Answer
strictly increasing
Work Step by Step
Given $$\left\{ \frac{n}{2n+1}\right\}_{1}^{\infty}$$
Since $a_n=\dfrac{n}{2n+1},\ \ a_{n+1}=\dfrac{n+1}{2n+3}$, then
\begin{align*}
a_{n+1}-a_n&=\dfrac{n+1}{2n+3}-\dfrac{n}{2n+1}\\
&=\dfrac{ (n+1)(2n+1)-n(2n+3)}{(2n+3)(2n+1)}\\
&=\dfrac{ 1}{(2n+3)(2n+1)}>0
\end{align*}
So, the sequence is eventually strictly increasing
