The Ratio Test
Whether a chip can compute a value like e-to-the-x, or how likely a rare event is to take
many tries, often comes down to adding up terms built from powers and factorials — and asking
whether that endless sum totals a finite number. The ratio test is the quick
tool that decides, by watching how fast each term shrinks compared with the one before.
Interrogate a series with a single question: how fast are you shrinking? Don't ask
for the value of any particular term — ask each term how it compares to the one before it. If,
eventually, every term is smaller than r times its predecessor for
some fixed r < 1, then the series is being outpaced by a
geometric series of
ratio r — and a geometric series with ratio below one converges. A
series that shrinks faster than a convergent series has no choice but to converge too.
That is the entire idea of the ratio test: it is
comparison with a geometric
series, automated. Instead of hunting for a comparison series by hand, you compute
one limit — the limiting ratio of consecutive terms —
L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|,
and read off the verdict:
- L < 1 — the series converges absolutely (a geometric series beats it).
- L > 1 (or L = \infty) — it diverges (the terms eventually grow).
- L = 1 — the test is silent. Not "converges slowly", not "probably fine" — no information at all.
Why "absolutely"? The test looks only at |a_n|, so when it says yes
it is certifying the stronger property of
absolute
convergence — convergence that survives even if you strip away all the minus signs.
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. Notice the asymmetry: L < 1 proves
convergence by comparison, but L > 1 proves divergence for the
cruder reason that the terms don't even die out.
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| (assuming the limit exists, possibly \infty). 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
test or comparison test instead; the ratio test only sees geometric-speed
change.
The factorial-algebra survival kit
Nearly every ratio-test computation lives or dies on three cancellations. Master them once and
the test becomes a two-line routine. The key identity is the definition of the factorial read
backwards:
(n+1)! = (n+1) \cdot n!
A factorial doesn't restart from scratch at each step — it is the previous factorial times one
new factor. So when a factorial meets a factorial in a fraction, almost everything cancels:
\frac{(n+1)!}{n!} = n+1, \qquad \frac{n!}{(n+1)!} = \frac{1}{n+1}, \qquad \frac{c^{\,n+1}}{c^{\,n}} = c.
Try one with a bigger gap: \dfrac{(n+1)!}{(n-1)!}. Peel off factors
one at a time — (n+1)! = (n+1) \cdot n \cdot (n-1)! — so the
quotient is (n+1)\,n. For n = 5 that's
720/24 = 30 = 6 \cdot 5. Check.
One golden habit: always write the ratio as a product, "next term times the reciprocal
of the current term",
\frac{a_{n+1}}{a_n} = a_{n+1} \cdot \frac{1}{a_n},
with every piece of a_{n+1} written out with
n+1 substituted everywhere n
appears. Then cancel piece by piece. Most ratio-test errors are made in this line, not in the
limit.
-
Forcing a verdict at L = 1. When
L = 1, the ratio test tells you nothing — writing
"L = 1, so it converges" (or "so it diverges") is wrong every
time it's written. The test is mute on every p-series, so the harmonic series
\sum 1/n (diverges) and \sum 1/n^2
(converges) both land on L = 1. The only legal conclusion is
"inconclusive — use another test".
-
Botching a_{n+1}. The ratio is
a_{n+1}/a_n, and building
a_{n+1} means replacing every
n by n+1: from
\frac{n^2}{3^n n!} you must get
\frac{(n+1)^2}{3^{n+1} (n+1)!} — brackets, exponent, factorial,
all three. Writing (n+1)! = n! + 1 or
3^{n+1} = 3^n + 3 is where marks actually die: most ratio-test
errors are algebra, not analysis.
Three worked examples
Example 1 — \displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n}.
The numerator grows, the denominator explodes — who wins? Build the ratio as a product and
cancel:
\left| \frac{a_{n+1}}{a_n} \right| = \frac{n+1}{2^{\,n+1}} \cdot \frac{2^{\,n}}{n} = \frac{n+1}{n} \cdot \frac{2^{\,n}}{2^{\,n+1}} = \left(1 + \frac{1}{n}\right) \cdot \frac{1}{2} \;\xrightarrow[n\to\infty]{}\; \frac{1}{2}.
L = \tfrac12 < 1: converges absolutely. The
polynomial factor n contributed only the harmless
1 + 1/n \to 1; the exponential set the ratio. (In fact the sum is
exactly 2 — but the ratio test neither knows nor cares about the
value; it only certifies that a value exists.)
Example 2 — \displaystyle\sum_{n=0}^{\infty} \frac{2^n}{n!}.
