已知 n>1且n属于N* ,求证logn(n+1)>logn+1(n+2)

已知 n>1且n属于N* ,求证logn(n+1)>logn+1(n+2)
已知 n>1且n属于N* ,求证logn(n+1)>logn+1(n+2)

已知 n>1且n属于N* ,求证logn(n+1)>logn+1(n+2)
logn(n+1)=lg(n+1)/lgn
lg(n+1)(n+2)=lg(n+2)/lg(n+1)
显然
验证
lg(n+1)/lgn 与 lg(n+2)/lg(n+1)大小即可
同时减去1
(lg(n+1)-lg n)/lgn （1
与(lg(n+2)-lg(n+1))/lg(n+1) （2
(lg(n+1)-lg n)=lg((n+1)/n)
(lg(n+2)-lg(n+1))=lg((n+2)/(n+1))
1+1/n>1+1/(n+1)
所以lg((n+1)/n)>lg((n+2)/(n+1))
又n+1>n
lgn（2
故得证