The Ratio Test

Given a series of positive numbers, how fast must its terms shrink before we can guarantee convergence? A natural instinct is to look at the ratio of consecutive terms. If each term is roughly a fixed fraction r<1r<1 of the one before it, the series should behave like a Geometric Series. It should converge. The Ratio Test makes this instinct precise. It turns a limiting ratio into a rigorous convergence criterion. This criterion requires no explicit formula for partial sums. This matters because most series encountered in practice — power series, series with factorials, series built from exponentials — do not have partial sums with closed forms. The Comparison Test is powerful but requires you to already know a comparable series. The Ratio Test manufactures its own comparison series internally. It uses only the algebraic structure of the terms themselves. It is often the first tool reached for when a series involves n!n!, nnn^n, or cnc^n. This is because ratios of such expressions simplify dramatically while the terms themselves do not. The big picture is that the Ratio Test is a local convergence detector. It examines the asymptotic behavior of ∣an+1/an∣|a_{n+1}/a_n|. From this, it infers global behavior of ∑an\sum a_n from a single limiting number LL. This is both its strength and its limitation. When L≠1L \ne 1, it gives a clean verdict. When L=1L = 1, the test is silent. It is genuinely inconclusive, not merely difficult. This failure mode is exactly the traffic the test cannot direct. This article states and proves the Ratio Test in full generality (absolute convergence, general real or complex terms). It explains why L=1L=1 is an unavoidable blind spot. It works two asymmetric examples. It shows via counterexample why the limit of the ratio matters. An inequality on finitely many terms is not enough. It closes by pointing toward the test's most important application. That application is determining the Radius of Convergence of a power series.

Key takeaways
  • The Ratio Test examines L=lim⁡n→∞∣an+1/an∣L = \lim_{n\to\infty} |a_{n+1}/a_n| to decide convergence of ∑an\sum a_n.
  • If L<1L < 1 the series converges absolutely; if L>1L > 1 (or the ratio →∞\to \infty) it diverges.
  • If L=1L = 1 the test gives no information — series on both sides of convergence produce L=1L=1.
  • The proof works by comparing the tail of the series to a genuine Geometric Series with ratio strictly between LL and 11.
  • The test is most effective for terms built from factorials, exponentials, or nthn^{th} powers, where ratios telescope algebraically.
  • Key idea: compare the tail behavior of a series to geometric decay, using the ratio of consecutive terms as a proxy for the common ratio.

Where this lives today

The Ratio Test is the primary computational engine for finding the Radius of Convergence of a power series: given ∑cn(x−a)n\sum c_n (x-a)^n, one computes L(x)=∣x−a∣⋅lim⁡∣cn+1/cn∣L(x) = |x-a| \cdot \lim |c_{n+1}/c_n| and solves L(x)<1L(x) < 1, directly yielding the radius R=1/lim⁡∣cn+1/cn∣R = 1/\lim|c_{n+1}/c_n| whenever that limit exists. This underlies the convergence theory of Taylor series in calculus, complex power series in complex analysis, and generating functions in combinatorics. In probability and statistics, ratio-test-style reasoning appears in verifying convergence of moment generating functions and in analyzing tail probabilities of distributions built from factorial or exponential terms (e.g. Poisson probabilities λne−λ/n!\lambda^n e^{-\lambda}/n!). In numerical analysis, ratio tests inform the study of convergence rates of iterative algorithms and truncation error bounds for series approximations of special functions (Bessel functions, hypergeometric series). In theoretical computer science, similar ratio-of-successive-terms reasoning appears in analyzing the convergence of recursive generating functions counting combinatorial structures. Cryptographic and coding-theoretic constructions relying on formal power series over finite fields also invoke ratio-style convergence radius arguments when working with analytic analogues.

Ask the tutor: How do I decide between the Ratio Test and the Root Test when a series' terms mix factorials and nthn^{th} powers?

DefinitionThe Ratio Test

Let (an)n=1∞(a_n)_{n=1}^\infty be a sequence of nonzero real or complex numbers, and consider the series

∑n=1∞an.\sum_{n=1}^\infty a_n.

