已知函数f(x)=（2^n-1)/(2^n+1),求证：对任意不小于3的自然数n,都有f(n)＞n/(n+1)

题目

已知函数f(x)=（2^n-1)/(2^n+1),求证：对任意不小于3的自然数n,都有f(n)＞n/(n+1)

答案

f(x)=1-2/(2^x+1)
f(n)=1-2/(2^n+1)
n/(n+1)=1-1/(n+1)
当n>3时,f(n)-n/(n+1)=(2^n-2n-1)/[(2^n+1)(n+1)]>0
所以f(n)>n/(n+1)