Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 16



Work Step by Step

If a series is convergent, then $\lim\limits_{n \to \infty}a_{n}=0$ $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}ln(\frac{n}{3n+1})$ $=ln\frac{1}{3}\ne 0$ Hence, the series is divergent since the limit does not equal $0$, by the Test for Divergence.
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