Suppose the limit

L=lim⁡n→∞∣an+1an∣L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|

exists in the extended sense (i.e. L∈[0,∞]L \in [0,\infty] is allowed). The Ratio Test asserts: 1. If L<1L < 1, the series ∑an\sum a_n converges absolutely. 2. If L>1L > 1 (including L=∞L = \infty), the series ∑an\sum a_n diverges. 3. If L=1L = 1, the test is inconclusive: no conclusion about convergence or divergence can be drawn from LL alone.

Intuition

The core idea is disarmingly simple: if ∣an+1/an∣|a_{n+1}/a_n| settles down near a number rr, then far out in the tail, each term is approximately rr times the previous one — exactly the recursive structure of a geometric sequence with ratio rr. If r<1r < 1, geometric decay wins and the tail sums to something finite; adding on the finitely many terms before the tail changes nothing about convergence, since a series' convergence depends only on its tail. The subtlety is turning "approximately geometric" into an actual comparison. We cannot assume ∣an+1∣=r∣an∣|a_{n+1}| = r|a_n| exactly — only that the ratio converges to LL. The proof's trick is to sacrifice precision for a small margin: pick any rr strictly between LL and 11. Because the ratio converges to LL, it eventually stays below rr forever, and that one-sided bound is enough to dominate the tail by an actual geometric series with ratio rr. This is why the test needs a genuine limit, not merely infinitely many terms with ratio less than 11 — a ratio that creeps toward 11 without ever committing to a fixed value below it can still permit divergence, as the counterexample below shows.

Figure
Comparing a series' tail to a geometric bound when L < 1
Ratio test: |a_{n+1}/a_n| → L = 0.6, pick r = 0.75 n N N+2 N+5 |a_N| r^{k} (dashed bound) |a_{N+k}| (actual terms) Since L=0.6 < r=0.75, eventually |a_{n+1}| ≤ r|a_n|, so the tail is dominated by a convergent geometric series.

Where the test aims

Once LL is computed, the verdict is immediate — but LL must actually be computed as a genuine limit. The next section proves this rigorously, in the general (complex or signed) setting, via direct comparison to a geometric series.

TheoremThe Ratio Test

Let (an)(a_n) be a sequence of nonzero complex numbers such that

L=lim⁡n→∞∣an+1an∣L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|

exists in [0,∞][0,\infty]. Then:

(a) If L<1L < 1, the series ∑n=1∞an\sum_{n=1}^\infty a_n converges absolutely.

(b) If L>1L > 1, the series ∑n=1∞an\sum_{n=1}^\infty a_n diverges.

(c) If L=1L = 1, the test yields no conclusion: there exist convergent series and divergent series both satisfying L=1L=1.

Proof

We prove (a), (b), and (c) in turn. The technique for (a) is a direct comparison to a geometric series; the technique for (b) is the divergence test (terms not tending to 00); (c) is settled by explicit counterexamples.

(a) Suppose L<1L < 1. Choose a number rr with L<r<1L < r < 1; such rr exists because L<1L < 1 (e.g. r=L+12r = \frac{L+1}{2}). Let ε=r−L>0\varepsilon = r - L > 0. Since ∣an+1an∣→L\left|\dfrac{a_{n+1}}{a_n}\right| \to L, by the definition of limit there exists N∈NN \in \mathbb{N} such that for all n≥Nn \geq N,

∣∣an+1an∣−L∣<ε,\left| \left| \frac{a_{n+1}}{a_n} \right| - L \right| < \varepsilon,

which in particular gives, for all n≥Nn \ge N,

∣an+1an∣<L+ε=r.\left| \frac{a_{n+1}}{a_n} \right| < L + \varepsilon = r.

Since an≠0a_n \neq 0 for all nn, this means

∣an+1∣<r ∣an∣for all n≥N.|a_{n+1}| < r\,|a_n| \qquad \text{for all } n \geq N.

We now show by induction that for every integer k≥0k \geq 0,

∣aN+k∣≤∣aN∣ rk.|a_{N+k}| \leq |a_N|\, r^{k}.

