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