Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.2 Monotone Sequences - Exercises Set 9.2 - Page 613: 23

Answer

Eventually strictly increasing.

Work Step by Step

We have: $a_n=\dfrac{n !}{3^n}$ and $a_{n+1} =\dfrac{(n+1)!}{3^{n+1}}$ Then $\dfrac{a_{n+1}}{a_n}=\dfrac{\dfrac{(n+1)!}{3^{n+1}}}{\dfrac{n !}{3^n}}\\=\dfrac{ 3^n (n+1)!}{3^{n+1}n!}\\=\dfrac{n+1}{3} \gt 1 \ for \ n \gt 2$ This implies that the given sequence is eventually strictly increasing.
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