Answer
Strictly increasing
Work Step by Step
Given $$\left\{1-\frac{1}{n}\right\}_{1}^{\infty}$$
Since $a_n=1-\dfrac{1}{n},\ \ a_{n+1}=1-\dfrac{1}{n+1}$, then
\begin{align*}
a_{n+1}-a_n&=1-\dfrac{1}{n+1}-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 increasing