University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 2

Answer

Divergent

Work Step by Step

Consider $f(x)=\dfrac{1}{x^{0.2}}$. This is positive, continuous and decreasing for $x \geq 1$ Now, take the integral test to find the convergence and divergence for the sequence. We have $\int_1^\infty \dfrac{1}{x^{0.2}} dx= \lim\limits_{a \to \infty} \int_1^a \dfrac{1}{x^{0.2}} dx$ or, $ \lim\limits_{a \to \infty} [1.25x^{0.8}]_1^a=\lim\limits_{a \to \infty} [1.25a^{0.8}-1.25]=\infty$ Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^{0.2}}$ is Divergent
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