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*}
\frac{a_{n+1}}{a_n}&=\dfrac{n+1}{2n+3}\frac{2n+1}{n}\\
&= \dfrac{2n^2+3n+1}{2n^2+3n}\\
&=1+\dfrac{ 1}{2n^2+3n}>1
\end{align*}
so, the sequence is eventually strictly increasing