Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises: 49


$a_{n}$ = $\ln(2n^{2} +1)$ - $\ln (n^{2}+1)$ = $\ln2$

Work Step by Step

Use the logarithm quotient rule, we have $a_{n}$= $\ln\frac{(2n^{2}+1)}{n^{2}+1}$. $\lim\limits_{n \to \infty} $$\ln\frac{(2n^{2}+1)}{n^{2}+1}$ = ln $\lim\limits_{n \to \infty}$ $\frac{2n^{2}+1}{n^{2}+1}$. Then we divide the top and bottom by $n^{2}$. We have $\ln$$\lim\limits_{n \to \infty}$ $\frac{2+\frac{1}{n^{2}}}{1+\frac{1}{n^{2}}}$ = $\ln\frac{2+0}{1+0}$ = $\ln2$.
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