Ratio Test for Convergence BC only

With $L = \lim \left|\frac{a_{n+1}}{a_n}\right|$, the series converges absolutely when $L < 1$, diverges when $L > 1$ or $L = \infty$, and is undecided when $L = 1$. The mechanism is comparison with a geometric series of ratio $L$, which is why the undecided case sits exactly at the endpoint 10.2 excluded: at $r = 1$ the comparison series neither shrinks nor grows, and everything then depends on how the ratio approaches $1$, which the limit has discarded. That $L = 1$ genuinely decides nothing is provable rather than cautionary, since $\sum \frac{1}{n}$ gives $\frac{n}{n+1} \to 1$ and diverges while $\sum \frac{1}{n^{2}}$ gives $\left(\frac{n}{n+1}\right)^{2} \to 1$ and converges.

Every $p$-series gives $L = 1$, so the test is blind to that whole family, which makes a bare power of $n$ in the denominator a reason to choose a different test rather than a hard case for this one. What the ratio test is built for is factorials, where $\frac{(n+1)!}{n!} = n+1$, and constants raised to the $n$, where the ratio collapses to that constant: $\sum \frac{2^{n}}{n!}$ gives $\frac{2}{n+1} \to 0$, $\sum \frac{n!}{2^{n}}$ gives $\frac{n+1}{2} \to \infty$, and $\sum \frac{n!}{n^{n}}$ gives $\left(\frac{n}{n+1}\right)^{n} \to \frac{1}{e}$, which no other test in the unit reaches. Because $L$ is defined on absolute values, a successful ratio test proves absolute convergence, the stronger conclusion 10.9 develops and the reason this is the standard tool for the radius of convergence in 10.13.