Answer
Divergent
Work Step by Step
$a_n=\Sigma \frac{\sqrt{1+n}}{2+n}$
$b_n=\Sigma \frac{\sqrt{n}}{n}=\Sigma \frac{1}{\sqrt{n}}$
$\lim\limits_{n \to \infty} \frac{a_n}{b_n}=\lim\limits_{n \to \infty}\Sigma \frac{\frac{\sqrt{1+n}}{2+n}}{\frac{1}{\sqrt{n}}}=\lim\limits_{n \to \infty}\frac{\sqrt{1+n}}{2+n}\frac{\sqrt{n}}{1}=\lim\limits_{n \to \infty}\frac{\sqrt{n+n^2}}{2+n}=\lim\limits_{n \to \infty}\frac{\sqrt{\frac{1}{n}+1}}{\frac{2}{n}+1}=\frac{\sqrt{0+1}}{0+1}=1\ne0\ne\infty$
So the series is divergent.