Answer
{$a_{n}$} is a decreasing sequence.
{$a_{n}$} is bounded above $\frac{1}{e}$ and bounded below by 0.
Work Step by Step
$a_{n} = ne^{-n}$
$f(x) =xe^{-x}$
$f'(x)=-xe^{-x}+e^{-x}$
$=(1-x)e^{-x} \leq 0$ for all $x \geq 1$
This is a decreasing function, so
$a_{n+1} \lt a_{n}$
therefore {$a_{n}$} is a decreasing sequence
For $n \geq 1$
$a_{n} \leq a_{1}$
$a_{n} \leq \frac{1}{e}$
since for $n \geq 1$, $a_{n} \geq 0$
Therefore {$a_{n}$} is bounded above $\frac{1}{e}$ and bounded below by 0.
