Answer
{$a_{n}$} is a bounded sequence.
Work Step by Step
$a_{n}=\frac{1}{2n+3}$
$a_{n+1}=\frac{1}{2(n+1)+3}=\frac{1}{2n+5}$
Since $2n+5 \gt 2n+3$
$\frac{1}{2n+5} \lt \frac{1}{2n+3}$ for all $n \geq 1$
Then $a_{n+1} \lt a_{n}$ for all $n \geq 1$
Since $n \geq 1$
Then $2n \geq 2$
$2n+3 \geq 5$
$\frac{1}{2n+3} \leq \frac{1}{5}$
$a_{n} \leq \frac{1}{5}$
Therefore {$a_{n}$} is a bounded sequence.
