Sn=a(n+1)-1
S(n-1)=an-1
an=Sn-S(n-1)=a(n+1)-an
a(n+1)=2an
故{an}是公比为2的等比数列
an=a1*2^(n-1)=2^(n-1)
a(n+1)=2^n
bn=2^n/[a(n+1)][a(n+1)+1]=1/(2^n+1)
因bnbn=2^n/[(an)+1][a(n+1)+1]bn=2^n/[(an)+1][a(n+1)+1]=2^n/[2^(n-1)+1][2^n+1]=2{1/[2^(n-1)+1]-1/(2^n+1)}Tn=2{(1/2-1/3)+(1/3-1/4)+...+1/[2^(n-1)+1]-1/(2^n+1)}=2[1/2-1/(2^n+1)]=1-2/(2^n+1)可见Tn<1因n=1,2,..所以Tn=1-2/(2^n+1)≥1-2/(2+1)=1/3得证