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: 95



Work Step by Step

Since, we are given that $a_{n+1}=\sqrt {5a_n}$ and $a_1=5$ Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {5a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$ Thus, $l=5l$ or, $l^2-5l=0$ or, $l=0 $ or, $5$ Since, we have $a_1=5$, which is $5 \gt 1$ and $a_2=\sqrt {(5)(5)} \gt 1$ As per the Principal of Mathematical Induction $a_n \geq 1$, which means that there exists n for all natural numbers N. Hence, the limit of the sequence is $5$
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