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#