 1+b 2+b 3+…+b n= (1-
1
2 )+(
1
2 -
1
3 )+…+(
1
n -
1
n+1 ) = 1-
1
n+1 <1 .
又∵ T n+1 - T n =
n+1
n+2 -
n
n+1 =
1
(n+2)(n+1) >0 ,∴ T n+1 > T n > T n-1 >…> T 1 =
1
2 ,
综上所述:不等式
1
2 ≤ T n <1 成立.