题目

比较n的n次方与n+1的n-1次方的大小,并证明

提问时间：2020-11-08

答案

n^n≥(n+1)^(n-1),当n=1时,左边=1,右边=1,上式成立；当n=2时,左边=4,右边=3,上式成立；设当n=k时,上式成立,k^k≥(k+1)^(k-1),两边取对数得：klogk≥(k-1)log(k+1),log(k+1)+klogk≥klog(k+1),klog(k+1)+log(k+1)≥2klog(k+1)-klogk,左边=(k+1)log(k+1)=log[(k+1)^(k+1)],右边=k[log(k+1)²-logk]=klog[(k+1)²/k]=klog[(k+2)+1/k]>[(k+1)-1]log[(k+1)+1]=log[(k+1)+1]^[(k+1)-1],则log[(k+1)^(k+1)]>log[(k+1)+1]^[(k+1)-1],[(k+1)^(k+1)]>[(k+1)+1]^[(k+1)-1],即当n=k+1时,上式成立,n^n≥(n+1)^(n-1).