Use the limit comparison test to determine whether ${\displaystyle \underset{n=12}{\overset{}{}}}{a}_n={\displaystyle \underset{n=12}{\overset{}{}}}\frac{6n^3-9n^2+12}{3+4n^4}$ converges or

diverges.

$($a$)$ Choose a series ${\displaystyle \underset{n=12}{\overset{}{}}}{b}_n$ with terms of the form $b_n=\frac{1}{n^p}$ and apply the limit comparison test. Write your

answer as a fully simplified fraction. For $n12,$

$\underset{n}{lim}\frac{a_n}{b_n}=\underset{n}{lim}$

$($b$)$ Evaluate the limit in the previous part. Enter $$ as infinity and $-$ as $-$infinity. If the limit does not exist,

enter DNE.

$\underset{n}{lim}\frac{a_n}{b_n}=$

$($c$)$ By the limit comparison test, does the series converge, diverge, or is the test inconclusive?