Answer
Diverges
Work Step by Step
Let $u_n=\dfrac{n+2}{n^2-n}$
When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing.
$\implies \dfrac{1}{ n-1} \geq \dfrac{1}{ n}$
so, $u_n \geq \dfrac{1}{n}$
Here, $\Sigma_{n=2}^\infty \dfrac{1}{n}$ diverges by the p-series with $p=1$
Hence, the series diverges by the comparison test.