Which test is the first to try on any series?

Series Convergence — All Convergence Tests

01 — From Sequences to Series

A series is what you get when you add the terms of a sequence. Given a sequence {aₙ}, the corresponding series is:

Infinite Series

Σ_n=1^∞ aₙ = a₁ + a₂ + a₃ + ⋯

The n-th partial sum is Sₙ = a₁ + a₂ + ⋯ + aₙ.

The series converges to S if and only if lim_n→∞ Sₙ = S (a finite number). Otherwise it diverges.

Convergence is a question about the partial sums {Sₙ}: does this new sequence converge? Every test in this page answers that question by a different route.

02 — Test 1: The Divergence Test

Always try this first. It takes five seconds and can immediately rule out convergence.

TEST 1 Divergence Test (nth-Term Test)

If lim_n→∞ aₙ ≠ 0, then Σ aₙ diverges.

DIVERGES if lim aₙ ≠ 0

INCONCLUSIVE if lim aₙ = 0

Critical warning: The converse is false. lim aₙ = 0 does not guarantee convergence. The harmonic series Σ 1/n has terms → 0 yet diverges.

03 — Known Series: Geometric and p-Series

Before testing, check if the series matches a known form.

KNOWN Geometric Series & p-Series

Geometric: Σ arⁿ → converges to a/(1−r) if |r| < 1; diverges if |r| ≥ 1.
p-Series: Σ 1/nᵖ → converges if p > 1; diverges if p ≤ 1.
Harmonic (p=1): Σ 1/n → DIVERGES. This is the canonical example of lim aₙ = 0 yet diverges.

04 — Test 2: The Integral Test

TEST 2 Integral Test

If f is continuous, positive, and decreasing on [1, ∞) with f(n) = aₙ, then:

Σ aₙ converges ⟺ ∫₁^∞ f(x) dx converges

The test gives convergence/divergence only — it does NOT give the sum of the series.

CONVERGES if the integral converges

DIVERGES if the integral diverges

Best for: Series whose general term is a nice function you can integrate, especially p-series, log terms, and terms involving eˣ.

05 — Test 3: The Comparison Test

TEST 3 Direct Comparison Test (DCT)

Suppose 0 ≤ aₙ ≤ bₙ for all n ≥ N. Then:

If Σ bₙ converges → Σ aₙ converges (smaller than a convergent series).
If Σ aₙ diverges → Σ bₙ diverges (larger than a divergent series).

Strategy: compare with a geometric series or p-series of similar behaviour. You need the inequality to go the right way, this takes practice.

06 — Test 4: The Limit Comparison Test

TEST 4 Limit Comparison Test (LCT)

Suppose aₙ, bₙ > 0. Compute the limit:

L = lim_n→∞ aₙ / bₙ

If 0 < L < ∞ (finite and positive): Σ aₙ and Σ bₙ both converge or both diverge.

Advantage: no need to find the inequality direction — just compute a limit.

SAME behaviour as Σ bₙ when 0 < L < ∞

INCONCLUSIVE when L = 0 or L = ∞

Best for: Rational functions, or any aₙ that "looks like" a known series for large n. Find the dominant terms in numerator and denominator.

07 — Test 5: The Ratio Test

TEST 5 Ratio Test (d'Alembert)

Compute the limit of consecutive term ratios:

L = lim_n→∞ |aₙ₊₁ / aₙ|

L < 1 → converges absolutely

L > 1 (or L = ∞) → diverges

L = 1 → inconclusive — try another test

Best for: series involving factorials (n!), exponentials (aⁿ), or products, because consecutive ratios simplify beautifully. Fails at L = 1 for all p-series.

08 — Test 6: The Root Test

TEST 6 Root Test (Cauchy)

Compute the n-th root of |aₙ|:

L = lim_n→∞ |aₙ|^(1/n)

L < 1 → converges absolutely | L > 1 → diverges | L = 1 → inconclusive

Best for: series where aₙ is itself an n-th power, e.g. aₙ = (f(n))ⁿ. The n-th root collapses the exponent cleanly.

