Question

How do you use the direct Comparison test on the infinite series `sum_(n=2)^oon^3/(n^4-1)` ?

Answer 1

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.