用数学归纳法:
(1)n=1时,左=1,右=1/2,成立
(2)设n=k时不等式成立,即1+1/2+1/3+...+1/(2^(k-1))>k/2,
则n=k+1时左=1+1/2+1/3+...+1/2^k={1+1/2+1/3+...+1/(2^(k-1))}+{1/(2^(k-1)+1)+...+1/2^k}>
k/2+))}+{1/(2^(k-1)+1)+...+1/2^k}>k/2+1/2^k+...+1/2^k(一共有2^(k-1)个1/2^k)=k/2+1/2=(k+1)/2
所以1+1/2+1/3+...+1/2^k>(k+1)/2
综上,有1+1/2+1/3+...+1/(2^n-1)>n/2