Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 26

Answer

Converges

Work Step by Step

Let $u_n=\dfrac{1}{\sqrt{n^3+2}}$ and $v_n=\dfrac{1}{ \sqrt{n^3}}$ Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{\sqrt{n^3+2}}}{1/\sqrt{n^3}}$ $ \implies \lim\limits_{n \to \infty} \sqrt{\dfrac{n^3}{(n^3+2)}}=1$ Here, $v_n=\Sigma_{n=1}^\infty \dfrac{1}{ \sqrt{n^3}}=\dfrac{1}{ n^{3/2}}$ is a convergent p-series with $p=\dfrac{3}{2}$ Thus, the given series converges by the limit comparison test.
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