Ratio Test

The statement

For a series ∑ aₙ with non-zero terms, define:

L = lim_{n→∞} |a_{n+1} / aₙ|

Then:

• If L < 1, the series ∑ aₙ converges absolutely.
• If L > 1 (or the limit is +∞), the series diverges.
• If L = 1, the test is inconclusive — convergence can go either way.

The number L measures how fast consecutive terms shrink. L < 1 means terms eventually shrink at least as fast as a geometric series with common ratio less than one — and geometric series converge. L > 1 means terms eventually grow geometrically, so they don't even tend to zero, and the divergence test kills the series.

Why it works — geometric domination

Suppose L < 1. Pick any r with L < r < 1. By definition of limit, there exists N such that for all n ≥ N:

|a_{n+1} / aₙ| < r ⇒ |a_{n+1}| < r · |aₙ|

Iterating from N gives |a_{N+k}| < r^k · |a_N|. So the tail of ∑ |aₙ| is bounded above by a geometric series with ratio r < 1, which converges. Hence ∑ |aₙ| converges, which is exactly absolute convergence.

The L > 1 case is symmetric. Pick r with 1 < r < L; for n past some threshold |a_{n+1}| > r |aₙ|, so |aₙ| grows at least geometrically and certainly does not tend to zero. Divergence follows from the divergence test.

Worked examples

1. ∑ n!/nⁿ. Compute:

a_{n+1}/aₙ = ((n+1)! / (n+1)^(n+1)) · (nⁿ / n!)
           = (n+1) · nⁿ / (n+1)^(n+1)
           = nⁿ / (n+1)ⁿ
           = 1 / (1 + 1/n)ⁿ
                ↓
                1 / e ≈ 0.368  <  1

L = 1/e < 1, so ∑ n!/nⁿ converges absolutely. The factorial in the numerator is decisively beaten by nⁿ in the denominator. This example is canonical because two rapidly growing sequences race, and the ratio test produces a clean tie-breaker.

2. ∑ 2ⁿ / n!. Compute:

a_{n+1}/aₙ = (2^(n+1) / (n+1)!) · (n! / 2ⁿ)
           = 2 / (n + 1) → 0  < 1

L = 0, far less than 1, so the series converges absolutely. (Its sum is e² − 1.) Whenever a series has a factorial in the denominator and only polynomial-or-exponential growth in the numerator, the ratio test decides instantly.

3. ∑ nⁿ / n!. Compute:

a_{n+1}/aₙ = ((n+1)^(n+1) / (n+1)!) · (n! / nⁿ)
           = (n+1)ⁿ / nⁿ
           = (1 + 1/n)ⁿ → e ≈ 2.718  > 1

L = e > 1, so the series diverges. Reverse the previous example and the verdict reverses cleanly.

4. ∑ 1/n (the harmonic series). Compute:

a_{n+1}/aₙ = n/(n+1) → 1

L = 1, inconclusive. The ratio test cannot decide. Independent argument: the integral test compares ∑ 1/n to ∫ dx/x = ln x, divergent.

5. ∑ 1/n². Compute:

a_{n+1}/aₙ = n²/(n+1)² → 1

L = 1 again, inconclusive. The integral test gives ∫ dx/x² = −1/x|₁^∞ = 1, finite, so the series converges. Both p-series ∑ 1/n and ∑ 1/n² hit L = 1 with opposite verdicts — the canonical demonstration that L = 1 carries no information.

Ratio test on a power series

Apply the ratio test to ∑ cₙ xⁿ:

|c_{n+1} x^(n+1)| / |cₙ xⁿ| = |x| · |c_{n+1}/cₙ|

Suppose |c_{n+1}/cₙ| → 1/R. Then the limit becomes |x|/R, and the ratio test says:

• |x|/R < 1, i.e. |x| < R: convergence.
• |x|/R > 1, i.e. |x| > R: divergence.
• |x| = R: ratio test says nothing.

R is the radius of convergence. Behaviour on the boundary circle |x| = R must be checked separately, often by other tests.

Example — the Taylor series of e^x is ∑ xⁿ/n!. Compute:

|x^(n+1) / (n+1)!| / |xⁿ / n!| = |x| / (n + 1) → 0 for every x

L = 0 < 1 for all x, so the series converges absolutely for every real or complex x. Radius of convergence is infinite. By contrast, the geometric series ∑ xⁿ has |x^(n+1)/xⁿ| = |x|, so R = 1 — converges for |x| < 1, diverges for |x| > 1.

