Answer
strictly decreasing
Work Step by Step
Given $$\left\{ ne^{-n}\right\}_{1}^{\infty}$$
Since $a_n=\dfrac{ n}{e^n },\ \ a_{n+1}= \dfrac{ n+1 }{e^{n+1}}$, then
\begin{align*}
\frac{a_{n+1}}{a_n}&=\dfrac{ n+1 }{e^{n+1}}\dfrac{e^n }{ n}\\
&= \dfrac{n+1}{ne}\\
&=\frac{1}{e}\left(1+\frac{1}{n}\right)\\
&<1
\end{align*}
So, the sequence is eventually strictly decreasing
