Calculus of a Single Variable, Chapter 9, 9.4, Section 9.4, Problem 13

Limit comparison test is applicable, if suma_n and sumb_n are series with positive terms. If lim_(n->oo)a_n/b_n=L where L is a finite number and L>0 , then either both series converge or both diverge.
Given series is sum_(n=1)^oon/(n^2+1)
We can compare the series with sum_(n=1)^oon/n^2=sum_(n=1)^oo1/n
The comparison series sum_(n=1)^oo1/n is a divergent harmonic series.
a_n/b_n=(n/(n^2+1))/(1/n)=n^2/(n^2+1)
lim_(n->oo)a_n/b_n=lim_(n->oo)n^2/(n^2+1)
=lim_(n->oo)n^2/(n^2(1+1/n^2))
=lim_(n->oo)1/(1+1/n^2)
=1>0
Since the comparison series sum_(n=1)^oo1/n diverges, so the series sum_(n=1)^oon/(n^2+1) diverges as well, by the limit comparison test.