Calculus of a Single Variable, Chapter 9, 9.4, Section 9.4, Problem 21

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=1)^oon^(k-1)/(n^k+1),k>=2
Let the comparison series be sum_(n=1)^oo1/n
The comparison series sum_(n=1)^oo1/n is a p-series with p=1.
As per the p-series test,sum_(n=1)^oo1/n^p is convergent if p>1 and divergent if 0So, the comparison series is a divergent series.
Now let's use the limit comparison test with: a_n=n^(k-1)/(n^k+1)
b_n=1/n
a_n/b_n=(n^(k-1)/(n^k+1))/(1/n)
a_n/b_n=(n(n^(k-1)))/(n^k+1)
a_n/b_n=n^k/(n^k+1)
a_n/b_n=n^k/(n^k(1+1/n^k))
a_n/b_n=1/(1+1/n^k)
lim_(n->oo)a_n/b_n=lim_(n->oo)1/(1+1/n^k)
=1>0
Since the comparison series sum_(n=1)^oo1/n diverges, the series sum_(n=1)^oon^(k-1)/(n^k+1) as well, diverges as per the limit comparison test.