Ratio Test & Root Test for Series

§ 01The Shared Idea — Geometric Series as a Template

Both tests work by comparing your series to a geometric series. If the ratio between consecutive terms (or the nᵗʰ root of each term) is eventually less than 1, the series behaves like a convergent geometric series.

Why two tests that do the same job? The Root Test is theoretically stronger — whenever the Ratio Test works, the Root Test also works, but not vice versa. However, the Ratio Test is usually far easier to apply when factorials or products are present, because cancellation happens naturally in the ratio. The Root Test shines for series involving explicit nᵗʰ powers like (f(n))ⁿ.

§ 02The Ratio Test (D'Alembert)

Ratio Test — Full Statement

Let Σaₙ be a series of non-zero terms. Compute:

L = lim_n→∞ |a_n+1| / |aₙ|

The test then gives three possible outcomes based on L:

L < 1

Series converges absolutely

Terms shrink fast enough — like a geometric series with ratio L < 1

L > 1 (or L = ∞)

Series diverges

Terms grow — eventually |aₙ₊₁| > |aₙ|, so aₙ ↛ 0

L = 1

Test is inconclusive

Could converge or diverge — need another test

Why L = 1 is the Dead Zone

When L = 1, the Ratio Test provides no information. As a striking illustration: both the harmonic series Σ 1/n (divergent) and Σ 1/n² (convergent) give L = 1 in the Ratio Test, yet they have opposite behaviour. The boundary L = 1 is where the geometric comparison breaks down.

The Ratio Test always gives L = 1 for p-series For Σ 1/nᵖ: L = lim (n/(n+1))ᵖ = 1 for every p. This is why the Ratio Test is useless for p-series and you must use the p-series theorem directly or the Integral Test instead.

Factorial Simplification — The Key Technique

The Ratio Test's power comes from factorial cancellation. When aₙ involves n!, forming the ratio a_n+1/aₙ causes almost everything to cancel, leaving a simple expression.

Factorial Cancellation Pattern

^(n+1)!⁄_n! = n+1 ^n!⁄_(n+1)! = ¹⁄_n+1 ^2ⁿ⁺¹⁄_2ⁿ = 2 ^rⁿ⁺¹⁄_rⁿ = r

Memorise these: they appear in nearly every Ratio Test calculation.

Worked Examples — Ratio Test

Example 1 — Series with n!

Test Σ_n=0^∞ n! / 5ⁿ for convergence.

Set up the ratio: |a_n+1/aₙ| = [(n+1)! / 5^(n+1)] / [n! / 5ⁿ]
= (n+1)! · 5ⁿ / (n! · 5^(n+1)) = (n+1) / 5
Take the limit: L = lim (n+1)/5 = ∞ > 1.
Since L > 1, the series diverges. (The terms blow up, n! grows faster than 5ⁿ.)
Diverges (L = ∞)

Example 2 — Exponential over factorial

Test Σ_n=0^∞ 3ⁿ / n! for convergence.

Ratio: |a_n+1/aₙ| = [3^(n+1) / (n+1)!] / [3ⁿ / n!]
= 3^(n+1) · n! / (3ⁿ · (n+1)!) = 3 / (n+1)
L = lim 3/(n+1) = 0 < 1.
Series converges absolutely. (This is the Maclaurin series for e³ = Σ 3ⁿ/n!, which sums to e³ ≈ 20.09.)
Converges (L = 0)

Example 3 — Power over power

Test Σ_n=1^∞ (2n)! / (n!)² for convergence.

Ratio: a_n+1/aₙ = [(2n+2)! / ((n+1)!)²] / [(2n)! / (n!)²]
= (2n+2)! · (n!)² / ((n+1)!)² · (2n)!)
Simplify: (2n+2)!/(2n)! = (2n+2)(2n+1). And (n!)²/((n+1)!)² = 1/(n+1)². = (2n+2)(2n+1)/(n+1)² = 2(n+1)(2n+1)/(n+1)² = 2(2n+1)/(n+1)
L = lim 2(2n+1)/(n+1) = lim (4n+2)/(n+1) = 4 > 1.
Diverges (L = 4)

Example 4 — Mixed polynomial and exponential

Test Σ_n=1^∞ n² / 2ⁿ for convergence.

