Answer
strictly increasing
Work Step by Step
Given $$\left\{\frac{n}{2n+1}\right\}_{1}^{\infty} $$
Consider $$f(x)=\frac{x}{2x+1}$$ Then
\begin{align*}
f'(x)&= \frac{\frac{d}{dx}\left(x\right)\left(2x+1\right)-\frac{d}{dx}\left(2x+1\right)x}{\left(2x+1\right)^2}\\
&=\frac{1\cdot \left(2x+1\right)-2x}{\left(2x+1\right)^2}\\
&=\frac{1}{\left(2x+1\right)^2}>0
\end{align*}
Then, the given sequence is strictly increasing