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🎯⭐ INTERACTIVE LESSON

Convergence Tests Summary

Learn step-by-step with interactive practice!

Convergence Tests Summary - Complete Interactive Lesson

Part 1: Core Concepts

The Master Convergence Test Guide

Part 1 of 7 — Overview of All Tests

Every Convergence Test You Need

TestApplies toConclusion
DivergenceAny ∑an\sum a_nIf lim⁡an≠0\lim a_n \neq 0, diverges
Geometric∑arn\sum ar^n$
pp-Series∑1/np\sum 1/n^pp>1p > 1: converges
Integralff positive, decreasing∑\sum and ∫\int same behavior
Comparison0≤an≤bn0 \le a_n \le b_n∑bn\sum b_n conv ⇒∑an\Rightarrow \sum a_n conv
Limit Comparisonan,bn>0a_n, b_n > 0lim⁡an/bn=L>0\lim a_n/b_n = L > 0: same behavior
AST∑(−1)nbn\sum (-1)^n b_nbn↓0b_n \downarrow 0: converges
RatioAny, best for n!n!L<1L < 1: abs conv; L>1L > 1: div
RootAny, best for bnnb_n^nSame thresholds as Ratio

The First Question: Does an→0a_n \to 0?

lim⁡n→∞an≠0⟹∑an diverges\boxed{\lim_{n \to \infty} a_n \neq 0 \quad \Longrightarrow \quad \sum a_n \text{ diverges}}

Key Fact: Always check the Divergence Test first. It's free, fast, and catches many series immediately.

Decision Flowchart

Step 1: Is lim⁡an≠0\lim a_n \neq 0? → Diverges (Divergence Test)

Step 2: Is it a known type?

  • Geometric: ∑arn\sum ar^n → check ∣r∣|r|
  • pp-Series: ∑1/np\sum 1/n^p → check pp
  • Telescoping: ∑(bn−bn+1)\sum (b_n - b_{n+1}) → evaluate

Step 3: Does it alternate?

  • Yes → AST (check bnb_n decreasing, →0\to 0)

Step 4: Contains factorials or rnr^n?

  • Yes → Ratio Test

Step 5: Form (bn)n(b_n)^n?

  • Yes → Root Test

Step 6: Looks like a known series?

  • Yes → Limit Comparison or Direct Comparison

Step 7: Positive, decreasing, integratable?

  • Yes → Integral Test

AP Tip: This flowchart covers 95%+ of AP exam series. Memorize it.

Test Identification

Quick Test Selection

Divergence Test

Summary

  • 9 convergence tests, each with specific strengths
  • Always start with Divergence Test, then look for known types
  • The flowchart guides you to the right test quickly
  • Divergence Test proves divergence only, never convergence

Next: Part 2 — Comparison Tests Deep Dive.

Part 2: Worked Examples

Comparison Tests Deep Dive

Part 2 of 7 — Direct & Limit Comparison

Direct Comparison Test (DCT)

For 0≤an≤bn0 \le a_n \le b_n for all nn (eventually):

∑bn converges⇒∑an converges\sum b_n \text{ converges} \Rightarrow \sum a_n \text{ converges} ∑an diverges⇒∑bn diverges\sum a_n \text{ diverges} \Rightarrow \sum b_n \text{ diverges}

Limit Comparison Test (LCT)

For an,bn>0a_n, b_n > 0: if lim⁡n→∞an/bn=L\lim_{n \to \infty} a_n/b_n = L where 0<L<∞0 < L < \infty, then ∑an\sum a_n and ∑bn\sum b_n have the same behavior.

When to Use Which

SituationUse
Easy to compare term-by-termDCT
Hard to prove an≤bna_n \le b_n directlyLCT
Series "looks like" a pp-seriesLCT with 1/np1/n^p

Key Fact: LCT is usually easier on the AP exam because you don't need to prove an inequality — just compute a limit.

LCT Example: ∑n+1n3−5\sum \frac{n+1}{n^3 - 5}

Looks like 1/n21/n^2 for large nn. Compare with bn=1/n2b_n = 1/n^2:

lim⁡(n+1)/(n3−5)1/n2=lim⁡n2(n+1)n3−5=lim⁡n3+n2n3−5=1\lim \frac{(n+1)/(n^3-5)}{1/n^2} = \lim \frac{n^2(n+1)}{n^3 - 5} = \lim \frac{n^3 + n^2}{n^3 - 5} = 1

Since 0<1<∞0 < 1 < \infty and ∑1/n2\sum 1/n^2 converges (p=2p = 2), ∑(n+1)/(n3−5)\sum (n+1)/(n^3-5) converges.