Ratio: |(n+1)² / 2^(n+1)| / |n² / 2ⁿ| = (n+1)² / (2 · n²)
L = lim (n+1)² / (2n²) = lim (n² + 2n + 1)/(2n²) = 1/2 < 1.
Converges absolutely.
Converges (L = 1/2)

Example 5 — Ratio test giving L = 1 (inconclusive)

Apply the Ratio Test to Σ 1/n³. What happens?

Ratio: |1/(n+1)³| / |1/n³| = n³/(n+1)³ = (n/(n+1))³
L = lim (n/(n+1))³ = 1³ = 1.
L = 1 ⟹ Ratio Test is inconclusive.
Use the p-series test instead: p = 3 > 1 ⟹ converges.
Inconclusive (use p-series)

Example 6 — Ratio test for power series radius

Find the radius of convergence of Σ_n=0^∞ xⁿ / n!.

Ratio: |x^(n+1)/(n+1)!| / |xⁿ/n!| = |x| · n!/(n+1)! = |x|/(n+1).
L = lim |x|/(n+1) = 0 for any fixed x.
Since L = 0 < 1 for all x ∈ ℝ, the series converges for every real number.
Radius of convergence R = ∞. (This is the Taylor series for eˣ.)
Converges for all x (R = ∞)

§ 03The Root Test (Cauchy)

Root Test — Full Statement

Let Σaₙ be any series. Compute the limsup (or limit if it exists):

L = lim_n→∞ |aₙ|^1/n

Same three-way conclusion as the Ratio Test: L < 1 ⟹ converges, L > 1 ⟹ diverges, L = 1 ⟹ inconclusive.

L < 1

Converges absolutely

|aₙ| < rⁿ eventually for some r < 1 — geometric bound

L > 1 (or ∞)

Diverges

|aₙ|^(1/n) > 1 infinitely often ⟹ |aₙ| ↛ 0

L = 1

Inconclusive

Same dead zone as the Ratio Test

Key Limit for the Root Test

The most important limit needed for the Root Test is:

Essential Limits

lim_n→∞ n^1/n = 1 lim_n→∞ (f(n))^1/n = lim_n→∞ e^ln(f(n))/n lim_n→∞ (aⁿ)^1/n = a for any constant a > 0

Polynomial factors like nᵏ contribute n^(k/n) → 1 and can be ignored in the limit.

Worked Examples — Root Test

Example 7 — Pure nᵗʰ power

Test Σ_n=1^∞ (3/4)ⁿ using the Root Test.

|aₙ|^(1/n) = |(3/4)ⁿ|^(1/n) = 3/4.
L = 3/4 < 1.
Series converges absolutely. (This is a geometric series; sum = (3/4)/(1 − 3/4) = 3.)
Converges (L = 3/4)

Example 8 — Polynomial base to the nᵗʰ power

Test Σ_n=1^∞ (2n/(3n+1))ⁿ for convergence.

|aₙ|^(1/n) = |(2n/(3n+1))ⁿ|^(1/n) = 2n/(3n+1).
L = lim 2n/(3n+1) = 2/3 < 1.
Series converges absolutely.
Converges (L = 2/3)

Example 9 — Root test: divergence

Test Σ_n=1^∞ (n/(n+1))^(n²) for convergence.

|aₙ|^(1/n) = [(n/(n+1))^(n²)]^(1/n) = (n/(n+1))ⁿ.
Rewrite: (n/(n+1))ⁿ = (1 − 1/(n+1))ⁿ.
As n → ∞: (1 − 1/(n+1))ⁿ → e^(−1) = 1/e ≈ 0.368 < 1.
L = 1/e < 1 ⟹ converges absolutely.
Converges (L = 1/e)

Example 10 — Root test: the 1/eⁿ limit

Test Σ_n=1^∞ n^n / eⁿ² for convergence. (Hint: use the Root Test.)

|aₙ|^(1/n) = [n^n / e^(n²)]^(1/n) = n^(n/n) / e^(n²/n) = n / eⁿ.
L = lim n/eⁿ. Since eⁿ grows much faster than n, this limit is 0 < 1.
Converges absolutely.
Converges (L = 0)

