Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.4 The Comparison Tests - 11.4 Exercises - Page 771: 7

Answer

$\displaystyle\frac{9^n}{3+10^n}$ is convergent

Work Step by Step

Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{9^n}{3+10^n}\] \[\Rightarrow a_n=\frac{9^n}{3+10^n}\] \[\frac{9^n}{3+10^n}\leq\frac{9^n}{10^n}=\left(\frac{9}{10}\right)^n\] Also $\displaystyle\sum_{n=1}^{\infty}\left(\frac{9}{10}\right)^n$ is geometric series with common ratio $\displaystyle\frac{9}{10}(<1)$ so $\displaystyle\sum_{n=1}^{\infty}\left(\frac{9}{10}\right)^n$ is convergent. By using Comparison Test, $\displaystyle\frac{9^n}{3+10^n}$ is convergent. Hence, $\displaystyle\frac{9^n}{3+10^n}$ is convergent.
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