数学归纳法:
当n=2时,有1/3*4/3=4/9>1/2*3/4=3/8
假设当n=K时成立;则有1/(k+1)(1+1/3+1/5+...+1/2k-1)>1/k*(1/2+1/4+...+1/2k)
则1/(k+2)*(1+1/3+...1/2k+1)
>(k+1)/(k+2)*1/k(1/2+...+1/2k)+1/(k+2)*1/(2k+1)
>1/(k+2)[(k+1)/k(1/2+...+1/2k)+1/(2k+1)]
显然(k+1)/k>1,1/(2k+1)>1/(2k+2)
则上式>1/(k+2)(1/2+...1/(2k+1)+1/(2k+2))
即证当n=k+1时也成立。
故对于所有的n>=2时结论成立!
1楼错解