Two pairings conclude, and two conclude nothing BC only

Suppose $0 \le a_n \le b_n$ for all large $n$. Being underneath $b_n$ helps only if $b_n$ is small enough to be worth being under, and being above helps only if what you are above is already unbounded:

1. $a_n \le b_n$ and $\sum b_n$ converges $\Rightarrow \sum a_n$ converges. A smaller pile of positive terms cannot outgrow a bounded one.
2. $a_n \ge b_n$ and $\sum b_n$ diverges $\Rightarrow \sum a_n$ diverges. Sitting above something unbounded leaves nowhere to settle.
3. $a_n \le b_n$ and $\sum b_n$ diverges $\Rightarrow$ nothing.
4. $a_n \ge b_n$ and $\sum b_n$ converges $\Rightarrow$ nothing.

The last two are not weaker conclusions; they are the absence of one. Being smaller than something that runs away tells you nothing, because everything convergent is also smaller than something that runs away. $\frac{1}{2^{n}} \le \frac{1}{n}$ is perfectly true and $\sum \frac{1}{n}$ diverges, and yet $\sum \frac{1}{2^{n}}$ converges. The same pairing applied to $\frac{1}{2n} \le \frac{1}{n}$ sits over a divergent series. One useless pairing, two opposite outcomes.

A working memory for it: a comparison only helps when it pushes you the way you want to go. To prove convergence, get underneath something convergent. To prove divergence, get above something divergent.

The stock of known series is small and sufficient: geometric from 10.2, and $p$-series from 10.5. Picking from it is a matter of reading dominant behaviour:

1. Rational terms: subtract degrees. $\frac{2n+1}{n^{3}+5}$ behaves like $\frac{2}{n^{2}}$, so compare with a $p$-series at $p = 2$.
2. Terms with an exponential: the exponential dominates everything polynomial. $\frac{n^{2}}{3^{n}}$ compares with a geometric series.
3. Added constants: drop them. $\frac{1}{n^{2}+1}$ behaves like $\frac{1}{n^{2}}$, and here the inequality even runs the convenient way, since $\frac{1}{n^{2}+1} < \frac{1}{n^{2}}$.

Then check that the inequality you need actually holds. This is where direct comparison gets awkward: $\frac{1}{n^{2}-1}$ behaves like $\frac{1}{n^{2}}$ but is larger than it, so the convenient inequality points the useless way. You can repair it by comparing against $\frac{2}{n^{2}}$ instead, and the fact that repairs are often necessary is what motivates the next section.

Rather than establishing an inequality, compute a limit. For positive $a_n$ and $b_n$, let

$$L = \lim_{n \to \infty} \frac{a_n}{b_n}.$$

If $L$ is finite and positive, then $\sum a_n$ and $\sum b_n$ do the same thing: both converge or both diverge. That is a two-way conclusion and it needs no inequality at all, which is why it is usually the easier tool.

The two boundary cases are one-directional and are where CA8 lives:

1. $L = 0$: $a_n$ is eventually much smaller than $b_n$. If $\sum b_n$ converges, so does $\sum a_n$. If $\sum b_n$ diverges, nothing follows.
2. $L = \infty$: $a_n$ is eventually much larger. If $\sum b_n$ diverges, so does $\sum a_n$. If $\sum b_n$ converges, nothing follows.

Notice these are the same two useful pairings as before, since $L = 0$ is a statement that $a_n$ is smaller and $L = \infty$ that it is larger. The limit comparison test is the direct comparison test with the inequality established automatically, and it inherits exactly the same restrictions on what can be concluded.

Both tests require positive terms, which rules out anything alternating until 10.9 provides a way in. Beyond that, a complete direct comparison has three parts and a complete limit comparison has three different ones:

1. Direct: name $b_n$, state its verdict with a reason, and prove the inequality in the direction you need.
2. Limit: name $b_n$, state its verdict, and compute $L$, saying explicitly that it is finite and positive.

The most common incomplete answer names a $b_n$ and asserts that the two series "behave the same" without either an inequality or a limit. That is the intuition the tests exist to make rigorous, and it is not itself an argument.

One practical note: with a rational $a_n$, limit comparison against $\frac{1}{n^{p}}$ where $p$ is the difference of degrees will always give a finite positive $L$, namely the ratio of leading coefficients. That makes the choice mechanical and the limit a formality, which is why it is the default move on any quotient of polynomials.