Answer
{$a_{n}$} is a decreasing sequence and is bounded above.
Work Step by Step
$a_{n} = \frac{n}{n^{2}+1}$
$f(x)=\frac{x}{x^{2}+1}$
$f'(x) = \frac{(x^{2}+1)(1)-x(2x)}{(x^{2}+1)^{2}}$
$=\frac{1-x^{2}}{(x^{2}+1)^{2}} \lt 0$ when $x \gt 1$
Therefore $f$ is decreasing on $(1, \infty)$ and so {$a_{n}$} is a decreasing sequence.
{$a_{n}$} is decreasing for $n \geq 1$
$a_{n} \leq a_{1}$
$ \frac{n}{n^{2}+1} \leq \frac{1}{2}$
$a_{n} \leq \frac{1}{2}$
Thus {$a_{n}$} is bounded above.