Ratio Test for Convergence: Study Guide (Calculus BC 10.8)

BC-only material. The ratio test is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

Prerequisite	Topic	What you use it for here
Geometric series	10.2	The ratio test is built on them
nth term test	10.3	Explains why L > 1 means divergence
p-series	10.5	The usual fallback when L = 1
Comparison tests	10.6	The other fallback, and the idea behind the proof
Limits at infinity	1.15	Finding L

Notation on this page: Σ from n = 1 to ∞ of aₙ is the series a₁ + a₂ + a₃ + …, and n! = 1 · 2 · 3 · … · n (with 0! = 1). Read n! as “n factorial”.

The idea: compare each term with the one before

In a geometric series every term is the previous one times a fixed number r. For example, in Σ 5(2/3)ⁿ, each term is 2/3 of the one before. Such a series converges exactly when |r| < 1.

Most series do not have a fixed ratio. But many have a ratio that settles down as n grows. If, far along the series, each term is roughly 1/3 of the one before, then the tail of the series behaves like a geometric series with ratio 1/3, and it should converge. The ratio test turns that idea into a test.

The ratio test

Ratio test. For a series Σ aₙ whose terms are not zero, find L = lim (n → ∞) |aₙ₊₁ / aₙ|.

• If L < 1, the series converges (in fact, Σ |aₙ| converges too).
• If L > 1, or the ratio grows without bound (L = ∞), the series diverges.
• If L = 1, the test gives no conclusion.

Why each case holds:

• L < 1. Pick a number r between L and 1. From some n on, every ratio |aₙ₊₁/aₙ| is below r. So from that point each |aₙ| is at most a fixed constant times rⁿ. The geometric series Σ rⁿ converges (r < 1), so by comparison Σ |aₙ| converges, and so does Σ aₙ.
• L > 1. From some n on, each |aₙ₊₁| is bigger than |aₙ|. The sizes grow, so aₙ cannot tend to 0. The series diverges by the nth term test.
• L = 1. The terms shrink (or grow) more slowly than any geometric series, so the comparison gives no information.

Other tests exist (for example the root test), but on this course you only need the ratio test alongside the tests from Topics 10.3 to 10.7.

Algebra you will need

• Powers: 3ⁿ⁺¹/3ⁿ = 3, and 2²⁽ⁿ⁺¹⁾/2²ⁿ = 4.
• Factorials: (n + 1)! = (n + 1) · n!, so (n + 1)!/n! = n + 1.
• Factorials of 2n: (2(n + 1))! = (2n + 2)! = (2n + 2)(2n + 1) · (2n)!, so (2n + 2)!/(2n)! = (2n + 2)(2n + 1).
• Polynomials: a ratio such as (n + 1)²/n² → 1. Any ratio of polynomials of the same degree tends to the ratio of leading coefficients.

Write aₙ₊₁/aₙ as aₙ₊₁ × (1/aₙ), group the powers together and the factorials together, then cancel.

Worked example 1: powers of n against an exponential

Question. Determine whether Σ from n = 1 to ∞ of (n² + 1)/3ⁿ converges. Justify your answer.

1. Write aₙ and aₙ₊₁. aₙ = (n² + 1)/3ⁿ and aₙ₊₁ = ((n + 1)² + 1)/3ⁿ⁺¹ = (n² + 2n + 2)/3ⁿ⁺¹.
2. Form the ratio. All terms are positive, so the absolute value changes nothing: aₙ₊₁/aₙ = (n² + 2n + 2)/3ⁿ⁺¹ × 3ⁿ/(n² + 1) = (n² + 2n + 2) / [3(n² + 1)].
3. Find the limit. Divide the top and bottom by n²: L = lim (1 + 2/n + 2/n²) / [3(1 + 1/n²)] = 1/3.
4. Conclude. L = 1/3 < 1, so the series converges by the ratio test.

What the graph shows. The ratio test is about the long run. The first few ratios are not 1/3, and that does not matter. What matters is the value they settle towards and which side of 1 it is on.

Background (not needed for the exam): this series happens to sum to 2. The ratio test does not give the sum.

Worked example 2: factorials

Question. Decide whether each series converges or diverges.

