Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 21


the series $\sum_{n=1}^{\infty} \frac{ 1}{3n^4+12n}$ converges

Work Step by Step

Use the limit comparison test with the convergent p-series $\sum_{n=1}^{\infty} \frac{1}{n^4}$ Since we have $$L=\lim_{n\to \infty } \frac{1/(3n^4+12n)}{1/n^4}=\lim_{n\to \infty } \frac{ n^4}{3n^4+12n}\\ =\lim_{n\to \infty } \frac{ 1}{3+12/n^3}=1/3\gt0$$ then the series $\sum_{n=1}^{\infty} \frac{ 1}{3n^4+12n}$ converges because the p-series $\sum_{n=1}^{\infty} \frac{ 1}{ n^4 }$, $p=4\gt1$, converges.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.