consider
for n>=3
1/(2^n-1) < (2/3) [1/2^(n-1)]
an = 1/[2^(n+1) -1]
a1=1/3
1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)
=Sn
=a1+a2+..+an
= 1/3 + 1/(2^3-1)-1/(2^4-1)+...+1/(2^n-1)这种方法我已经得到了。谢谢大神，确实很厉害。期待其他方法