09 — Test 7: The Alternating Series Test

TEST 7 Alternating Series Test (Leibniz)

For series of the form Σ (−1)ⁿ bₙ (or Σ (−1)ⁿ⁺¹ bₙ) with bₙ > 0:

bₙ is decreasing: bₙ₊₁ ≤ bₙ for all n ≥ N
lim_n→∞ bₙ = 0

If both conditions hold, the series converges.

CONVERGES (conditionally, at minimum)

Error bound: |S − Sₙ| ≤ bₙ₊₁, the error is bounded by the first omitted term. Useful for numerical estimates.

Absolute vs conditional: If Σ |aₙ| also converges, the series converges absolutely. If Σ |aₙ| diverges but Σ aₙ converges, it is conditionally convergent.

10 — Decision Flowchart: Which Test to Use?

How to choose a convergence test

11 — Quick-Reference Summary Table

Test	Condition	Conclusion	Best used for
Divergence	lim aₙ ≠ 0	Diverges	First check always
Geometric	\|r\| < 1	Converges = a/(1−r)	Recognising the form
p-Series	p > 1	Converges	Recognising Σ 1/nᵖ
Integral	∫ f(x)dx converges	Σ aₙ converges	Integrable terms, log
Direct Comparison	aₙ ≤ bₙ, Σbₙ conv.	Σ aₙ converges	When inequality is clear
Limit Comparison	lim aₙ/bₙ = L ∈ (0,∞)	Same as Σ bₙ	Rational-like terms
Ratio	lim\|aₙ₊₁/aₙ\| < 1	Converges absolutely	Factorials, exponentials
Root	lim\|aₙ\|^(1/n) < 1	Converges absolutely	n-th power terms
Alternating Series	bₙ↓0, alternating signs	Converges	(−1)ⁿ series

12 — Twelve Worked Examples

EX 01Divergence Test — immediate divergence

Test Σ n/(n+1) for convergence.

Compute: lim n/(n+1) = lim 1/(1 + 1/n) = 1 ≠ 0

Since the limit of terms ≠ 0, the Divergence Test applies.

DIVERGES — Divergence Test.

EX 02Geometric series — direct formula

Find the sum of Σ_n=0^∞ 4·(3/5)ⁿ.

a = 4, r = 3/5. |r| = 0.6 < 1 ✓

S = 4/(1 − 3/5) = 4/(2/5) = 10

CONVERGES, S = 10.

EX 03p-Series test

Does Σ 1/n³ converge?

This is a p-series with p = 3.

p = 3 > 1, so the series converges.

CONVERGES — p-series with p = 3 > 1.

EX 04Integral Test — harmonic-like series

Test Σ 1/(n ln n) for convergence (n ≥ 2).

f(x) = 1/(x ln x) is positive, continuous, decreasing on [2, ∞) ✓

Integrate: ∫ dx/(x ln x) = ln|ln x| + C

∫₂^∞ dx/(x ln x) = [ln(ln x)]₂^∞ = ∞

DIVERGES — Integral Test (integral diverges).

EX 05Direct Comparison Test

Test Σ 1/(n² + n) for convergence.

Note: n² + n > n² for all n ≥ 1.

Therefore: 1/(n² + n) < 1/n² for all n ≥ 1.

Σ 1/n² converges (p-series, p = 2 > 1).

By DCT: our series ≤ convergent series → converges.

CONVERGES — Direct Comparison with Σ 1/n².

EX 06Limit Comparison Test

Test Σ (3n + 1)/(n³ − 2n) for convergence.

Dominant terms for large n: numerator ~ 3n, denominator ~ n³. Compare with bₙ = 1/n².

Compute: L = lim [(3n+1)/(n³−2n)] / [1/n²] = lim (3n³+n²)/(n³−2n)

Divide by n³: L = lim (3 + 1/n)/(1 − 2/n²) = 3

0 < L = 3 < ∞. Σ 1/n² converges (p = 2). By LCT, our series also converges.

CONVERGES — Limit Comparison with Σ 1/n², L = 3.

EX 07Ratio Test — factorial series

Test Σ n!/nⁿ for convergence.

Compute ratio: |(n+1)!/(n+1)ⁿ⁺¹| / |n!/nⁿ| = (n+1)·n! / [(n+1)·(n+1)ⁿ] · nⁿ/n!

