## Thomas' Calculus 13th Edition

The sequence $a_n$ is monotonic and bounded.
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 $n \in N$ we can see that $\dfrac{3n+1}{n+1} \lt \dfrac{3n+4}{n+2}$ for all $n \in N$ This means that $a_n \lt a_{n+1}$ for all $n \in N$ .Thus $a_n$ increases and is monotonic. Now, to check whether the sequence is bounded or not we will consider $0 \lt \dfrac{3n+1}{n+1}=2(\dfrac{n}{n+1})+1$ Also, $\dfrac{n}{n+1} \lt 1$ for all $n \in N$ This implies $0 \lt (2)(1) +1=3$ for all $n \in N$ so, $a_n$ is bounded. Therefore, we can see that the sequence $a_n$ is monotonic and bounded.