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

Question

4506 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-15T04:06:37

The series:

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

is convergent.

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.