Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises - Page 743: 16


Absolutely convergent

Work Step by Step

Given: $a_{n}=\Sigma_{n=1 }^{\infty}\frac{n^{10}}{(-10)^{n+1}}$ $a_{n+1}=\Sigma_{n=1 }^{\infty}\frac{(n+1)^{10}}{(-10)^{n+2}}$ $L=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(n+1)^{10}}{(-10)^{n+2}} }{\frac{n^{10}}{(-10)^{n+1}}}|$ $=\lim\limits_{n \to \infty}|(\frac{n+1}{n})^{10}\frac{1}{(-10)}|$ $=\lim\limits_{n \to \infty}(\frac{n+1}{n})^{10}\frac{1}{(10)}$ $=\lim\limits_{n \to \infty}(1+\frac{1}{n})^{10}\frac{1}{(10)}$ $=(1+0)^{10}\frac{1}{(10)}$ $=\frac{1}{10}$ $=\frac{1}{10}\lt 1$ Hence, the series converges absolutely by ratio test.
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