# How do you use the limit comparison test to determine if Sigma 1/(n(n^2+1)) from [1,oo) is convergent or divergent?

Jan 15, 2017

The series:

${\sum}_{n = 1}^{\infty} \frac{1}{n \left({n}^{2} + 1\right)}$

is convergent.

#### Explanation:

The limit comparison test states that if we have two series with positive terms:

${\sum}_{n = 0}^{\infty} {a}_{n}$ and ${\sum}_{n = 0}^{\infty} {b}_{n}$

where the limit:

${\lim}_{n \to \infty} {a}_{n} / {b}_{n}$

exists and is finite, then if one series converges, the other is also convergent.

Given:

${a}_{n} = \frac{1}{n \left({n}^{2} + 1\right)}$

we can choose:

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

that we know is convergent based on the p-series test and verify that:

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = \frac{\frac{1}{n \left({n}^{2} + 1\right)}}{\frac{1}{{n}^{3}}} = {n}^{3} / \left(n \left({n}^{2} + 1\right)\right) = {n}^{3} / \left({n}^{3} + n\right) = 1$

So the series:

${\sum}_{n = 1}^{\infty} \frac{1}{n \left({n}^{2} + 1\right)}$

is convergent.