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?
Best Answer
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 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.