已知数列an中,a1=3,an+1=2an-1,设bn=2n/anan+1,求证：数列bn的前n项和Sn＜1/3

数列an中,a1=3,a=2an-1,
∴a-1=2(an-1),
∴an-1=(a1-1)*2^(n-1)=2^n,
∴an=2^n+1,
∴bn=2n/{(2^n+1)(2^(n+1)+1]},
∴Sn=2/(3*5)+4/(5*9)+……+2n/{(2^n+1)(2^(n+1)+1]}
=1/3-1/5+1/5-1/9+……+[n/2^(n-1)][1/(2^n+1)-1/[2^(n+1)+1],
当n∈N+时n