What to do when L = 1

Situation	Better tool	Reason
Rational function of n, like ∑ (n+1)/(n³+5)	Limit comparison with a p-series	Limit-comparison preserves the polynomial-degree intuition
aₙ = f(n) with f decreasing, integrable	Integral test	Reduces to a calculus integral you can evaluate
p-series-like, ∑ 1/(n · (ln n)ᵖ)	Cauchy condensation or Raabe's test	Pulls out the slow log factor that the ratio test misses
Alternating, |aₙ| → 0 monotonically	Alternating series test	Cancellation that the ratio test ignores can produce convergence
Pure n^k powers, like ∑ (n/(n+1))ⁿ²	Root test	Root test extracts the underlying geometric ratio more cleanly

Raabe's test is the most useful follow-up when L = 1. Compute:

R = lim_{n→∞} n · (|aₙ / a_{n+1}| − 1)

If R > 1 the series converges; R < 1 it diverges. Raabe's test handles many of the p-series-style problems where the ratio test fails.

Ratio test vs other convergence tools

	Ratio test	Root test	Integral test	Alternating series
Quantity computed	lim |a_{n+1}/aₙ|	lim ⁿ√|aₙ|	∫ f(x) dx convergence	monotone bₙ → 0?
Best at	Factorials, exponentials, n^k	Pure powers of n; nested powers	Slowly-decaying or monotone aₙ	Alternating signs
Inconclusive case	L = 1	L = 1	None (clean dichotomy)	bₙ not eventually decreasing
Strength	Weaker than root test	Stronger than ratio test	Hard for arbitrary aₙ	Very specific, but proves convergence quickly
Computation cost	Cheap (algebra)	Sometimes harder (n-th roots)	Need to evaluate integral	Cheap (check monotonicity, limit)
Tells you absolute convergence?	Yes	Yes	Yes (positive aₙ implies absolute)	No — only ordinary convergence

The root test logically dominates the ratio test: any L < 1 verdict from the ratio test is also reached by the root test, and the root test sometimes succeeds where the ratio test fails (e.g. ratios that oscillate but n-th roots that converge). Practically, the ratio test is preferred when factorials are involved, because algebraically the cancellations are cleaner.

Where the ratio test is the workhorse

• Taylor series radius. Computing the radius of convergence of cos x, sin x, e^x, ln(1+x), arctan x, the binomial series — all standard ratio-test exercises. The ratio of consecutive coefficients comes from the derivatives of the function at the expansion point, and limits of those ratios decide where the polynomial approximation is valid.
• Special-function series. The hypergeometric function ₂F₁(a, b; c; x), Bessel functions Jₙ(x), Legendre polynomials Pₙ(x) — all defined as series whose convergence radii come from ratio-test calculations. Mathematicians and physicists derive these radii once and look them up forever after.
• Generating functions in combinatorics. ∑ aₙ xⁿ has a radius of convergence governed by lim |a_{n+1}/aₙ| (when the limit exists). Whether this radius is positive or zero distinguishes "the sequence grows at most geometrically" from "the sequence grows super-exponentially". For Catalan numbers Cₙ ~ 4ⁿ/(n^(3/2) √π), the ratio test gives R = 1/4 immediately.
• Probability distributions. Computing whether an infinite sum of probabilities is finite (so it can be normalised to a distribution) often reduces to a ratio test on factorial-laden expressions. The Poisson distribution's normalisation ∑ λⁿ/n! = e^λ is verified by the ratio test giving L = 0 for any fixed λ.

Common mistakes

• Concluding anything when L = 1. The ratio test gives no information at L = 1. ∑ 1/n diverges, ∑ 1/n² converges, both with L = 1. Falling back on a different test is mandatory; never write "since L = 1, the series converges".
• Forgetting the absolute value. The test is L = lim |a_{n+1}/aₙ|, not lim a_{n+1}/aₙ. Without absolute values, alternating signs leave you with a limit that does not even exist or is negative — and the geometric-series argument requires the magnitude.
• Applying ratio test to series with zero terms. If aₙ = 0 for some n, the ratio a_{n+1}/aₙ is undefined. Either restrict to the non-zero subsequence or use a different test.
• Treating limsup as the limit. If the ratios oscillate (e.g., aₙ alternates between 2ⁿ and 3ⁿ patterns), no plain limit exists. The strict ratio test does not apply; the root test usually does, since limsup ⁿ√|aₙ| still exists.
• Using ratio-test convergence without checking the boundary. For a power series ∑ cₙ xⁿ with radius R, ratio test gives convergence for |x| < R and divergence for |x| > R, but says nothing at |x| = R. The Taylor series of ln(1+x) has R = 1, converges at x = 1 (alternating-series test), and diverges at x = −1 (harmonic series). You need separate analysis at the boundary.