logn(n+1)=lg(n+1)/lgnlg(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))...