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 489: 111


The sequence $a_n$ is monotonic and bounded.

Work Step by Step

Our aim is to find out if the sequence $a_n$ is monotonic and bounded. In order to this, we will consider $a_n=\dfrac{3n+1}{n+1}$ for all $n \in N$ Also, $a_{n+1}=\dfrac{3(n+1)+1}{(n+1)+1}=\dfrac{3n+4}{n+2}$ for all $n \in N$ Thus, we find that $\dfrac{3n+1}{n+1} \lt \dfrac{3n+4}{n+2}$ for all $n \in N$ This implies that $a_n \lt a_{n+1}$ for all $n \in N$ and so, $a_n$ is increasing and thus, monotonic. Now, we will check for a bounded sequence. In order to this we wiil take $0 \lt \dfrac{3n+1}{n+1}=2(\dfrac{n}{n+1})+1$ and $\dfrac{n}{n+1} \lt 1$ for all $n \in N$ or, $0 \lt (2)(1) +1=3$ for all $n \in N$ Thus, $a_n$ is bounded. Hence, the sequence $a_n$ is monotonic and bounded.
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