University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.5 - Absolute Convergence; The Ratio and Root Tests - Exercises - Page 516: 23

Answer

Converges

Work Step by Step

Consider $a_n=\dfrac{2+(-1)^n}{1.25^{n}}$ Here, $\Sigma_{n=1}^\infty \dfrac{2+(-1)^n}{1.25^{n}} \leq \Sigma_{n=1}^\infty \dfrac{3}{1.25^{n}}=3 \Sigma_{n=1}^\infty (\dfrac{1}{1.25})^{n} $ we see that $3 \Sigma_{n=1}^\infty (\dfrac{1}{1.25})^{n} $ is a geometric convergent series. Hence, the given series converges by the direct comparison test.
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