§ 04Ratio vs Root — When to Use Each

Feature	Ratio Test	Root Test
Best for	Series with n! (factorials), products of n terms, or explicit exponentials like aⁿ or rⁿ	Series of the form (f(n))ⁿ — anything raised to the nᵗʰ power
Key operation	Compute a_n+1/aₙ and cancel	Compute \|aₙ\|^(1/n) and simplify
Factorial handling	Excellent — (n+1)!/n! = n+1 cancels cleanly	Poor — n!^(1/n) requires Stirling's approximation
nᵗʰ power handling	Works but messier	Excellent — ((f(n))ⁿ)^(1/n) = f(n) collapses immediately
Theoretical strength	Weaker — Root Test always gives same or better result	Stronger — if Ratio Test gives L, Root Test gives same L
Fails for (L = 1)	All p-series, many rational-term series	All p-series, many rational-term series
Power series radius	Both work: R = 1/L	Both work: R = 1/L

Quick decision rule See n! → use Ratio Test. See (something)ⁿ → try Root Test first. See only polynomials in n → try comparison tests or p-series instead (both Ratio and Root give L = 1 and are inconclusive).

Two More Worked Examples — Mixed Situations

Example 11 — Ratio test for power series

Find the radius of convergence of Σ_n=1^∞ n · xⁿ / 3ⁿ.

aₙ = n · xⁿ / 3ⁿ. Apply Ratio Test: |a_n+1/aₙ| = |(n+1)x^(n+1)/3^(n+1)| / |nxⁿ/3ⁿ| = (n+1)|x| / (3n)
L = lim (n+1)|x|/(3n) = |x|/3.
Converges when L < 1: |x|/3 < 1 iff |x| < 3.
Radius of convergence R = 3. At |x| = 3 (the boundary), check separately with other tests.
Converges for |x| < 3 (R = 3)

Example 12 — Choose wisely: Root or Ratio?

Test Σ_n=1^∞ (n/(n+1))ⁿ · (1/2ⁿ) for convergence using the Root Test.

aₙ = (n/(n+1))ⁿ · (1/2)ⁿ = [(n/(n+1)) · (1/2)]ⁿ = (n/(2(n+1)))ⁿ.
Root Test: |aₙ|^(1/n) = n/(2(n+1)).
L = lim n/(2(n+1)) = 1/2 < 1.
Converges absolutely.
Converges (L = 1/2 via Root Test)

Interactive — ratio |a_n+1/aₙ| converging to L for selectable series

Series:

Terms shown: 30

Rust curve: |a_{n+1}/a_n|. Gold dashed: L value. When the curve settles below 1, the series converges.

Interactive — Root Test: |aₙ|^(1/n) converging to L

Series:

Teal curve: |aₙ|^(1/n). Gold dashed: L. Settles below/above 1 to determine convergence/divergence.

§ 05Common Mistakes

Mistake 1 — Concluding convergence from L = 1 L = 1 is strictly inconclusive. Both Σ 1/n (diverges) and Σ 1/n² (converges) give L = 1 in the Ratio Test. If you get L = 1, you must switch to a different test.

Mistake 2 — Forgetting absolute values The limit must use |a_n+1/aₙ|, not just the ratio. For alternating series or series with negative terms, ignoring absolute values can give a negative L — which is meaningless. The ratio must always be non-negative.

Mistake 3 — Applying Ratio Test to p-series As shown in Example 5, the Ratio Test always gives L = 1 for p-series, regardless of p. Never use the Ratio or Root Test on series with only polynomial terms in the denominator — use p-series or comparison tests instead.

Mistake 4 — Misapplying n^(1/n) → 1 In the Root Test, polynomial factors like nᵏ contribute |nᵏ|^(1/n) = n^(k/n) → 1, and so they vanish in the limit. Students sometimes forget this and leave polynomial factors in the answer, getting an incorrect L.

Mistake 5 — Confusing the radius formula For power series Σ aₙxⁿ, if the Ratio Test gives lim |a_n+1/aₙ| = C, then the radius of convergence is R = 1/C (not C). Always state: converges when C|x| < 1, i.e. |x| < 1/C = R.

Ratio & Root Tests

§ 01The Shared Idea — Geometric Series as a Template

§ 02The Ratio Test (D'Alembert)

Why L = 1 is the Dead Zone

Factorial Simplification — The Key Technique

Worked Examples — Ratio Test

§ 03The Root Test (Cauchy)

Key Limit for the Root Test

Worked Examples — Root Test

§ 04Ratio vs Root — When to Use Each

Two More Worked Examples — Mixed Situations

§ 05Common Mistakes

§ 06Related Pages

§ 07Quiz — Test Your Understanding