Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.1 Exercises - Page 724: 51

Answer

converges to $\frac{\pi}{2}$

Work Step by Step

Given: $a_n=arctan(ln(n))$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}arctan(ln(n))$ Now, $\lim\limits_{n \to \infty}(ln(n))=\infty$ Therefore, $\lim\limits_{n \to \infty}arctan(ln(n))=\lim\limits_{n \to \infty}arctan(\infty)=\frac{\pi}{2}$ Hence, the sequence converges to $\frac{\pi}{2}$.
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