原不等式等价于(2^2-2)/(2^2-1)+(2^3-2)/(2^3-1)+...+(2^(n+1)-2)/(2^(n+1)-1) > n-2/3.
左端 = (1-1/(2^2-1))+(1-1/(2^3-1))+...(1-1/(2^(n+1)-1)) = n-(1/(2^2-1)+1/(2^3-1)+...+1/(2^(n+1)-1)).
故不等式进一步等价于1/(2^2-1)+1/(2^3-1)+...+1/(2^(n+1)-1) < 2/3.
当k > 2,有2^(k-2) > 1,故2^k-1 > 2^k-2^(k-2) = 3·2^(k-2).
因此1/(2^2-1)+1/(2^3-1)+...+1/(2^(n+1)-1)
< 1/3+1/(3·2)+...+1/(3·2^(n-1))
= 2/3-1/(3·2^(n-1))
< 2/3.