Answer
Strictly decreasing
Work Step by Step
Given $$\left\{\frac{1}{n}\right\}_{1}^{\infty}$$
Since $a_n=\dfrac{1}{n},\ \ a_{n+1}=\dfrac{1}{n+1}$, then
\begin{align*}
a_{n+1}-a_n&=\dfrac{1}{n+1}-\dfrac{1}{n }\\
&=\dfrac{n-n-1}{n(n+1)}\\
&=\frac{-1}{n^2+n}<0
\end{align*}
So, the sequence is eventually strictly decreasing