Finding series convergence by using the Comparison test

Question

I want to find whether the Series diverges using the Comparison test $$ \sum_{n=1}^\infty \left(\frac{n+1}{n^2 - n}\right) $$ would it be correct to compare it to $\left(\frac{n}{n^2}\right)$ so I would get $$ \left(\frac{n+1}{n^2 - n}\right) \geq \left(\frac{n}{n^2}\right) $$ and since $$ \left(\frac{n}{n^2}\right) = \left(\frac{1}{n}\right) $$ and $\left(\frac{1}{n}\right)$ is a harmonic series and is divergent so the original series also diverges?

Answer 1

Your intuition is correct. For me, it's simpler to use some elementary asymptotic analysis:

A polynomial is asymptotically equivalent to its leading term, hence $$\frac{n+1}{n^2-n}\sim_\infty \frac n{n^2}=\frac1n,$$ which diverges.

Answer 2

Yes your way is fine, as an alternative we can use that

$$\frac{n+1}{n^2 - n}=\frac{n-1+2}{n^2 - n}=\frac{n-1}{n^2 - n}+\frac{2}{n^2 - n}=\frac1n+\frac{2}{n^2 - n}$$