Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 557: 77



Work Step by Step

Given $$\sum_{n=2}^{\infty} \frac{4 n^{2}+15 n}{3 n^{4}-5 n^{2}-17}$$ Compare with $\sum \frac{4}{3n^2}$, which is a convergent p-series with $p>1$. Using the limit comparison test gives: \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{4 n^{2}+15 n}{3 n^{4}-5 n^{2}-17} \cdot \frac{3 n^{2}}{4}\\ &=\lim _{n \rightarrow \infty} \frac{12 n^{4}+45 n^{3}}{12 n^{4}-20 n^{2}-68}\\ &=\lim _{n \rightarrow \infty} \frac{12+45 / n}{12-20 / n^{2}-68 / n^{4}}\\ &=1 \end{align*} Thus $\sum_{n=2}^{\infty} \frac{4 n^{2}+15 n}{3 n^{4}-5 n^{2}-17} $ also converges.
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