设n属于N,n>1,求证logn (n+1)>logn+1 (n+2)

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设n属于N,n>1,求证logn(n+1)>logn+1(n+2)设n属于N,n>1,求证logn(n+1)>logn+1(n+2)设n属于N,n>1,求证logn(n+1)>logn+1(n+2)换

设n属于N,n>1,求证logn (n+1)>logn+1 (n+2)
设n属于N,n>1,求证logn (n+1)>logn+1 (n+2)

设n属于N,n>1,求证logn (n+1)>logn+1 (n+2)
换底公式 loga(b)=logn(b)/logn(a)
logn(n+1)=lg(n+1)/lgn
logn+1 (n+2) =lg(n+2)/lg(n+1)
logn(n+1)-logn+1(n+2)={lg(n+1)^2-lg(n^2+2n)}/lgnlg(n+1)
由于n>1,故lgnlg(n+1)>0,研究lg(n+1)^2-lg(n^2+2n)的符号
lg(n+1)^2-lg(n^2+2n)=lg{(n+1)^2/n(n+2)},因(n+1)^2-n(n+2)=1>0,故lg(n+1)^2-lg(n^2+2n)>0
即logn(n+1)-logn+1(n+2)>0