构造函数法
证明:
记f(x)=ln(x+1)-x/(1+x),(x>0)
求导
f'(x)=1/(x+1)-[x+1-x]/(x+1)^2=x/(x+1)^2>0
即有f(x)在x>0上单调递增
又f(x)可在x=0处连续,则f(x)>f(0)=0,x>0
即ln(x+1)-x/(1+x)>0
亦即ln(x+1)>x/(1+x),x>0
当k>=2,k∈N+,取1/k(>0)替换x得
ln[(1/k)+1]>(1/k)/(1+1/k),
整理得ln[(k+1)/k]>1/(k+1),命题得证.
k是大于等于2的正整数.证明:ln[(k+1)/k]>1/(k+1),
k是大于等于2的正整数.证明:ln[(k+1)/k]>1/(k+1),
数学人气:141 ℃时间:2019-12-29 15:17:38
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