Answer
Convergent
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{\sqrt{n+2}}{2 n^{2}+n+1}$$
Use the Limit Comparison Test with $a_n =\dfrac{\sqrt{n+2}}{2 n^{2}+n+1}$ and $b_n=\dfrac{1}{n^{3/2}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^{3/2}\sqrt{n+2}}{2 n^{2}+n+1}\\
&=\lim _{n \rightarrow \infty} \frac{n^{3/2}/n^{2}\sqrt{n+2}}{2 n^{2}/n^{2}+n/n^{2}+1/n^{2}}\\
&=1/2
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ is convergent ( $p$ -series with $p=\frac{3}{2}>1$) , then $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n+2}}{2 n^{2}+n+1}$ is also convergent.