Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 812: 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 $ln(8)-ln(5)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.