(Solved): how to solve? Use the limit comparison test to determine whether \( \sum_{n=16}^{\infty} a_{n}=\sum_ ...

Use the limit comparison test to determine whether \( \sum_{n=16}^{\infty} a_{n}=\sum_{n=16}^{\infty} \frac{6 n^{3}-4 n^{2}+16}{8+2 n^{4}} \) converges or diverges. (a) Choose a series \( \sum_{n=16}^{\infty} b_{n} \) 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 \( n \geq 16 \), \[ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\lim _{n \rightarrow \infty} \] (b) Evaluate the limit in the previous part. Enter \( \infty \) as infinity and \( -\infty \) as -infinity. If the limit does not exist, enter DNE. \[ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= \] (c) By the limit comparison test, does the series converge, diverge, or is the test inconclusive?