The limit comparison test states that if #L=lim_(nrarroo)a_n/b_n# where #L# is a positive, finite value, then #suma_n# and #sumb_n# either both converge or both diverge.
When working with the limit comparison test, it's often helpful to let the series we are working with be #suma_n#. Thus, let #a_n=1/(nsqrt(n^2+1))#.
The trickier thing to do is choose #b_n#. The best thing to do is make #b_n# mimic whatever #a_n# does at infinity.
As #1/(nsqrt(n^2+1))# gets larger and larger, the constant becomes more and more irrelevant and we see that #a_n~~1/(nsqrt(n^2))=1/(n(n))=1/n^2#.
So, let #b_n=1/n^2#.
Now going to the limit comparison test, let #L=lim_(nrarroo)a_n/b_n#.
Then #L=lim_(nrarroo)(1/(nsqrt(n^2+1)))/(1/n^2)=lim_(nrarroo)n^2/(nsqrt(n^2+1))#
#=lim_(nrarroo)n/sqrt(n^2+1)#
#=lim_(nrarroo)n/(nsqrt(1+1/n^2)#
#=lim_(nrarroo)1/sqrt(1+1/n^2#
#=1/sqrt(1+0)#
#=1#
So #L=1#. Since this is positive and finite, we know that #suma_n# and #sumb_n# have the same convergence or divergence.
We know that #sumb_n=sum_(n=1)^oo1/n^2# converges through the p-series test, so this means that #suma_n=sum_(n=1)^oo1/(nsqrt(n^2+1))# also converges through the limit comparison test.
