Question

How do you use the limit comparison test on the series `sum_(n=1)^oo1/sqrt(n^3+1)` ?

Answer 1

The Limit Comparison Test tells you that if
`0 < lim_(n->oo)a_n/b_n < oo`, with `a_n>0 and b_n >0,`
then `sum_(n=1)^(oo)a_n` converges if and only if `sum_(n=1)^(oo)b_n` converges.

We want to find a series whose terms `b_n` compare to our
`a_n = 1/(sqrt(n^3+1))`; let's choose `b_n = 1/(sqrt(n^3))=1/(n^(3/2))`.

So `sum_(n=1)^(oo)b_n` converges by the Integral p-test; `p = 3/2 > 1.`
`lim_(n->oo)a_n/b_n=lim_(n->oo)(sqrt(n^3))/(sqrt(n^3+1))=sqrt(lim_(n->oo)(n^3)/(n^3+1))=1`

Since `0 < 1 < oo,` our series converges:
1. by the Limit Comparison Test, and
2. by the \ illustrious dansmath / ;-}