已知数列{an}的通项公式an=n/n^2+1,判断函数的单调性

Question

已知数列{an}的通项公式an=n/n^2+1,判断函数的单调性

Answer 1

an=n/(n²+1) >0,数列各项均为正
a(n+1)/an=[(n+1)/[(n+1)²+1]]/[n/(n²+1)]
=(n+1)(n²+1)/[n[(n+1)²+1]]
=(n³+n²+n+1)/(n³+2n²+n+1)
=(n³+2n²+n+1-n²)/(n³+2n²+n+1)
=1- n²/(n³+2n²+n+1)