University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 40



Work Step by Step

Since, we have $\Sigma_{n=1}^\infty \dfrac{2^n+3^n}{3^{n}+4^n} \leq \Sigma_{n=1}^\infty \dfrac{3^n+3^n}{3^{n}+4^n} \leq \Sigma_{n=1}^\infty\dfrac{3^{n}+3^n}{4^n}$ and $\Sigma_{n=1}^\infty\dfrac{3^{n}+3^n}{4^n}= \Sigma_{n=1}^\infty 2(\dfrac{3}{4})^n$ The series $\Sigma_{n=1}^\infty 2(\dfrac{3}{4})^n$ is a convergent geometric series with common ratio $r \lt 1$. Thus, the given series is convergent as well.
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