How do you determine if #sum n^3/((n^4)-1)# from n=2 to #n=oo# is convergent? Calculus Tests of Convergence / Divergence Limit Comparison Test for Convergence of an Infinite Series 1 Answer Massimiliano Jun 14, 2015 It is divergent. Explanation: Since #n^3/(n^4-1)~n^3/n^4~1/n# that is the harmonic series, that diverges. (#~# means asymptotic). Answer link Related questions How do you use the limit comparison test on the series #sum_(n=1)^oon/(2n^3+1)# ? How do you use the limit comparison test on the series #sum_(n=1)^oo(n+1)/(n*sqrt(n))# ? How do you use the limit comparison test on the series #sum_(n=2)^oosqrt(n)/(n-1)# ? How do you use the limit comparison test on the series #sum_(n=1)^oo(n^2-5n)/(n^3+n+1)# ? How do you use the limit comparison test on the series #sum_(n=1)^oo1/sqrt(n^3+1)# ? What is the Limit Comparison Test? How do I use the Limit Comparison Test on the series #sum_(n=1)^oosin(1/n)# ? How do I know when to use limit comparison test vs the direct comparison test? How do you use the comparison test (or the limit comparison test) for #(1+sin(x))/10^x#? How do you determine whether #1/(n!)# convergence or divergence with direct comparison test? See all questions in Limit Comparison Test for Convergence of an Infinite Series Impact of this question 7629 views around the world You can reuse this answer Creative Commons License