Base case (k=0k=0): ∣aN+0∣=∣aN∣=∣aN∣r0|a_{N+0}| = |a_N| = |a_N| r^0, true with equality. Inductive step: suppose ∣aN+k∣≤∣aN∣rk|a_{N+k}| \leq |a_N| r^k for some k≥0k \geq 0. Applying the bound ∣an+1∣<r∣an∣|a_{n+1}| < r |a_n| with n=N+k≥Nn = N+k \geq N gives

∣aN+k+1∣<r∣aN+k∣≤r⋅∣aN∣rk=∣aN∣rk+1.|a_{N+k+1}| < r|a_{N+k}| \leq r \cdot |a_N| r^k = |a_N| r^{k+1}.

This completes the induction, so ∣aN+k∣≤∣aN∣rk|a_{N+k}| \leq |a_N| r^k for all k≥0k \geq 0. Now consider the series ∑k=0∞∣aN∣rk\sum_{k=0}^\infty |a_N| r^k. This is a geometric series with ratio r∈(0,1)r \in (0,1) (if L=0L=0 we may take any r∈(0,1)r\in(0,1); the argument is unchanged), and since ∣r∣<1|r| < 1 it converges, with sum ∣aN∣1−r\dfrac{|a_N|}{1-r}. This is a standard fact about geometric series, established independently of the Ratio Test. By the Comparison Test applied to the nonnegative sequences ∣aN+k∣|a_{N+k}| and ∣aN∣rk|a_N| r^k, since 0≤∣aN+k∣≤∣aN∣rk0 \le |a_{N+k}| \le |a_N| r^k for all k≥0k \geq 0 and ∑k∣aN∣rk\sum_k |a_N| r^k converges, the series

∑k=0∞∣aN+k∣=∑n=N∞∣an∣\sum_{k=0}^\infty |a_{N+k}| = \sum_{n=N}^\infty |a_n|

converges. Finally, ∑n=1∞∣an∣=∑n=1N−1∣an∣+∑n=N∞∣an∣\sum_{n=1}^\infty |a_n| = \sum_{n=1}^{N-1} |a_n| + \sum_{n=N}^\infty |a_n| is a finite sum plus a convergent series, hence converges. Therefore ∑an\sum a_n converges absolutely.

(b) Suppose L>1L > 1 (allowing L=∞L = \infty). Choose rr with 1<r<L1 < r < L if L<∞L < \infty (or simply r=2r = 2 if L=∞L = \infty). As above, there exists NN such that for all n≥Nn \geq N,

∣an+1an∣>r>1,\left|\frac{a_{n+1}}{a_n}\right| > r > 1,

so ∣an+1∣>r∣an∣|a_{n+1}| > r |a_n| for n≥Nn \ge N. By induction exactly as before (with the inequality reversed),

∣aN+k∣≥∣aN∣rkfor all k≥0.|a_{N+k}| \geq |a_N| r^k \quad \text{for all } k \geq 0.

Since r>1r > 1, rk→∞r^k \to \infty as k→∞k \to \infty, and ∣aN∣>0|a_N| > 0 (as aN≠0a_N \ne 0), so ∣aN+k∣→∞|a_{N+k}| \to \infty. In particular ∣an∣↛0|a_n| \not\to 0 as n→∞n \to \infty. But a basic necessary condition for convergence of ∑an\sum a_n is that an→0a_n \to 0 (if ∑an=S\sum a_n = S converges with partial sums Sn→SS_n \to S, then an=Sn−Sn−1→S−S=0a_n = S_n - S_{n-1} \to S - S = 0; this uses only the algebra of limits, not the Ratio Test itself). Since an↛0a_n \not\to 0 here, the series ∑an\sum a_n diverges.

(c) L=1L = 1 is inconclusive. We exhibit two series, both with L=1L=1, one convergent and one divergent. Take an=1n2a_n = \dfrac{1}{n^2}. Then

∣an+1an∣=n2(n+1)2=(nn+1)2→12=1.\left|\frac{a_{n+1}}{a_n}\right| = \frac{n^2}{(n+1)^2} = \left(\frac{n}{n+1}\right)^2 \to 1^2 = 1.

