Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - Review Exercises - Page 677: 50

Answer

Diverges

Work Step by Step

Here, we have the given series as $\Sigma_{n=1}^{\infty} \dfrac{1}{ \sqrt {n^3+3n}}$ Let us consider that $a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{ \sqrt {n^3+3n}}$ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{ \sqrt{n}}$ We can see that $a_n \leq b_n$ and $b_n$ shows a p-series with $p=\dfrac{1}{2}$ and $p=\dfrac{1}{2} \lt 1$ This implies that the series $b_n$ by the p-series test. Hence, the given series also diverges by the direct comparison test.
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