Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.7 - Strategy for Testing Series - 11.7 Exercises - Page 746: 31

Answer

Divergent

Work Step by Step

$\Sigma_{n=1}^{\infty}\frac{5^{k}}{3^{k}+4^{k}}\gt \Sigma_{n=1}^{\infty}\frac{5^{k}}{4^{k}+4^{k}}$ $=\Sigma_{n=1}^{\infty}\frac{5}{8} (\frac{5}{4})^{k-1}$ This is a geometric series with a common ratio $r=\frac{5}{2}$, hence diverging.
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