Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 75


$ln(1+\frac{3}{5})$ or $ln(8)-ln(5)$

Work Step by Step

Given: $\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{3^{n}}{5^{n}(n)}$ $=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{3^{n}(1/5^{n})}{n}$ $=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{(\frac{3}{5})^{n}}{n}$ As we know: $ln(1+x)=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{x^{n}}{n}$ Thus, $ln(1+\frac{3}{5})=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{(\frac{3}{5})^{n}}{n}$ or $ln(1+\frac{3}{5})=ln(\frac{8}{5})=ln(8)-ln(5)$ Hence, $ln(1+\frac{3}{5})$ or $ln(8)-ln(5)$
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