Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.7 Strategy for Testing Series - 11.7 Exercises - Page 786: 11



Work Step by Step

$$\sum_{n=1}^\infty\left(\frac{1}{n^3}+\frac{1}{3^n}\right)=\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)+\sum_{n=1}^\infty\left(\frac{1}{3^n}\right)$$Split the series and solve one at a time. The series $\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)$ is a $p$-series with $p=3>1$ Thus, the series $\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)$ is convergent. The series $\sum_{n=1}^\infty\left(\frac{1}{3^n}\right)=\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$ $\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$ is a Geometric Series with $[|r|=\frac{1}{3}<1]$ Thus the series $\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$ is convergent. The sum of two convergent series is also convergent, so the series $\sum_{n=1}^\infty\left(\frac{1}{n^3}+\frac{1}{3^n}\right)$ is convergent.
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