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: 97

Answer

$1 +\sqrt 2 $

Work Step by Step

According to the Principle of Mathematical Induction $a_n \geq 0$ this implies that there exist $n$ for all natural numbers N. Given: $a_{n+1}=2+(\dfrac{1}{a_n})$ Suppose $l=\lim\limits_{n \to \infty} a_n ;\\l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}2+(\dfrac{1}{a_n})$ Then, $l=2+\dfrac{1}{l}$ or, $l^2-2l-1=0 \implies l=1 +\sqrt 2 $ or, $1 -\sqrt 2$ As we are given that $a_n\gt 0$ for all $n \geq 1$ Thus, the limit of the sequence is $l=1 +\sqrt 2 $
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