Answer
Diverges
Work Step by Step
Let $u_n=\sqrt {\dfrac{ n+1}{n^2+2}}$ and $v_n=\dfrac{ 1}{\sqrt n}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\sqrt {\dfrac{( n+1)}{(n^2+2)}}}{\dfrac{ 1}{\sqrt n}}$
$\implies \lim\limits_{n \to \infty} \dfrac{\sqrt {n^2+n}}{\sqrt {n^2+2}}=\lim\limits_{n \to \infty} \dfrac{\sqrt {1+1/n}}{\sqrt {1+2/n^2}}=1$
But $v_n=\dfrac{ 1}{\sqrt n}=\dfrac{ 1}{n^{1/2}}$ is a divergent series
Thus, the series Diverges due to the limit comparison test.