calculus - determining absolute or conditional convergence using limit comparison test

I'm currently learning determining absolute or conditional convergence on Khan Academy.

One of the exercise question is

The answer provided is

I don't understand how the sum function does not converge by using the LIMIT COMPARISON TEST, I must be missing some of the required knowledge (specifically, the reason why limit comparison test can be used to prove the sum function does not converge), can someone please help me fill in the gap?

Answer

We have
$$\lim_n \frac{\frac{n}{n^2+1}}{\frac{1}{n}} = \lim_n \frac{n^2}{n^2+1} = 1,$$
which implies that $\sum_n \frac{n}{n^2+1}$ and $\sum_n \frac{1}{n}$ either both converge or both diverge (by the limit comparison test). Since we know that $\sum_n \frac{1}{n}$ diverges, it follows that $\sum_n \frac{n}{n^2+1}$ diverges.

The limit comparison test can be applied (as above) since, for all $n$, we have $\frac{n}{n^2+1} \geq 0$ and $\frac{1}{n} > 0$.