$L = 1$ is not a verdict, it is no information BC only

For a series with nonzero terms, let

$$L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|.$$

1. $L < 1$: the series converges, and converges absolutely, since the test is built on absolute values.
2. $L > 1$, including $L = \infty$: the series diverges.
3. $L = 1$: no conclusion. The test has failed and another is needed.

The idea behind the first two is geometric comparison. If the ratio settles near some $L < 1$, then far out the series behaves like a geometric series with that ratio, and 10.2 says such a series converges. If it settles above $1$ the terms are eventually growing, so they cannot tend to zero and 10.3 finishes it.

That also explains the third case. At $L = 1$ the comparison series is geometric with $r = 1$, which is the endpoint 10.2 excluded: the terms neither shrink geometrically nor grow. Everything then depends on how the ratio approaches $1$, which this limit has thrown away.

Take the two most familiar series in the unit.

For $\sum \frac{1}{n}$:

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n}{n+1} \longrightarrow 1,$$

and this series diverges. For $\sum \frac{1}{n^{2}}$:

$$\left|\frac{a_{n+1}}{a_n}\right| = \left(\frac{n}{n+1}\right)^{2} \longrightarrow 1,$$

and this series converges. Same limit, opposite outcomes. No rule can turn $L = 1$ into a verdict, because a verdict would have to be wrong about one of these two.

Generalise it: every $p$-series gives $L = 1$, whatever $p$ is, since $\left(\frac{n}{n+1}\right)^{p} \to 1$ for every fixed $p$. So the ratio test is blind to the entire $p$-series family, including the boundary case the family is famous for. That is worth knowing as a selection fact rather than a curiosity: seeing a power of $n$ in the denominator and nothing else is a reason not to reach for this test at all.

The ratio test earns its place where the ratio simplifies dramatically, which happens with two ingredients:

1. Factorials. $\frac{(n+1)!}{n!} = n+1$, so a factorial collapses to a single factor. Nothing else in this unit handles factorials at all.
2. Constants raised to the $n$. $\frac{c^{n+1}}{c^{n}} = c$, so an exponential collapses to a constant.

Three worked ratios, each a line:

1. $\sum \frac{2^{n}}{n!}$ gives $\frac{2}{n+1} \to 0$. Converges.
2. $\sum \frac{n!}{2^{n}}$ gives $\frac{n+1}{2} \to \infty$. Diverges.
3. $\sum \frac{n!}{n^{n}}$ gives $\left(\frac{n}{n+1}\right)^{n} \to \frac{1}{e} \approx 0.368$. Converges, and no other test in the unit would have got there.

Conversely, polynomials alone give $L = 1$ every time, so a rational term is a comparison-test problem rather than a ratio-test one. Reading the shape of the term before choosing is the whole of test selection, and it is faster than trying tests in turn.

$L = 1$ is not a dead end, it is a redirection. The series still has a verdict; this test simply cannot see it. What to reach for:

1. A power of $n$ in the denominator: it is a $p$-series or behaves like one. Use 10.5, or limit comparison from 10.6.
2. Alternating signs: 10.7's test, which is unaffected by the ratio being $1$.
3. Something integrable: 10.4.

One further note on what a successful ratio test gives you. Because $L$ is defined with absolute values, $L < 1$ establishes absolute convergence, which is stronger than convergence and is the subject of 10.9. This is the only test so far that produces the stronger conclusion directly, and it is why the ratio test is the standard tool for finding the radius of convergence in 10.13.