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 10.1 - Sequences - Exercises 10.1 - Page 570: 86


converges to $0$

Work Step by Step

Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} [\dfrac{(\ln n)^{5}}{\sqrt n}]$ Here, $ \lim\limits_{n \to \infty} [\dfrac{(\ln n)^{5}}{\sqrt n}]=\dfrac{\infty}{\infty}$ This shows that the limit has an Indeterminate form so, we will use L-Hospital's rule. This implies that $\lim\limits_{n \to \infty} [ \dfrac{5(\dfrac{\ln n)^{4}}{n})}{\dfrac{1}{2}( \sqrt n)}]=\lim\limits_{n \to \infty} \dfrac{(5)(2) (\ln n)^4}{\sqrt n}=\dfrac{\infty}{\infty}$ Now, use L-Hospital's rule again. Then, $\lim\limits_{n \to \infty} \dfrac{(5!) (2^5)}{\sqrt n}=0$ Hence, {$a_n$} is Convergent and converges to $0$.
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