Why L < 1 forces convergence
The argument bottles the series inside a geometric one. Suppose
L = \lim_{n\to\infty} |a_{n+1}/a_n| < 1.
Step 1 — slip a number between L and
1. Since L < 1, there is room
to pick a ratio r with
L < r < 1.
Step 2 — the ratios eventually drop below r. The
ratios converge to L, which is below r,
so from some index N onward they never climb back above
r:
\left| \frac{a_{n+1}}{a_n} \right| \le r \quad\Longrightarrow\quad |a_{n+1}| \le r\,|a_n| \qquad \text{for all } n \ge N.
Step 3 — iterate the bound into a geometric decay. Apply that inequality
repeatedly, starting from the anchor term |a_N|:
|a_{N+1}| \le r\,|a_N|, \quad |a_{N+2}| \le r\,|a_{N+1}| \le r^2 |a_N|, \quad \dots, \quad |a_{N+k}| \le r^{k}\,|a_N|.
Step 4 — dominate the tail by a geometric series. Sum the tail from
N onward and bound it term by term:
\sum_{k=0}^{\infty} |a_{N+k}| \;\le\; \sum_{k=0}^{\infty} r^{k}\,|a_N| = |a_N| \sum_{k=0}^{\infty} r^{k} = \frac{|a_N|}{1 - r}.
Step 5 — conclude absolute convergence. Because
0 \le r < 1, that geometric series converges, so by direct
comparison the tail \sum |a_{N+k}| is finite. Adding the finitely
many terms before N keeps it finite, so
\sum |a_n| < \infty — the original series converges
absolutely:
\sum_{n=1}^{\infty} |a_n| < \infty.
If instead L > 1, the ratios eventually exceed
1, so the terms grow — they cannot tend to zero, and the
divergence test
already condemns the series.
For a series \sum a_n with
a_n \ne 0 eventually, let
L = \displaystyle\lim_{n\to\infty} \left| \dfrac{a_{n+1}}{a_n} \right|. Then:
-
L < 1 \Rightarrow
\sum a_n converges absolutely (the tail is
dominated by a geometric series of ratio r \in (L, 1)).
-
L > 1 or L = \infty
\Rightarrow \sum a_n
diverges (the terms grow, so they cannot tend to zero).
-
L = 1 \Rightarrow
inconclusive — both convergent and divergent series achieve
L = 1.
The ratio test shines whenever terms carry factorials or
nth powers, because those telescope under division. Take
a_n = \dfrac{x^n}{n!} (the series for
e^x):
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n} \right| = \frac{|x|}{n + 1} \;\xrightarrow[n\to\infty]{}\; 0 < 1,
so it converges for every x — one line, no contest.
But the test is blind to p-series. For
a_n = 1/n^p,
\left| \frac{a_{n+1}}{a_n} \right| = \left( \frac{n}{n+1} \right)^{p} \;\xrightarrow[n\to\infty]{}\; 1,
giving L = 1 for all p —
convergent (p > 1) and divergent
(p \le 1) cases alike. When polynomial decay is the whole story,
reach for the integral or comparison test instead; the ratio test only sees geometric-speed
change.