(1)当n=1时,S 1=2a 1-1,得a 1=1.(1分)
∵S n=2a n-n,
∴当n≥2时,S n-1=2a n-1-(n-1),
两式相减得:a n=2a n-2a n-1-1,
∴a n=2a n-1+1.(3分)
∴a n+1=2a n-1+2=2(a n-1+1),(5分)
∴{a n+1}是以a 1+1=2为首项,2为公比的等比数列.(6分)
(2)由(1)得a n+1=2•2 n-1=2 n,∴a n=2 n-1,n∈N *.(8分)
∴b n=log 2(a n+1)=log 22 n=n,n∈N *. (10分)
(3) c n =
2 n
a n a n+1 , c n+1 =
2 n+1
a n+1 a n+2 ①
c n+1 - c n =
2 n+1
( 2 n+1 -1)( 2 n+2 -1) -
2 n
( 2 n -1)( 2 n+1 -1)
=
-2× 4 n - 2 n
( 2 n+1 -1)( 2 n+2 -1)( 2 n -1) <0
∴数列{c n}单调递减.(12分)
∴①n=1时数列{c n}的最大值为 c 1 =
2
3 .(14分)
②由 c n =
2 n
( 2 n -1)( 2 n+1 -1) =
1
2 n -1 -
1
2 n+1 -1 ,(16分)
所以c 1+c 2+…+c n= 1-
1
2 n+1 -1 .∴
lim
n→∞ ( c 1+c 2+…+c n)=1.(18分)