(a) Σ from n = 1 to ∞ of (2n)! / [(n!)² · 5ⁿ]

1. Write aₙ₊₁. aₙ₊₁ = (2n + 2)! / [((n + 1)!)² · 5ⁿ⁺¹].
2. Group like parts. aₙ₊₁/aₙ = [(2n + 2)!/(2n)!] × [(n!)²/((n + 1)!)²] × [5ⁿ/5ⁿ⁺¹].
3. Cancel each group.
   • (2n + 2)!/(2n)! = (2n + 2)(2n + 1).
   • (n!)²/((n + 1)!)² = 1/(n + 1)².
   • 5ⁿ/5ⁿ⁺¹ = 1/5.
4. Simplify. aₙ₊₁/aₙ = (2n + 2)(2n + 1) / [5(n + 1)²] = 2(n + 1)(2n + 1) / [5(n + 1)²] = 2(2n + 1) / [5(n + 1)].
5. Limit. L = lim (4n + 2)/(5n + 5) = 4/5.
6. Conclude. L = 4/5 < 1, so the series converges by the ratio test.

Check with numbers: a₁ = 2/5, a₂ = 6/25 and a₂/a₁ = 3/5, which matches 2(3)/(5 · 2) = 3/5 from the formula. ✓

(b) Σ from n = 1 to ∞ of n!/10ⁿ

1. aₙ₊₁/aₙ = [(n + 1)!/n!] × [10ⁿ/10ⁿ⁺¹] = (n + 1)/10.
2. As n → ∞, (n + 1)/10 → ∞. So L = ∞, which counts as L > 1.
3. Conclude. The series diverges by the ratio test.

Interpretation. The first terms of (b) are tiny (a₁₀ ≈ 0.0004), which can fool you into thinking the series converges. But once n ≥ 10, each ratio (n + 1)/10 is above 1, so every term is bigger than the one before; by n = 30, aₙ is about 265. Factorials eventually beat any exponential.

When the ratio test fails: L = 1

Try the ratio test on two p-series:

Series	aₙ₊₁/aₙ	L	Truth
Σ 1/n	n/(n + 1)	1	diverges (harmonic)
Σ 1/n²	n²/(n + 1)²	1	converges (p = 2 > 1)

Both give L = 1, but one converges and one diverges. So L = 1 tells you nothing. This happens for every series whose terms are built only from powers of n, such as (3n + 1)/(n³ + 2): the ratio of two such expressions at n + 1 and n always tends to 1.

When L = 1, switch to the test that fits: a p-series, a comparison or limit comparison test, the integral test, or (for alternating signs) the alternating series test.

Choosing the ratio test

Terms contain …	Ratio test?	Usually better
n! or (2n)!	Yes, first choice	—
A power with n in the exponent: 3ⁿ, (0.9)ⁿ, 2²ⁿ	Yes	Geometric test, if the series is exactly geometric
Only powers of n (rational functions, roots)	No: L = 1	p-series, comparison, limit comparison
ln n, or forms like 1/(n ln n)	Usually L = 1	Integral or comparison tests

Common misconceptions

• Flipping the ratio. The ratio is aₙ₊₁/aₙ (next over current). Using aₙ/aₙ₊₁ gives 1/L and reverses the conclusion.
• “L = 1 means the series diverges.” L = 1 gives no conclusion. Σ 1/n² has L = 1 and converges.
• “L < 1 means the terms tend to 0, so that is all it shows.” L < 1 proves the whole series converges, not just that the terms shrink.
• Using early ratios. The first few ratios can be above 1 even when L < 1, and the reverse. Only the limit counts.
• Factorial errors. (n + 1)! is (n + 1) · n!, not n! + 1, and (2n + 2)! is not (2n)! · 2.
• Dropping the absolute value for terms with signs, such as (−2)ⁿ. Without the bars, the “ratio” is negative and you may compare the wrong number with 1.
• Forgetting to conclude. End with the value of L, its comparison with 1, and “by the ratio test”.

Where this leads

Next, Topic 10.9 uses the ratio test on Σ |aₙ| to show absolute convergence: see the absolute and conditional convergence study guide. Topic 10.13 applies the same ratio to power series, where L depends on x, to find a radius and interval of convergence. The previous topic, the alternating series test, covers one of the cases where the ratio test gives L = 1. Use the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.