Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 548: 46


See the proof below.

Work Step by Step

We have $$ \sum_{n=0}^{\infty} \frac{9^{n}+2^{n}}{5^{n}}=\sum_{n=0}^{\infty} \frac{9^{n}}{5^{n}}+\sum_{n=0}^{\infty} \frac{2^{n}}{5^{n}} $$ One can see that the series $\sum_{n=0}^{\infty} \frac{9^{n}}{5^{n}}$ is a geometric series with $r=\frac{9}{5}\gt 1$. Thus, it is divergent and hence the given series is divergent.
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