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 10.3 - The Integral Test - Exercises 10.3 - Page 586: 27

Answer

Divergent

Work Step by Step

Let us consider $a_n=\lim\limits_{n \to \infty} \dfrac{\sqrt n}{\ln n}$ Here, $\lim\limits_{n \to \infty} \dfrac{\sqrt n}{\ln (n)}=\dfrac{\infty}{\infty}$ Thus, the limit has an indeterminate form so we will take the help of L-Hospital's rule. Now, $\lim\limits_{n \to \infty} \dfrac{\dfrac{1}{2 \sqrt n}}{\dfrac{1}{n}} =\lim\limits_{n \to \infty} \dfrac{(\sqrt n)}{(2)}=\infty $ Thus, the series is divergent.
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