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

## (Use the appropriate test)

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 $\sum {b}_{n} ,$ then if

$c = {\lim}_{n \to \infty} {a}_{n} / {b}_{n} > 0 \ne \infty$, then both series either converge or diverge.

Here, ${a}_{n} = {n}^{2} / \left({n}^{3} + 1\right) .$

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

${b}_{n} = {n}^{2} / {n}^{3} = \frac{1}{n}$

Now, ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges as it's a harmonic series, and the $p -$series test also tells us since it's in the form $\sum \frac{1}{n} ^ p$ where $p = 1 ,$ it diverges.

So,

$c = {\lim}_{n \to \infty} \frac{{n}^{2} / \left({n}^{3} + 1\right)}{\frac{1}{n}} = {\lim}_{n \to \infty} \frac{n \left({n}^{2}\right)}{{n}^{3} + 1} = {\lim}_{n \to \infty} {n}^{3} / \left({n}^{3} + 1\right) = 1 > 0 \ne \infty$

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 $\sum {a}_{n}$ diverges, $\sum | {a}_{n} |$ will also diverge.