University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 98


$\dfrac{1 +\sqrt 5}{2} $

Work Step by Step

Since, we are given that $a_{n+1}=\sqrt {1+a_n}$ Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {1+a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$ Thus, $l=\sqrt {1+l}$ or, $l^2-l-1=0$ or, $l=\dfrac{1 +\sqrt 5}{2} , \dfrac{1 -\sqrt 5}{2}$ Since, we have $a_n\gt 0$ for all $n \geq 1$ Hence, the limit of the sequence is $\dfrac{1 +\sqrt 5}{2} $
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