## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - Chapter Review Exercises - Page 592: 86

Diverges

#### Work Step by Step

Given $$\sum_{n=2}^{\infty} \frac{n^{3}-2 n^{2}+n-4}{2 n^{4}+3 n^{3}-4 n^{2}-1}$$ Compare with the divergent series $\displaystyle\sum_{n=2}^{\infty} \frac{1}{2 n }$; then by using the limit comparison test \begin{align*} \lim_{n\to \infty } \frac{a_n}{b_n}&= \lim_{n\to \infty } \frac{n^{4}-2 n^{3}+n^2-4n}{2 n^{4}+3 n^{3}-4 n^{2}-1}\\ &= \lim_{n\to \infty } \frac{1-2/n+1/n^2-4/n}{2+3/n-4 /n^{2}-1/n^4}\\ &=\frac{1}{2}\gt 0 \end{align*} Thus the given series also diverges.

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