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: 28

Answer

Divergent

Work Step by Step

$\sum_{n=2}^{\infty}a_{n}=\sum_{n=2}^{\infty}(\dfrac{-2n}{n+1})^{5n}$ $|a_{n}|=(\frac{2n}{n+1})^{5n}$ $\lim\limits_{n \to \infty} \sqrt[n] |a_{n}|=\lim\limits_{n \to \infty} \sqrt[n] {(\frac{2n}{n+1})^{5n}}$ $=\lim\limits_{n \to \infty} {(\frac{2n}{n+1})^{5}}$ $=\lim\limits_{n \to \infty} {(\frac{2n}{n})^{5}}$ $=2^{5}$ $=32 \gt 1$ The series is divergent.
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