求证log(n)(n+1)>log(n+1)(n+2),其中n∈N,且n>1n和n+1都是底数
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求证log(n)(n+1)>log(n+1)(n+2),其中n∈N,且n>1n和n+1都是底数
求证log(n)(n+1)>log(n+1)(n+2),其中n∈N,且n>1
n和n+1都是底数
求证log(n)(n+1)>log(n+1)(n+2),其中n∈N,且n>1n和n+1都是底数
要证:log(n)(n+1)>log(n+1)(n+2),
即要证log(n+1)/logn>log(n+2)/log(n+1)(换底公式)
即要证log(n+1)log(n+1)>logNlog(n+2)
而由基本不等式
lognlog(n+2)
log(n)(n+1)>log(n+1)(n+2)
证明函数
f(x)=log x (x+1) 在(1,正无穷)上单调减
f'(x)=(ln(x+1)/ln x)'
=(ln x /(x+1)-ln(x+1)/x)/(lnx)^2<0
所以log(n)(n+1)>log(n+1)(n+2)
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