DCT Example: ∑sin⁡2nn2\sum \frac{\sin^2 n}{n^2}

0≤sin⁡2n≤10 \le \sin^2 n \le 1, so 0≤sin⁡2n/n2≤1/n20 \le \sin^2 n/n^2 \le 1/n^2.

∑1/n2\sum 1/n^2 converges → ∑sin⁡2n/n2\sum \sin^2 n/n^2 converges by DCT.

DCT Example: ∑1n−ln⁡n\sum \frac{1}{n - \ln n}

For large nn: n−ln⁡n<nn - \ln n < n, so 1/(n−ln⁡n)>1/n1/(n - \ln n) > 1/n.

Since ∑1/n\sum 1/n diverges and our series is term-by-term larger, ∑1/(n−ln⁡n)\sum 1/(n - \ln n) diverges by DCT.

AP Tip: For LCT, pick the comparison series by looking at the dominant terms in numerator and denominator.

Comparison Practice

Comparison Selection

LCT Limit

Summary

  • DCT: prove an≤bna_n \le b_n directly; used when comparison is obvious
  • LCT: compute lim⁡an/bn\lim a_n/b_n; easier, more flexible
  • Pick comparison by looking at dominant terms
  • L=0L = 0 or L=∞L = \infty: partial results (one direction only)
  • 0<L<∞0 < L < \infty: both series have same behavior

Next: Part 3 — Integral Test and Unusual Series.

Part 3: Problem-Solving Patterns

Integral Test & Special Series

Part 3 of 7 — When Other Tests Fail

The Integral Test

If f(x)f(x) is positive, continuous, and decreasing for x≥Nx \ge N, and an=f(n)a_n = f(n), then:

∑n=N∞an and ∫N∞f(x) dx converge or diverge together\boxed{\sum_{n=N}^{\infty} a_n \text{ and } \int_N^{\infty} f(x)\,dx \text{ converge or diverge together}}

When to Use the Integral Test

  • an=1/(nln⁡n)a_n = 1/(n \ln n) → ∫dx/(xln⁡x)\int dx/(x \ln x), easy substitution
  • an=1/(n(ln⁡n)2)a_n = 1/(n(\ln n)^2) → converges
  • an=ne−n2a_n = ne^{-n^2} → ∫xe−x2 dx\int xe^{-x^2}\,dx, substitution
  • Any series where you can anti-differentiate f(x)f(x) easily

Important: The Test Does NOT Give the Sum

The integral gives the same convergence/divergence behavior, but:

∑an≠∫f(x) dx\sum a_n \neq \int f(x)\,dx

The integral provides bounds, not the exact sum.

Key Fact: The Integral Test is the "test of last resort" for positive series that don't match other patterns. It's also how we PROVE the pp-Series Test.

Example 1: ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n}

f(x)=1/(xln⁡x)f(x) = 1/(x \ln x): positive, decreasing for x≥2x \ge 2.

∫2∞dxxln⁡x\int_2^{\infty} \frac{dx}{x \ln x}: let u=ln⁡xu = \ln x, du=dx/xdu = dx/x:

∫duu=ln⁡u∣ln⁡2∞=∞\int \frac{du}{u} = \ln u \Big|_{\ln 2}^{\infty} = \infty

Diverges. So ∑1/(nln⁡n)\sum 1/(n \ln n) diverges.

Example 2: ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}

∫2∞dxx(ln⁡x)2:∫duu2=−1u∣ln⁡2∞=1ln⁡2\int_2^{\infty} \frac{dx}{x(\ln x)^2}: \quad \int \frac{du}{u^2} = -\frac{1}{u}\Big|_{\ln 2}^{\infty} = \frac{1}{\ln 2}

Converges. So ∑1/(n(ln⁡n)2)\sum 1/(n(\ln n)^2) converges.

General Pattern

∑1n(ln⁡n)p:{convergesp>1divergesp≤1\sum \frac{1}{n(\ln n)^p}: \quad \begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \le 1 \end{cases}

This is like a "log-pp-series."

AP Tip: Integral Test problems on the AP exam usually involve ln⁡n\ln n in the denominator where other tests fail.

