要证f(n)>n/(n+1)即证1-2/(2^n+1)>1-1/(n+1)即证1/(n+1)>2/(2^n+1)即证2^n+1>2n+2即证2^n>2n+1数学归纳法：当n=3时2^3=8>7=2*3+1设n-1时成立即2^(n-1)>2n-1两边同时乘以2，即2^n>4n-2=2n+1+2n-3因为n-1>=3即n>=4,所以2n-3>0所以2^n>2n+1固原式成立
