Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 735: 44


The series is divergent.

Work Step by Step

$\Sigma^{\infty}_{n=1} \ln (\frac{n}{n+1}) = \Sigma^{\infty}_{n=1}(\ln(n) - \ln(n+1)) = S$ $S_{1}= \ln(n) = \ln 1, \ln 2, \ln 3,..., \ln (\infty -1)$ $S_{2}= -\ln(n+1) = -\ln2, -\ln3,...,-\ln(\infty-1), -\ln\infty$ $S_{1} + S_{2} = \ln1 -\ln\infty = -\infty$ $\lim\limits_{n \to \infty}S=-\infty$ so $S$ diverges.
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