Integral Test Practice

Integral Test Decisions

Integral Test Evaluation

Summary

  • Integral Test: same convergence behavior as the improper integral
  • Use when other tests fail, especially for series with ln⁡n\ln n
  • Log-pp-series: ∑1/(n(ln⁡n)p)\sum 1/(n(\ln n)^p) converges iff p>1p > 1
  • The integral gives bounds, not the sum

Next: Part 4 — Absolute vs. Conditional Convergence.

Part 4: Graphs and Interpretation

Absolute vs. Conditional Convergence

Part 4 of 7 — Three Categories

Convergence Classification

Every series falls into exactly one category:

CategoryDefinitionExample
Absolutely convergent$\suma_n
Conditionally convergent∑an\sum a_n converges but $\suma_n
Divergent∑an\sum a_n diverges∑1/n\sum 1/n

The Hierarchy

Absolute convergence⇒Convergence\boxed{\text{Absolute convergence} \Rightarrow \text{Convergence}}

The converse is FALSE: convergence does NOT imply absolute convergence.

Testing Procedure

Step 1: Check ∑∣an∣\sum |a_n|.

  • If it converges → absolutely convergent (done!)

Step 2: If ∑∣an∣\sum |a_n| diverges, check ∑an\sum a_n.

  • If ∑an\sum a_n converges (usually by AST) → conditionally convergent
  • If ∑an\sum a_n diverges → divergent

Key Fact: "Absolute convergence" means you can rearrange the terms in any order and still get the same sum. Conditionally convergent series can be rearranged to sum to ANY value (Riemann's rearrangement theorem).

Example 1: ∑n=1∞(−1)nn3\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}

∑∣an∣=∑1/n3\sum |a_n| = \sum 1/n^3. pp-Series, p=3>1p = 3 > 1 → converges.

Absolutely convergent.

Example 2: ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (alternating harmonic)

∑∣an∣=∑1/n\sum |a_n| = \sum 1/n → diverges (harmonic).

∑an=∑(−1)n+1/n\sum a_n = \sum (-1)^{n+1}/n → converges by AST.

Conditionally convergent.

Example 3: ∑n=1∞(−1)nnn+1\sum_{n=1}^{\infty} \frac{(-1)^n n}{n+1}

lim⁡∣an∣=lim⁡n/(n+1)=1≠0\lim |a_n| = \lim n/(n+1) = 1 \neq 0.

Divergent (by Divergence Test — doesn't even converge).

Quick Classification Guide

| Series | ∑∣an∣\sum|a_n| | ∑an\sum a_n | Classification | |--------|-----------|----------|---------------| | ∑(−1)n/n2\sum (-1)^n/n^2 | Conv (p=2p = 2) | Conv | Absolute | | ∑(−1)n/n\sum (-1)^n/n | Div (harmonic) | Conv (AST) | Conditional | | ∑(−1)n/n3\sum (-1)^n/\sqrt[3]{n} | Div (p=1/3p = 1/3) | Conv (AST) | Conditional | | ∑(−1)n\sum (-1)^n | Div | Div | Divergent |

Classification Practice

Classify These Series

Classification

Summary

  • Three categories: absolute, conditional, divergent
  • Check ∑∣an∣\sum |a_n| first; if it converges, done (absolute)
  • If ∑∣an∣\sum |a_n| diverges but ∑an\sum a_n converges → conditional
  • Absolute ⇒ convergent, but not vice versa

Next: Part 5 — The Hardest AP Problems.

Part 5: Applications

AP Exam Strategies — Convergence

Part 5 of 7 — Test Selection Under Pressure

AP Exam Convergence Problem Types

TypeWhat to expectStrategy
"Determine convergence"Single series, pick a testUse the flowchart
"Which test and why?"Justify your choiceName the test, verify hypotheses
"Determine interval of convergence"Power seriesRatio Test for RR, test endpoints separately
"Absolute or conditional?"Alternating seriesCheck $\sum
"FRQ series justification"Part of larger problemBe precise: state theorem, verify conditions

The 30-Second Flowchart for MC

Divergence Test→Geometric/p?→Alternating?→Ratio/Root?→Comparison\boxed{\text{Divergence Test} \to \text{Geometric/p?} \to \text{Alternating?} \to \text{Ratio/Root?} \to \text{Comparison}}

  1. Quick check: lim⁡an≠0\lim a_n \neq 0? → Diverges
  2. Recognizable? Geometric or pp-series → formula
  3. Alternating? AST (verify bn↓0b_n \downarrow 0)
  4. Factorials or nnth powers? Ratio or Root Test
  5. Compare: DCT or LCT with known series

AP Tip: On FRQs, always state the test name, verify ALL conditions, and write a concluding statement.