Simplify: = nⁿ/(n+1)ⁿ = (n/(n+1))ⁿ = (1 − 1/(n+1))ⁿ

As n → ∞: (1 − 1/(n+1))ⁿ → e⁻¹ = 1/e ≈ 0.368 < 1

CONVERGES — Ratio Test, L = 1/e < 1.

EX 08Ratio Test — exponential vs factorial

Test Σ 5ⁿ/n! for convergence.

Ratio: |5ⁿ⁺¹/(n+1)!| / |5ⁿ/n!| = 5/(n+1)

L = lim 5/(n+1) = 0 < 1.

CONVERGES — Ratio Test, L = 0. (The sum is e⁵ by the Taylor series for eˣ.)

EX 09Root Test

Test Σ (2n/(n+1))ⁿ for convergence.

Take n-th root: |aₙ|^(1/n) = 2n/(n+1)

lim 2n/(n+1) = 2 > 1

DIVERGES — Root Test, L = 2 > 1.

EX 10Alternating Series Test

Test Σ (−1)ⁿ/√n for convergence.

bₙ = 1/√n > 0 ✓

Decreasing: 1/√(n+1) < 1/√n ✓

lim bₙ = lim 1/√n = 0 ✓

Both AST conditions satisfied. But Σ 1/√n is a p-series with p = 1/2 < 1, so it diverges → conditionally convergent.

CONVERGES conditionally — Alternating Series Test.

EX 11Ratio Test inconclusive — switch to LCT

Test Σ 1/n² using the Ratio Test, then fix it.

Ratio: L = lim (1/(n+1)²)/(1/n²) = lim n²/(n+1)² = 1

L = 1 → Ratio Test inconclusive. Switch approach.

Recognise as p-series with p = 2 > 1 → converges directly.

CONVERGES (p-series). The Ratio Test always gives L = 1 for p-series — avoid it for these.

EX 12Absolute vs conditional convergence

Classify Σ (−1)ⁿ/n²: absolutely convergent, conditionally convergent, or divergent?

Test |aₙ| = 1/n²: p-series with p = 2 > 1 → Σ |aₙ| converges.

Since Σ |aₙ| converges, the original series converges absolutely.

ABSOLUTELY CONVERGENT. (Contrast: Σ(−1)ⁿ/n converges only conditionally.)

13 — Common Mistakes

⚠️

Mistake 1 — Reversing the Divergence Test: If lim aₙ = 0, the series might converge, but it might not. The harmonic series Σ 1/n is the canonical counterexample. Do not conclude convergence from lim aₙ = 0.

⚠️

Mistake 2 — Using Ratio Test on p-series: The Ratio Test always gives L = 1 for p-series, which is inconclusive. Use the p-series rule directly (converges iff p > 1).

⚠️

Mistake 3 — Forgetting the Integral Test gives no sum: The Integral Test tells you whether a series converges — it does not give you what it converges to. The sum of Σ 1/n² is π²/6 (found by Euler), not ∫ 1/x² dx.

⚠️

Mistake 4 — Ratio Test with L = 1: When L = 1, the test is completely silent — you cannot conclude convergence or divergence. Pick a different test.

🔗

Next steps: For a deep dive on a single test, see Geometric Series, p-Series, Comparison Tests, and Alternating Series.

🔗 Related Topics — Series & Sequences

Sequences Geometric Series p-Series Test Comparison Tests Ratio & Root Tests Alternating Series Power Series Taylor Series Limits at Infinity Improper Integrals All Series Topics

Series Convergence
Every Test, One Page

01 — From Sequences to Series

02 — Test 1: The Divergence Test

03 — Known Series: Geometric and p-Series

04 — Test 2: The Integral Test

05 — Test 3: The Comparison Test

06 — Test 4: The Limit Comparison Test

07 — Test 5: The Ratio Test

08 — Test 6: The Root Test

09 — Test 7: The Alternating Series Test

10 — Decision Flowchart: Which Test to Use?

11 — Quick-Reference Summary Table

12 — Twelve Worked Examples

13 — Common Mistakes

🔗 Related Topics — Series & Sequences

14 — Quiz: Series Convergence (10 Questions)