Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 717: 64

Answer

The series is divergent.

Work Step by Step

$\Sigma^{\infty}_{n=1}a_{n} = \Sigma^{\infty}_{n=1}\ln (1+\frac{1}{n}) = \Sigma^{\infty}_{n=1} \ln (\frac{n+1}{n})$ $s_{n}=\ln2+\ln\frac{3}{2}+\ln\frac{4}{3}+\ln\frac{5}{4}+\ln\frac{6}{5}+\ln\frac{7}{6}+...+\ln\frac{n}{n-1}+\ln\frac{n+1}{n}$ $=\ln2+(\ln3-\ln2)+(\ln4-\ln3)+...+(\ln(n)-\ln(n-1))+(\ln(n+1)-\ln(n))$ $s_{n} = \ln(n+1)$ $s_{n} ā†’ \infty$ as $nā†’\infty$ So {$s_{n}$} is divergent.
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