How do you use the limit comparison test for $sum( 3n-2)/(n^3-2n^2+11)$ as n approaches infinity?

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Answer 1 · 2015-07-29T16:29:17

The series converges.

The terms of $sum(3n-2)/(n^3-2n^2+11)$ ,

in the limit look like those of

$sum3/n^2" "$ -- a series we know to be convergent.

To apply the limit comparison test, evaluate

$c = lim_(nrarroo)((3n-2)/(n^3-2n^2+11))/(3/n^2)$

$= lim_(nrarroo)((3n-2)/(n^3-2n^2+11))* (n^2/3)$

$= lim_(nrarroo)((3n^3-2n^2)/(3n^3-6n^2+33))$

$= lim_(nrarroo)((3-2/n)/(3-6/n+33/n^2))$

$=1$

Because $c$ is positive and finite, the series either both converge or both diverge.

$sum3/n^2" "$ converges, so we conclude that

$sum(3n-2)/(n^3-2n^2+11)" "$ also converges.

Note

We could have used the series $sum1/n^2$ for the limit comparison, (We would have gotten $c = 3$ .)

But I think it is more clear that the terms eventually behave like $3/n^2$

How do you use the limit comparison test for sum( 3n-2)/(n^3-2n^2+11)sum( 3n-2)/(n^3-2n^2+11) as n approaches infinity?