(1)解: {an}的前n项和Sn=a1+a2+…+an=f(1)=n2, 由an=Sn–Sn–1=n2–(n–1)2=2n–1(n≥2),又a1=S1=1满足an=2n–1. 故{an}通项公式为an=2n–1(n∈N*) ∴ (2)证明: ∵f()=1·+3·+…+(2n–1) ① ∴f()=1·+3·+…+(2n–3)+(2n–1) ② ①–②得:f()=1·+2·+2·+…+2·–(2n–1)· ∴f()=++++…+–(2n–1)=1– ∵ (n∈N*) ∴0<<1,∴0<1–<1,即0<f()<1 |