Question

如何证明y=(1+ 1/n)^n为单调增函数

Answer 1

因为均值不等式 

(a1a2...an)^(1/n)≤(a1+a2+...+an)/n

2边n次方 得到 a1a2...an)≤[(a1+a2+...+an)/n]^n

等号成立 当且仅当 a1=a2=……=an成立

后面详细的看图片吧，图片上很详细，这里不好打格式

(1+1/n)^n=(1+1/n)^n  *  1  =  (1+1/n)(1+1/n)...(1+1/n)*1

< {1/(n+1)[(1+1/n)+……+(1+1/n)+1] }^(n+1)

={1/(n+1)[(1+1/n)*n+1] }^(n+1)

=[1+1/(n+1)]^(n+1)

Answer 2

设函数Y=lgy=nlg(1+ 1/n),首先n≠0,当n＞0,a＞b,Ya/Yb＞1.n＜0,a＞b,Ya/Yb＜1,所以函数Y为增函数,Y=lgX为增函数，故y=(1+ 1/n)^n为单调增函数

Answer 3

(1+1/n)^n=(1+1/n)(1+1/n)...(1+1/n)*1 <= {[n*(1+1/n)+1]/(n+1)}^(n+1)

=[1+1/(n+1)]^(n+1)