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

1 Answer
Jan 15, 2017

The series:

sum_(n=1)^oo 1/(n(n^2+1))

is convergent.

Explanation:

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

sum_(n=0)^oo a_n and sum_(n=0)^oo b_n

where the limit:

lim_(n->oo) a_n/b_n

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

Given:

a_n = 1/(n(n^2+1))

we can choose:

b_n = 1/(n^3)

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

lim _(n->oo) a_n/b_n = (1/(n(n^2+1)))/(1/(n^3)) =n^3/(n(n^2+1))= n^3/(n^3+n) = 1

So the series:

sum_(n=1)^oo 1/(n(n^2+1))

is convergent.