求證1+1/2+1/3+…+1/n≥2n/n+1,n為正整數(shù) 用數(shù)學(xué)歸納法…^-^……

數(shù)學(xué)人氣：329 ℃時(shí)間：2020-05-11 07:34:30

優(yōu)質(zhì)解答

當(dāng)n=1時(shí),1>=1,不等式成立.
若n=k時(shí)式子成立,則有1+1/2+……+1/k>=2k/(k+1)=2-2/(k+1)
當(dāng)n=k+1時(shí),需證明1+1/2+……+1/k+1/(k+1)>=2(k+1)/(k+2)=2-2/(k+2)
,與上面的式子想減,只需證明1/(k+1)>=2(k+1)/(k+2)-2k/(k+1)=2/(k+1)-2/(k+2)
即只需證明1/(K+1)>=2/(k+1)-2/(k+2)=2/(k+1)(k+2)
通分后即為只需證K+2>=2,這是顯然的.因此有
1+1/2+……+1/k+1/(k+1)>=2(k+1)/(k+2)
即n=k+1成立.從而原不等式對任意n都成立.