How to use the comparison test with a series whose terms are given by a rational function? What is the proper way to find $B_n$ using the comparison test, do I have to compare $A_n$ to the largest term on the numerator/denominator?

Ex. $A_n:\sum\frac n{n^2-1}$ diverges.

If I compare $A_n:\sum\frac n{n^2-1}$ to $B_n = 1 / n^2$ then $A_n > B_n$ 
which means if $B_n$ diverges then $A_n$ will diverge (Does not Diverge)?
If I compare $A_n:\sum\frac n{n^2-1}$ to $B_n = n / n^2$ then $A_n > B_n$
which means if $B_n$ diverges then $A_n$ will diverge (correct as $B_n$ diverges)
If the proper method is to take highest term on numerator / denominator of $A_n$ to use as $B_n$ then why do I often see solutions where this is not the case?
Ex. https://www.youtube.com/watch?v=hpF3wqiPHQU
Ex. https://www.youtube.com/watch?v=SzE01kHg3Yo
 A: 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.
