Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 537: 25



Work Step by Step

Since we have $$\lim_{n\to \infty}a_n=\lim_{n\to \infty} \ln\left(\frac{12n+2}{-9+4n}\right) =\lim_{n\to \infty}\ln\left(\frac{12+(2/n)}{-(9/n)+4}\right)\\ =\ln\left(\frac{12+0}{0+4}\right)=\ln3,$$ then the sequence $a_n$ converges to $\ln3$.
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