Is the series #\sum_(n=1)^\inftyn^2/(n^3+1)# absolutely convergent, conditionally convergent or divergent?

(Use the appropriate test)

1 Answer
Apr 20, 2018

Diverges by the Limit Comparison Test

Explanation:

Due to the simplicity of the series, we can use the Limit Comparison Test, which tells us if we have some positive sequence #b_n# and we know the convergence or divergence of #sumb_n,# then if

#c=lim_(n->oo)a_n/b_n>0 ne oo#, then both series either converge or diverge.

Here, #a_n=n^2/(n^3+1).#

For the comparison sequence, we need one whose series' convergent or divergent behavior we know. So, we'll say

#b_n=n^2/n^3=1/n#

Now, #sum_(n=1)^oo1/n# diverges as it's a harmonic series, and the #p-#series test also tells us since it's in the form #sum1/n^p# where #p=1,# it diverges.

So,

#c=lim_(n->oo)(n^2/(n^3+1))/(1/n)=lim_(n->oo)(n(n^2))/(n^3+1)=lim_(n->oo)n^3/(n^3+1)=1>0neoo#

Thus, as both series must diverge because of this result, the series diverges by the Limit Comparison Test.

We don't need to check for absolute convergence -- if #suma_n# diverges, #sum|a_n|# will also diverge.