sum_(n=0)^oo 1/sqrt(n^2+1) Use the Limit Comparison Test to determine the convergence or divergence of the series.

Limit comparison test is applicable when 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=0)^oo1/sqrt(n^2+1)
Let's compare with the series sum_(n=1)^oo1/sqrt(n^2)=sum_(n=1)^oo1/n
The comparison series sum_(n=1)^oo1/n is divergent p-series.
a_n/b_n=(1/sqrt(n^2+1))/(1/n)=n/sqrt(n^2+1)
lim_(n->oo)a_n/b_n=lim_(n->oo)n/sqrt(n^2+1)
=lim_(n->oo)n/(nsqrt(1+1/n^2))
=lim_(n->oo)1/sqrt(1+1/n^2)
=1>0
Since the comparison series sum_(n=1)^oo1/n diverges, so the series sum_(n=0)^oo1/sqrt(n^2+1) diverges as well, by the limit comparison test.