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

Jul 29, 2015

The series converges.

#### Explanation:

The terms of $\sum \frac{3 n - 2}{{n}^{3} - 2 {n}^{2} + 11}$,

in the limit look like those of

$\sum \frac{3}{n} ^ 2 \text{ }$ -- a series we know to be convergent.

To apply the limit comparison test, evaluate

$c = {\lim}_{n \rightarrow \infty} \frac{\frac{3 n - 2}{{n}^{3} - 2 {n}^{2} + 11}}{\frac{3}{n} ^ 2}$

$= {\lim}_{n \rightarrow \infty} \left(\frac{3 n - 2}{{n}^{3} - 2 {n}^{2} + 11}\right) \cdot \left({n}^{2} / 3\right)$

$= {\lim}_{n \rightarrow \infty} \left(\frac{3 {n}^{3} - 2 {n}^{2}}{3 {n}^{3} - 6 {n}^{2} + 33}\right)$

$= {\lim}_{n \rightarrow \infty} \left(\frac{3 - \frac{2}{n}}{3 - \frac{6}{n} + \frac{33}{n} ^ 2}\right)$

$= 1$

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

$\sum \frac{3}{n} ^ 2 \text{ }$ converges, so we conclude that

$\sum \frac{3 n - 2}{{n}^{3} - 2 {n}^{2} + 11} \text{ }$ also converges.

Note

We could have used the series $\sum \frac{1}{n} ^ 2$ for the limit comparison, (We would have gotten $c = 3$.)

But I think it is more clear that the terms eventually behave like $\frac{3}{n} ^ 2$