Yet ∑1n2\sum \frac{1}{n^2} converges (a standard pp-series fact, p=2>1p=2>1, established by the Comparison Test against ∫1∞x−2 dx\int_1^\infty x^{-2}\,dx or by direct telescoping bounds). Take bn=1nb_n = \dfrac1n. Then

∣bn+1bn∣=nn+1→1.\left| \frac{b_{n+1}}{b_n}\right| = \frac{n}{n+1} \to 1.

Yet ∑1n\sum \frac1n (the harmonic series) diverges. Both series produce L=1L = 1, one convergent and one divergent, so the value L=1L=1 alone cannot determine the outcome. This proves the test gives no information when L=1L=1. ■\blacksquare

∎

Worked examples

Example 1 (asymmetric factorial/power series). Determine convergence of

∑n=1∞3n (n+2)n! 5 n.\sum_{n=1}^\infty \frac{3^n\,(n+2)}{n!\,5^{\,n}}.

Let an=3n(n+2)n! 5na_n = \dfrac{3^n (n+2)}{n! \, 5^n}. Note an>0a_n > 0 for all n≥1n\ge 1, so absolute value bars are unnecessary. Compute the ratio:

an+1an=3n+1(n+3)(n+1)! 5n+1⋅n! 5n3n(n+2)=35⋅1n+1⋅n+3n+2.\frac{a_{n+1}}{a_n} = \frac{3^{n+1}(n+3)}{(n+1)!\,5^{n+1}} \cdot \frac{n!\,5^n}{3^n(n+2)} = \frac{3}{5} \cdot \frac{1}{n+1} \cdot \frac{n+3}{n+2}.

As n→∞n \to \infty, n+3n+2→1\dfrac{n+3}{n+2} \to 1 and 1n+1→0\dfrac{1}{n+1} \to 0, so

L=lim⁡n→∞an+1an=35⋅0⋅1=0.L = \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac35 \cdot 0 \cdot 1 = 0.

Since L=0<1L = 0 < 1, the series converges absolutely by the Ratio Test. Notice how the factorial in the denominator forces L=0L=0 regardless of the polynomial factor (n+2)(n+2) or the exponential factors 3n,5n3^n, 5^n — factorial growth dominates every exponential and polynomial factor, a fact the Ratio Test extracts automatically without needing this dominance stated as a separate lemma.

Example 2 (a series where the ratio must be handled with lim sup⁡\limsup-free care, and the exponential base matters). Determine convergence of

∑n=1∞(2n)!(n!)2 xn,x=15.\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2}\, x^n, \qquad x = \frac{1}{5}.

This is the central binomial coefficient series ∑(2nn)xn\sum \binom{2n}{n} x^n evaluated at x=1/5x = 1/5. Let an=(2n)!(n!)2(15)na_n = \dfrac{(2n)!}{(n!)^2} \left(\dfrac15\right)^n; all terms are positive. Compute:

an+1an=(2n+2)!((n+1)!)2⋅(n!)2(2n)!⋅15=(2n+2)(2n+1)(n+1)2⋅15.\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!} \cdot \frac15 = \frac{(2n+2)(2n+1)}{(n+1)^2} \cdot \frac15.

Simplify the polynomial ratio:

(2n+2)(2n+1)(n+1)2=2(n+1)(2n+1)(n+1)2=2(2n+1)n+1.\frac{(2n+2)(2n+1)}{(n+1)^2} = \frac{2(n+1)(2n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1}.

So

an+1an=2(2n+1)5(n+1).\frac{a_{n+1}}{a_n} = \frac{2(2n+1)}{5(n+1)}.

Taking the limit:

L=lim⁡n→∞2(2n+1)5(n+1)=lim⁡n→∞4n+25n+5=45L = \lim_{n\to\infty} \frac{2(2n+1)}{5(n+1)} = \lim_{n\to\infty} \frac{4n+2}{5n+5} = \frac{4}{5}

(dividing numerator and denominator by nn). Since L=45<1L = \frac45 < 1, the series converges absolutely at x=1/5x=1/5. This example previews the mechanism behind Power Series and Radius of Convergence: repeating this computation for general xx gives L=4∣x∣L = 4|x|, so the series converges when 4∣x∣<14|x|<1, i.e. ∣x∣<1/4|x| < 1/4 — the radius of convergence is exactly 1/41/4, and at x=1/5<1/4x=1/5 < 1/4 we land safely inside the disk of convergence, matching the direct computation above.

