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