#### Answer

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.