用换底公式,只需证lg(n+1)/lgn>lg(n+2)/lg(n+1)
只需证lg(n+1)>根号[lgn*lg(n+2)]
事实上,2lg(n+1)=lg(n^2+2n+1)>lg(n^2+2n)=lgn+lg(n+2)>2根号[lgn*lg(n+2)]