Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 5

Answer

The series $\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{ (1.1)^n}$ converges absolutely.

Work Step by Step

We have the absolute series $$\sum_{n=0}^{\infty} |\frac{(-1)^{n-1}}{ (1.1)^n}|=\sum_{n=0}^{\infty} \frac{1}{ (1.1)^n}$$ which is a convergent geometric series as $\frac{1}{1.1}\lt 1$. Thus, the alternating series $\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{ (1.1)^n}$ converges absolutely.
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