(Solved): Use the limit comparison test to determine whether n=19an=n=192+4n49n35n2+1 ...

Use the limit comparison test to determine whether ${?}_{n=19}^{?}{a}_n={?}_{n=19}^{?}\frac{9n^3?5n^2+19}{2+4n^4}$ converges or diverges. (a) Choose a series ${?}_{n=19}^{?}{b}_n$ 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 $n?19$, ${\lim }_{n??}\frac{a_n}{b_n}={\lim }_{n??}$ (b) Evaluate the limit in the previous part. Enter $?$ as infinity and $??$ as -infinity. If the limit does not exist, enter $DNE$. ${\lim }_{n??}\frac{a_n}{b_n}=$ (c) By the limit comparison test, does the series converge, diverge, or is the test inconclusive?