e2=(1+1/2)^2,……,en=(1+1/n)^n.求证e(n)小于e(n+1).

证明：e(n)=(1+1/n)^n=1*(1+1/n)*(1+1/n)……*(1+1/n),(n项[1+1/n]相乘,加补乘的1,共有n+1项)利用均值不等式有：1+(1+1/n)+(1+1/n)……+(1+1/n)＞(n+1)*(1*(1+1/n)*(1+1/n)……*(1+1/n))^(1/(n+1)),(因为对任意n,1≠1+1/n,所以不会取等号)整理得：e(n)＜{[1+(1+1/n)+(1+1/n)……+(1+1/n)]/(n+1)}^(n+1)={[1+n*(1+1/n)]/(n+1)}^(n+1)=[(n+2)/(n+1)]^(n+1)=[1+1/(n+1)]^(n+1)=e(n+1)