Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 6



Work Step by Step

Given:$a_{n}=\frac{ln n}{\sqrt n}$ A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant. $\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{ln n}{\sqrt n}$ Use the property: $a ln b=lnb^{a}$ $=\lim\limits_{n \to \infty}\frac{ln (n^{1/2})^{2}}{\sqrt n}$ $=\lim\limits_{n \to \infty}\frac{2 ln \sqrt n}{\sqrt n}$ Substitute $ \sqrt n=t$ and note that $t \to \infty$ when $n \to \infty$ $=\lim\limits_{t \to \infty}\frac{2 ln t}{t}$ We can also use L-Hospital's rule because the limit is of the form $\frac{\infty}{\infty}$ After applying L-Hospital's rule we get $=\lim\limits_{t \to \infty}\frac{\frac{2}{t}}{1}$ $=\lim\limits_{t \to \infty}\frac{2}{t}$ $=0$ Because the limits exists , the sequence converges. Hence, the given sequence converges to $0$.
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