CounterexampleWhy the limit — and not merely "ratio less than 1 infinitely often" — is essential

The theorem requires ∣an+1an∣\left|\dfrac{a_{n+1}}{a_n}\right| to converge to some L<1L<1. It is tempting to weaken this to: "if ∣an+1/an∣<1|a_{n+1}/a_n| < 1 for all sufficiently large nn, the series converges." This weaker hypothesis is false, and the harmonic series itself is the counterexample. Let an=1na_n = \dfrac1n. For every n≥1n \geq 1,

an+1an=nn+1<1.\frac{a_{n+1}}{a_n} = \frac{n}{n+1} < 1.

So the ratio is strictly less than 11 for every single term, with no exceptions — a stronger hypothesis than merely "eventually less than 11." Yet ∑1n\sum \frac1n diverges. The reason the (correct) theorem is not violated is that the ratio nn+1→1\frac{n}{n+1} \to 1, not to any fixed r<1r<1; there is no single r<1r < 1 that bounds nn+1\frac{n}{n+1} from above for all large nn (since nn+1→1\frac{n}{n+1} \to 1, for any r<1r<1 eventually nn+1>r\frac n{n+1} > r). The proof of part (a) crucially used that the ratio stays below a fixed number r<1r<1 for all large nn — this is exactly what a limit L<1L<1 guarantees (via the margin ε=r−L\varepsilon = r-L), and exactly what "ratio <1<1, with the ratio drifting up toward 11" fails to guarantee. This shows the hypothesis "L=lim⁡∣an+1/an∣L = \lim |a_{n+1}/a_n| exists and is <1<1" cannot be relaxed to "the ratio is eventually (or always) less than 11" — the rate at which the ratio approaches its limit, or rather the existence of a uniform gap below 11, is indispensable.

Scope, edge cases & common traps

SituationWhat happensWhy
LL does not exist (ratio oscillates, no limit)Basic Ratio Test as stated does not applyRequires the stronger Root Test (via lim sup⁡\limsup) or another method
an=0a_n = 0 for infinitely many nnRatio an+1/ana_{n+1}/a_n undefinedTest requires an≠0a_n \ne 0; reformulate or use comparison directly
L=1L = 1Totally inconclusive — both outcomes occurSee part (c) of the theorem and the 1/n1/n vs 1/n21/n^2 example
Ratio <1<1 for all nn but →1\to 1Series can still divergeSee the counterexample above (harmonic series)
Series is a power series ∑cnxn\sum c_n x^nRatio Test gives radius of convergenceCompute L(x)L(x) as a function of xx; solve L(x)<1L(x)<1

The mathematician behind it

Origins · 1 minJean le Rond d'Alembert and how this was discovered
JL
Jean le Rond d'Alembert
1717–1783 · Enlightenment Paris, Encyclopédie & mathematical analysis

D'Alembert was a French mathematician, physicist, and co-editor of the Encyclopédie, whose broad program touched mechanics, wave equations, and the foundations of calculus during the early formalization of series and limits in the 18th century.

History
- 1768: D'Alembert investigates ratio-based criteria for series convergence in the course of studying the binomial series and related expansions, giving the test its alternate name, the "d'Alembert ratio test."
- 1821: Augustin-Louis Cauchy, in his Cours d'analyse, gives the first fully rigorous treatment of convergence criteria for series, placing the ratio test alongside the closely related root test on a modern epsilon-based footing.
- 1830s: The ratio and root tests become standard tools in the systematic study of power series, tying convergence radius directly to coefficient growth rates.
- 1893: Textbook treatments (e.g. following Cauchy and Weierstrass's rigor program) fix the modern statement distinguishing L<1L<1, L>1L>1, L=1L=1 explicitly, cementing the inconclusive case as a known and named limitation.
- 20th century: The test becomes a staple of undergraduate real analysis curricula worldwide, serving as the standard first computational tool before the Root Test and Integral Test are introduced.
Why does the Ratio Test fail to decide anything when L=1L=1?

