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

##### 1 Answer

Jul 29, 2015

The series converges.

#### Explanation:

The terms of

in the limit look like those of

To apply the limit comparison test, evaluate

# = 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

**Note**

We could have used the series

But I think it is more clear that the terms eventually behave like