University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Practice Exercises - Page 552: 39

Answer

Converges Absolutely

Work Step by Step

Consider $a_n=\dfrac{1}{\sqrt {n(n+1)(n+2)}}$ Since, we see that $\Sigma_{n=1}^\infty \dfrac{1}{\sqrt {n(n+1)(n+2)}} \lt \Sigma_{n=1}^\infty \dfrac{1}{\sqrt {(n)(n)(n)}}=\Sigma_{n=1}^\infty \dfrac{1}{{\sqrt [3] {n}}}$ Now, $\Sigma_{n=1}^\infty \dfrac{1}{{\sqrt [3] {n}}}=\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ Thus, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series. Hence, the given series Converges Absolutely by the direct comparison test.
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