logn(n+1)=ln(n+1)/ln(n)={ln(n)+ln[(n+1)/n]}/ln(n)=1+ln[(n+1)/n]/ln(n) 同样logn+1(n+2)=1+ln[(n+2)/(n+1)]/ln(n+1) (n+1)/n>(n+2)/(n+1) => ln[(n+1)/n]>ln[(n+2)/(n+1)] 又ln(n)1+ln[(n+2)/(n+1)]/ln(n+1) 则lo...