How do you use the direct Comparison test on the infinite series #sum_(n=2)^oon^3/(n^4-1)# ? Calculus Tests of Convergence / Divergence Direct Comparison Test for Convergence of an Infinite Series 1 Answer Wataru Sep 8, 2014 Since #n^3/{n^4-1} geq n^3/n^4=1/n# for all #n geq2# and #sum_{n=2}^infty1/n# is a harmonic series, which is known to be divergent, we may conclude that #sum_{n=2}^inftyn^3/{n^4-1}# also diverges by Direct Comparison Test. Answer link Related questions What is the Direct Comparison Test for Convergence of an Infinite Series? How do you use the direct comparison test for infinite series? How do you use the direct comparison test for improper integrals? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo5/(2n^2+4n+3)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooln(n)/n# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo9^n/(3+10^n)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo(1+sin(n))/(5^n)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooarctan(n)/(n^1.2)# ? How do you use basic comparison test to determine whether the given series converges or diverges... How do you use the Comparison Test to see if #1/(4n^2-1)# converges, n is going to infinity? See all questions in Direct Comparison Test for Convergence of an Infinite Series Impact of this question 3566 views around the world You can reuse this answer Creative Commons License