## Calculus 8th Edition

{$a_{n}$} is a bounded sequence.
$a_{n}=\frac{1}{2n+3}$ $a_{n+1}=\frac{1}{2(n+1)+3}=\frac{1}{2n+5}$ 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.