Answer
$\{a_{n}\}$ is monotonic and bounded
Work Step by Step
Bounded Sequences:
1. A sequence $\{a_{n}\}$ is bounded above when there is a real number $M$ such that $a_{n}\leq M$ for all $n$.
The number $M$ is called an upper bound of the sequence.
2. A sequence $\{a_{n}\}$ is bounded below when there is a real number $N$ such that $N\leq a_{n}$ for all $n$.
The number $N$ is called a lower bound of the sequence.
3. A sequence $\{a_{n}\}$ is bounded when it is bounded above and bounded below.
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Let $f(x)=\displaystyle \frac{3x}{x+2}$ , and $f(n)=a_{n}.$
The derivative of f, (quotient rule)
$f^{\prime}(x)=\displaystyle \frac{3(x+2)-3x(1)}{(x+2)^{2}}=\frac{6}{(x+2)^{2}} $ is always positive for positive x,
so, $f$ is increasing
so, $\{a_{n}\}$ is increasing ( monotonic).
Checking for monotony:
$a_{n}=\displaystyle \frac{3n}{n+2} < \frac{3n+6}{n+2} = \frac{3(n+2)}{n+2}=3$
so
$0< a_{n} < 3$, meaning that $\{a_{n}\}$ is bounded
$\{a_{n}\}$ is monotonic and bounded
Graph (see below) confirms our conclusions.