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: 95

Answer

$5$

Work Step by Step

According to the Principle of Mathematical Induction $a_n \geq 0$. Given: $a_{n+1}=\sqrt {5a_n}$ and $a_1=5$ Consider $l=\lim\limits_{n \to \infty} a_n$ ;$l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {5a_n}$ Then, $l^2=5l$ so, $l^2-5l=0$ or, $l(l-5)=0 \implies l=0 $ or, $5$ As we are given that $a_1=5$ and $5 \gt 1$ and $a_2=\sqrt {(5)(5)} \gt 1$. Thus, the limit of the sequence is $l=5$.
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