How do you use the limit comparison test to determine if $Sigma (2n^2-1)/(3n^5+2n+1)$ from $[1,oo)$ is convergent or divergent?

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Answer 1 · 2017-01-27T13:30:55

The series:

$sum_(n=1)^oo (2n^2-1)/(3n^5+2n+1)$

is convergent.

Given:

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

we have to find a convergent series $sum_(n=1)^oo b_n$ such that $a_n < b_n$

We can now consider that:

$2n^2-1 < 2n^2$

and

$(3n^5+2n+1) > 3n^5$

so that:

$(2n^2-1)/(3n^5+2n+1) < (2n^2)/(3n^5)$

$(2n^2-1)/(3n^5+2n+1) < 1/n^3$

The series $sum_(n=1)^oo 1/n^3$ is known to be convergent based on the p-series test, therefore also the series:

$sum_(n=1)^oo (2n^2-1)/(3n^5+2n+1)$

is convergent.

How do you use the limit comparison test to determine if Sigma (2n^2-1)/(3n^5+2n+1)Sigma (2n^2-1)/(3n^5+2n+1) from [1,oo)[1,oo) is convergent or divergent?