Answer
strictly decreasing
Work Step by Step
Given $$\left\{ \frac{n}{4n-1}\right\}_{1}^{\infty}$$
Since $a_n=\dfrac{n}{4n-1},\ \ a_{n+1}=\dfrac{n+1}{4n+3}$, then
\begin{align*}
a_{n+1}-a_n&=\dfrac{n+1}{4n+3}-\dfrac{n}{4n-1}\\
&=\dfrac{ (4n-1)(n+1)-n(4n+3)}{(4n+3)(4n-1)}\\
&=\dfrac{ -1}{(4n+3)(4n-1)}<0
\end{align*}
So, the sequence is eventually strictly decreasing
