For the direct comparison test you want to first determine whether or not you think the series converges.
If you think the series $\sum_{n=1}^{+\infty} A_n$ converges, then try to find a $B_n$ such that $B_n \ge A_n$ for all sufficiently large $n$ and $\sum_{n=1}^{+\infty} B_n$ converges. The idea is that the "larger" series $\sum_{n=1}^{+\infty} B_n$ converges, therefore so should the "smaller" series $\sum_{n=1}^{+\infty} A_n$.
If instead you think the series $\sum_{n=1}^{+\infty} A_n$ diverges, then try to find a $B_n$ such that $B_n \le A_n$ for all sufficiently large $n$ and $\sum_{n=1}^{+\infty} B_n$ diverges. The idea is that the "smaller" series $\sum_{n=1}^{+\infty} B_n$ diverges, therefore so should the "larger" series $\sum_{n=1}^{+\infty} A_n$.
In this case, since $A_n = \dfrac n{n^2-1}$, you can choose $B_n = \dfrac1n$, because
$$ n^2 - 1 < n^2 \iff \frac1n < \frac{n}{n^2-1}$$
for all $n > 1$. Therefore $B_n < A_n$ for all $n > 1$. Then since $\sum_{n=1}^{+\infty} B_n$ diverges (it's the harmonic series), we know that $\sum_{n=1}^{+\infty} A_n$ must also diverge.
For the limit comparison test your method for choosing $B_n$ is correct in your second case - choose $B_n$ to be what $A_n$ will "behave like" as $n$ gets very large. If $A_n = \dfrac n{n^2-1}$, then as $n$ gets very large, the $-1$ contributes nothing relevant, so $A_n$ will "behave like" $\dfrac n{n^2}$ for very large $n$. So you would choose your $B_n$ to be $B_n = \dfrac n{n^2} = \dfrac1n$, and since $\lim_{n \to +\infty} \frac{A_n}{B_n} = 1$, then $\sum_{n=1}^{+\infty} A_n$ and $\sum_{n=1}^{+\infty} B_n$ both converge or they both diverge. Well, we already know that $\sum_{n=1}^{+\infty} B_n$ diverges because it's the harmonic series. Therefore $\sum_{n=1}^{+\infty} A_n$ diverges as well.
Note: These two comparison tests will not work on alternating series.
If the proper method is to take highest term on numerator / denominator of An to use as Bn then why do I often see solutions where this is not the case?
If by "highest term" you mean "highest-order term" or "term with the highest degree" then that really only applies to functions whose numerator and denominator are polynomials (or at least algebraic functions with real powers of $n$). What you really want to think about is what $A_n$ "behaves like" as $n$ gets really large, as explained above, but remember that this is only how you want to choose $B_n$ for the limit comparison test, not the direct comparison test. Both videos you linked to are doing the direct comparison test.
The limit comparison states that for two series $\Sigma_n a_n$ and $\Sigma_n b_n$ with $a_n\geq 0, b_n > 0$ for all $n$ we have
$$L=\lim_{n \to \infty} \frac{a_n}{b_n}$$
- If $0<L<\infty$ (i.e., L is a positive, finite number) then either the series $\Sigma_n a_n$ and $\Sigma_n b_n$ both converge or both diverge.
- If $L=0$ and $\Sigma_n b_n$ converges, then $\Sigma_n a_n$ converges.
- If $L=\infty$ and $\Sigma_n b_n$ diverges, then $\Sigma_n a_n$ diverges.
The idea is that as $L=\lim_{n \to \infty} \dfrac{a_n}{b_n}$, eventually we will have $a_n\approx Lb_n$. So if one of the series converges (or diverges) so does the other since the two are essentially scalar multiples of each other.
In the second case, if
$$\lim_{n \to \infty} \dfrac{a_n}{b_n}=L=0$$
and we know that $\Sigma_n b_n$ converges, then $\Sigma_n a_n$ must also converge since $a_n\approx Lb_n$ and $L=0$. In the third case, if
$$\lim_{n \to \infty} \dfrac{a_n}{b_n}=L=\infty$$
and we know that $\Sigma_n b_n$ diverges, then $\Sigma_n a_n$ must also diverge since $a_n\approx Lb_n$ and $L=\infty$.
Best Answer
Hint:
If $\liminf b_n = \alpha >0 $ then there exists $N$ such that $ b_n > \alpha/2$ for all $n >N$