Answer
divergent
Work Step by Step
Given
$$ \sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{2+n }$$
Use the Limit Comparison Test with $a_{n}=\dfrac{\sqrt{n+1}}{2+n }$ and $b_{n}=\dfrac{1}{ \sqrt{n}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{\sqrt{n(n+1)}}{2+n }\\
&=\lim _{n \rightarrow \infty} \frac{\sqrt{ 1+1/n}}{2/n+1 }\\&=1
\end{align*}
since $\displaystyle\sum_{n=1}^{\infty} \frac{1}{ \sqrt{n}}$ is divergent $(p-\text { series } p<1),$ then $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{2+n }$ also divergent