Answer
Diverges
Work Step by Step
Consider $a_n=\sqrt {\dfrac{ n+1}{n^2+2}}$ and $b_n=\dfrac{ 1}{\sqrt n}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\sqrt {\dfrac{ n+1}{n^2+2}}}{\dfrac{ 1}{\sqrt n}}$
Thus, we have $ =\lim\limits_{n \to \infty} \dfrac{\sqrt {n^2+n}}{\sqrt {n^2+2}}$
or, $=1$
Hence, the series Diverges (as $b_n=\dfrac{ 1}{\sqrt n}$ is a divergent series) due to the limit comparison test.