Because at L=1L=1 the tail of the series is not forced to behave like any fixed geometric series — the comparison geometric series used in the proof has ratio rr strictly between LL and 11, and no such rr exists once L=1L=1 itself equals the boundary. Series with L=1L=1 can decay just fast enough to converge (like 1/n21/n^2) or just slowly enough to diverge (like 1/n1/n), and the ratio test's single real number LL cannot distinguish these finer growth rates. Resolving L=1L=1 cases requires finer tools: the Comparison Test with a specific reference series, the integral test, or refinements like Raabe's test that examine the rate at which the ratio approaches 11.

How does the Ratio Test relate to the Root Test?

Both tests infer convergence from asymptotic behavior, but the Root Test examines lim⁡∣an∣n\lim \sqrt[n]{|a_n|} rather than lim⁡∣an+1/an∣\lim |a_{n+1}/a_n|. A theorem of real analysis shows that whenever the ratio limit LL exists, the root limit also exists and equals the same LL; hence the Root Test is strictly more powerful (it applies to some sequences, like ones with oscillating ratios, where the Ratio Test's limit fails to exist but the root limit still does). In practice the Ratio Test is preferred computationally whenever the terms involve factorials or nice recursive ratios, since those simplify better under division than under nthn^{\text{th}} roots.

Does the Ratio Test require the terms to be positive?

No. The theorem as stated works for complex (hence real, signed) sequences, because the proof only ever uses absolute values ∣an∣|a_n| and concludes absolute convergence, which for real or complex series implies ordinary convergence via the completeness of R\mathbb{R} or C\mathbb{C} (every absolutely convergent series converges). The only structural requirement is an≠0a_n \ne 0 for all nn, so the ratio an+1/ana_{n+1}/a_n is defined.

What happens if the ratio ∣an+1/an∣|a_{n+1}/a_n| never converges to a single limit?

The Ratio Test as stated simply does not apply — it has no hypothesis to check. In that situation, one can try to use lim sup⁡n→∞∣an+1/an∣\limsup_{n\to\infty} |a_{n+1}/a_n| in place of the limit: if this limsup is <1<1, the same geometric-comparison proof goes through essentially unchanged (using a tail bound valid past some NN), giving convergence. If lim inf⁡∣an+1/an∣>1\liminf |a_{n+1}/a_n| > 1, divergence follows similarly. But if the ratio oscillates so that lim inf⁡<1<lim sup⁡\liminf < 1 < \limsup, no ratio-based test resolves the question, and one typically turns to the Root Test or a direct comparison.

Can the Ratio Test be used directly to find a radius of convergence for a power series?

Yes, and this is its most important application. For ∑cn(x−a)n\sum c_n (x-a)^n, treat xx as fixed and apply the test to an=cn(x−a)na_n = c_n(x-a)^n: compute L(x)=∣x−a∣⋅lim⁡∣cn+1/cn∣L(x) = |x-a|\cdot \lim|c_{n+1}/c_n| (assuming this limit exists). The series converges absolutely where L(x)<1L(x)<1 and diverges where L(x)>1L(x)>1, giving radius R=1/lim⁡∣cn+1/cn∣R = 1/\lim|c_{n+1}/c_n|. The boundary points x=a±Rx = a\pm R correspond exactly to L=1L=1, which is why convergence at the endpoints of a power series interval must always be checked separately by other means.

Is the harmonic series a counterexample to the Ratio Test?

No — it is a counterexample to a weaker, incorrect version of the test (checking ratio <1<1 pointwise rather than taking a genuine limit strictly below 11). The actual theorem is not violated: for an=1/na_n = 1/n, L=lim⁡∣an+1/an∣=1L = \lim |a_{n+1}/a_n| = 1 exactly, which falls into case (c), the inconclusive case, and the theorem correctly makes no claim there. The harmonic series's divergence is consistent with the theorem; it merely shows why the hypothesis L<1L<1 (a strict inequality on the limit) cannot be replaced by a weaker termwise condition.

Before this

Where this leads