Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.2 - Series - 11.2 Exercises: 31

Answer

Converges 9

Work Step by Step

$\sum ^{\infty }_{n=1}3^{n+1}4^{-n}=3\sum ^{\infty }_{n=1}\left( \dfrac {3}{4}\right) ^{n}\Rightarrow a_{1}=3\times \left( \dfrac {3}{4}\right) ^{1}=\dfrac {9}{4};r=\dfrac {3}{4} < 1\Rightarrow $ series converges $ S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {9}{4}}{1-\dfrac {3}{4}}=9$
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