Question

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

(Use the appropriate test)

Answer 1

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.