Answer
$\left\{ {\frac{{3{n^2}}}{{{n^2} + 2}}} \right\}$ is increasing for $n>0$.
The upper bound is 3.
Work Step by Step
Let $f$ be a function such that $f\left( x \right) = \frac{{3{x^2}}}{{{x^2} + 2}}$. So, $f\left( n \right) = {a_n}$.
The derivative of $f$ is $f'\left( x \right) = \frac{{12x}}{{{{\left( {{x^2} + 2} \right)}^2}}}$.
Since $f'$ is positive for $x>0$, $f\left( x \right)$ is increasing in this interval. It follows that ${a_n} = f\left( n \right)$ is increasing for $n>0$.
To find an upper bound, we notice that
${a_n} = \frac{{3{n^2}}}{{{n^2} + 2}} \le \frac{{3{n^2}}}{{{n^2} + 2}} + \frac{6}{{{n^2} + 2}}$ ${\ \ \ \ }$ for all $n$.
So,
${a_n} \le \frac{{3{n^2} + 6}}{{{n^2} + 2}}$,
${a_n} \le \frac{{3\left( {{n^2} + 2} \right)}}{{{n^2} + 2}} = 3$.
Since ${a_n} \le 3$ for all $n$, by definition, the sequence $\left\{ {{a_n}} \right\}$ is bounded above by 3.
