n为正整数,x=（1+1/n)^n.y=(1+1/n)^（n+1).比较x^y和y^x的大小

Question

n为正整数,x=（1+1/n)^n.y=(1+1/n)^（n+1).比较x^y和y^x的大小

Answer 1

x^y=(1+1/n)^[n*(1+1/n)^(n+1)]
y^x=(1+1/n)^[(n+1)*(1+1/n)^n]
则x^y/y^x=(1+1/n)^[n(1+1/n)^(n+1)-(n+1)(1+1/n)^n]=(1+1/n)^[(1+1/n)^n*(n(1+1/n)-(1+n))]
=(1+1/n)^[(1+1/n)^0]=1+1/n
因为n是正整数,所以1+1/n>1,所以x^y>y^x