# `sum_(n=1)^oo (2n^2-1)/(3n^5+2n+1)` Use the Limit Comparison Test to determine the convergence or divergence of the series.

Limit comparison test is applicable when `suma_n` and `sumb_n` are series with positive terms. If `lim_(n->oo)a_n/b_n =L` where L is a finite number and `L>0` , then either both series converge or both diverge.

Given series is `sum_(n=1)^oo(2n^2-1)/(3n^5+2n+1)`

Let the comparison series be `sum_(n=1)^oon^2/n^5=sum_(n=1)^oo1/n^3`

`a_n/b_n=((2n^2-1)/(3n^5+2n+1))/(1/n^3)`

`a_n/b_n=(n^3(2n^2-1))/(3n^5+2n+1)`

`a_n/b_n=(2n^5-n^3)/(3n^5+2n+1)`

`lim_(n->oo)a_n/b_n=lim_(n->oo)(2n^5-n^3)/(3n^5+2n+1)`

`=lim_(n->oo)(n^5(2-n^3/n^5))/(n^5(3+(2n)/n^5+1/n^5))`

`=lim_(n->oo)(2-1/n^2)/(3+2/n^4+1/n^5)`

`=2/3>0`

The comparison series `sum_(n=1)^oo1/n^3` is a p-series with `p=3`

As per p-series test `sum_(n=1)^oo1/n^p` is convergent if `p>1` and divergent if `0<p<=1`

Since the comparison series `sum_(n=1)^oo1/n^3` converges, so the series `sum_(n=1)^oo(2n^2-1)/(3n^5+2n+1)` as well ,converges by the limit comparison test.

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