Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{n+2}{n^2-n}$
When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase.
Thus, $\dfrac{1}{ n-1} \geq \dfrac{1}{ n}$
we get $a_n \geq \dfrac{1}{n}$
Here, $\Sigma_{n=2}^\infty \dfrac{1}{n},p=1$, a divergent p-series.
Thus the series diverges by the comparison test.
