当n=1时,1>=1,不等式成立.
若n=k时式子成立,则有1+1/2+……+1/k>=2k/(k+1)=2-2/(k+1)
当n=k+1时,需证明1+1/2+……+1/k+1/(k+1)>=2(k+1)/(k+2)=2-2/(k+2)
,与上面的式子想减,只需证明1/(k+1)>=2(k+1)/(k+2)-2k/(k+1)=2/(k+1)-2/(k+2)
即只需证明1/(K+1)>=2/(k+1)-2/(k+2)=2/(k+1)(k+2)
通分后即为只需证K+2>=2,这是显然的.因此有
1+1/2+……+1/k+1/(k+1)>=2(k+1)/(k+2)
即n=k+1成立.从而原不等式对任意n都成立.