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

Question

(Use the appropriate test)

Emily Ammerman · Accepted Answer

undefined To determine whether the series $ \sum_{n=1}^\infty \frac{n^2}{n^3+1} $ is absolutely convergent, conditionally convergent, or divergent, we need to analyze its convergence behavior.

First, consider the series $ \sum_{n=1}^\infty \frac{n^2}{n^3+1} $. As $ n $ approaches infinity, the terms of the series behave like $ \frac{n^2}{n^3} = \frac{1}{n} $.

Now, the series $ \sum_{n=1}^\infty \frac{1}{n} $ is a p-series with $ p = 1 $. It is a known result that this series diverges.

Since $ \frac{n^2}{n^3+1} $ is bounded by $ \frac{1}{n} $, and $ \sum_{n=1}^\infty \frac{1}{n} $ diverges, then by the Comparison Test, the series $ \sum_{n=1}^\infty \frac{n^2}{n^3+1} $ also diverges.

Thus, the series $ \sum_{n=1}^\infty \frac{n^2}{n^3+1} $ is divergent.

Emma Aldridge · Answer

Diverges by the Limit Comparison Test

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.