Writing Perfect Justifications

Bad answer (no credit): "It converges by comparison."

Good answer (full credit): "Since 0≤1n2+1≤1n20 \leq \frac{1}{n^2 + 1} \leq \frac{1}{n^2} for all n≥1n \geq 1, and ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges (pp-series, p=2>1p = 2 > 1), by the Direct Comparison Test, ∑n=1∞1n2+1\sum_{n=1}^{\infty} \frac{1}{n^2+1} converges."

FRQ Checklist

StepExample
State the test"By the Ratio Test..."
Verify conditions"Since an>0a_n > 0 and lim⁡n→∞an+1/an=L\lim_{n\to\infty} a_{n+1}/a_n = L..."
Compute the limit"L=lim⁡(n+1)!/2n+1n!/2n=lim⁡n+12=∞L = \lim \frac{(n+1)!/2^{n+1}}{n!/2^n} = \lim \frac{n+1}{2} = \infty"
Conclude"Since L>1L > 1, the series diverges by the Ratio Test."

Common AP Pitfalls

MistakeWhy it loses points
Not checking lim⁡an=0\lim a_n = 0 firstDivergence Test is always first
Saying "converges by Divergence Test"Div Test can only prove divergence
Forgetting endpoint checks for IOCRR alone is not the full answer
LCT: not choosing the right comparisonCompare 1/(n2+n)1/(n^2+n) to 1/n21/n^2, not 1/n1/n

AP Strategy Questions

Best Test Selection

Quick Decision

Summary

  • Use the flowchart: Div Test → recognize → alternating → ratio/root → comparison
  • FRQs: name the test, verify conditions, write a conclusion
  • Common errors: "converges by Divergence Test," forgetting endpoint tests, weak comparisons

Next: Part 6 — Problem-Solving Workshop.

Part 6: Exam Strategy

Problem-Solving Workshop

Part 6 of 7 — Mixed Practice

Work through these problems choosing the best test for each.

Workshop Set A — Identify & Classify

Workshop Set B — Test Selection

Classify Each Series

Compute the Limit

Workshop Complete

Key takeaways:

  • Always start with Divergence Test
  • Match the series structure to the best test
  • Classify: check ∑∣an∣\sum |a_n| first, then ∑an\sum a_n

Next: Part 7 — Comprehensive Review.

Part 7: Mixed Review

Comprehensive Review — Convergence Tests

Part 7 of 7 — Final Assessment

Master Reference

TestUse when...Conclusion
DivergenceAlways firstlim⁡an≠0⇒\lim a_n \neq 0 \Rightarrow diverges
Geometric∑arn\sum ar^n$
pp-Series∑1/np\sum 1/n^pp>1p > 1: conv; p≤1p \leq 1: div
AST∑(−1)nbn\sum (-1)^n b_nbn↓0b_n \downarrow 0: conv
RatioFactorials, nnth powers of constantsL<1L < 1: conv; L>1L > 1: div
Rootan=[f(n)]na_n = [f(n)]^nL<1L < 1: conv; L>1L > 1: div
DCTCan bound 0≤an≤bn0 \leq a_n \leq b_n∑bn\sum b_n conv ⇒ ∑an\sum a_n conv
LCTRational-type termslim⁡an/bn=L>0\lim a_n/b_n = L > 0: same behavior
IntegralPositive, decreasing, continuous∫1∞f\int_1^\infty f and ∑an\sum a_n agree
TelescopingPartial fractions collapseCompute lim⁡Sn\lim S_n

Flowchart: Div Test→Recognizable→Alternating→Ratio/Root→Comparison\boxed{\text{Flowchart: Div Test} \to \text{Recognizable} \to \text{Alternating} \to \text{Ratio/Root} \to \text{Comparison}}

Review Set A — Convergence/Divergence

Review Set B — Classification

Best Test Selection

Final Challenge

Convergence Tests Summary — Complete

You've mastered:

  • All 9 convergence tests and when to use each
  • Direct and Limit Comparison Tests
  • Integral Test and special series (log-pp)
  • Absolute vs. conditional convergence
  • AP exam strategy and justification writing

Master the flowchart. Verify all conditions. Write clear conclusions.\boxed{\text{Master the flowchart. Verify all conditions. Write clear conclusions.}}