How do you use the limit comparison test on the series sum_(n=1)^oo1/sqrt(n^3+1)∞∑n=11√n3+1 ?

1 Answer
dansmath
Sep 13, 2014

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 / ;-}

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