Exponential upstairs, factorial downstairs:
\left| \frac{a_{n+1}}{a_n} \right| = \frac{2^{\,n+1}}{(n+1)!} \cdot \frac{n!}{2^{\,n}} = \frac{2}{n+1} \;\xrightarrow[n\to\infty]{}\; 0.
L = 0 — the smallest limit the test can produce — so it converges
absolutely with room to spare. The moral is worth saying out loud:
factorials crush exponentials. An exponential multiplies by the same
factor 2 forever, but the factorial multiplies by
n+1, a factor that keeps growing. Past
n = 2 the factorial wins every single round. (This series is
e^2, by the way.)
Example 3 — \displaystyle\sum_{n=1}^{\infty} \frac{n^n}{n!}.
Now the factorial faces something even stronger. Substitute
n+1 everywhere and use (n+1)! = (n+1) \cdot n!:
\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^{\,n+1}}{(n+1)!} \cdot \frac{n!}{n^n} = \frac{(n+1)^{\,n+1}}{(n+1)\,n^n} = \frac{(n+1)^{\,n}}{n^n} = \left(1 + \frac{1}{n}\right)^{\!n} \;\xrightarrow[n\to\infty]{}\; e.
The famous limit (1 + 1/n)^n \to e \approx 2.718 appears out of
nowhere, and L = e > 1: diverges. So
n^n beats n! — decisively. The full
growth-rate league table, slowest to fastest, is
\ln n \;\ll\; n^p \;\ll\; c^n \;\ll\; n! \;\ll\; n^n,
and the ratio test is precisely the instrument that measures where a term sits on it: each
rung grows faster than the one before by more than any geometric factor, and the ratio test
detects exactly that geometric-speed difference.
Open any calculus text at the chapter on
power series
and you'll find the ratio test doing all the heavy lifting. A power series
\sum c_n x^n hands the test its favourite food — an
nth power. Run the ratio:
\left| \frac{c_{n+1} x^{\,n+1}}{c_n x^{\,n}} \right| = \left| \frac{c_{n+1}}{c_n} \right| \cdot |x| \;\longrightarrow\; \ell\,|x|,
and the verdict \ell |x| < 1 becomes
|x| < 1/\ell — a disc of values of
x where the series converges. That number
1/\ell is the
radius of
convergence: every radius you will ever compute is the ratio test in disguise.
Even the "interval of convergence" ritual — check the two endpoints separately — is forced by
this page's fine print: at the endpoints \ell|x| = 1 exactly, the
one place the ratio test refuses to speak.
And when the ratio itself misbehaves (terms that are sometimes zero, ratios that oscillate),
its sibling the
root test
takes over — same geometric idea, measured with an nth root
instead of a quotient.
The L = 1 stalemate, in full
The inconclusive case deserves to be seen, not just stated. Run the test on two series you
already know intimately — one divergent, one convergent — and watch it give the same
answer for both.
The harmonic series \sum 1/n (diverges — its
partial sums grow like \ln n):
\left| \frac{a_{n+1}}{a_n} \right| = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} \;\xrightarrow[n\to\infty]{}\; 1.
The Basel series \sum 1/n^2 (converges — to
\pi^2/6, no less):
\left| \frac{a_{n+1}}{a_n} \right| = \frac{n^2}{(n+1)^2} = \left( \frac{n}{n+1} \right)^{2} \;\xrightarrow[n\to\infty]{}\; 1.
Same L, opposite fates. This is the proof-by-example that
L = 1 carries zero information: knowing
L = 1 is compatible with either outcome, so no conclusion can
follow. What's really happening is that both series shrink at merely polynomial
speed — slower than every geometric series — so the geometric yardstick reads "1" and shrugs.
Distinguishing 1/n from 1/n^2 needs a
finer instrument: the
integral test or a
comparison.
A practical triage rule follows. Scan the term for its fastest-growing ingredient:
factorials or c^n present → ratio test first;
only powers of n (and logs) → don't bother, you'll get
L = 1, go straight to comparison or the integral test.
Watch the ratio find its limit
The ratio test is a story told along the n-axis, so watch it
unfold. Pick a series below and follow the consecutive-term ratio
|a_{n+1}/a_n| as it settles on its limit
L. The dashed line is the verdict threshold
1: end up below it and the series converges; sit above it and the
series diverges.
Each choice acts out one act of the drama. The factorial series
x^n/n! (shown at x = 2) has ratio
2/(n+1), which plunges toward
0 — the factorial crushing the exponential in real time. The
p-series 1/n^2 creeps up toward the fence at
1 and never crosses it — the stalemate, drawn. And
2^n holds flat at 2, comfortably in the
divergence zone. Notice what the early values of the ratio are worth: nothing. Only the
limit matters — a series may shrink erratically for the first fifty terms and still
have a perfectly crisp L.
See it explained