解;(1)∵an+1=f(an)=
,an
an+1
∴
=1+1 an+1
,1 an
即
-1 an+1
=1,又1 an
=11 a1
∴数列{
}是以1为首项,以1为公差的等差数列.1 an
(2)∵(1)得
=1+(n-1)×1=n,∴an=1 an
1 n
∵bn=anan+1,∴bn=
=1 n×(n+1)
-1 n
1 n+1
Sn=1-
+1 2
-1 2
+1 3
?1 3
+…+1 4
?1 n
=1-1 n+1
<11 n+1
又知{Sn}为递增数列,∴Sn≥S1=b1=
=1 1×2
1 2
∴
≤Sn<11 2