(Solved): How do I use a comparison test to solve this? I'm confused by the direction, would I use 7/2n? U ...

How do I use a comparison test to solve this?

I'm confused by the direction, would I use 7/2n?

Use the limit comparison test to determine whether \( \sum_{n=19}^{\infty} a_{n}=\sum_{n=19}^{\infty} \frac{7 n^{3}-4 n^{2}+19}{3+2 n^{4}} \) converges or diverges. (a) Choose a series \( \sum_{n=19}^{\infty} b_{n} \) 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 \( n \geq 19 \), \( \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\lim _{n \rightarrow \infty} \) (b) Evaluate the limit in the previous part. Enter \( \infty \) as infinity and \( -\infty \) as -infinity. If the limit does not exist, enter DNE. \( \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= \)