Skip to content
🎯⭐ INTERACTIVE LESSON

Infinite Series

Learn step-by-step with interactive practice!

Infinite Series - Complete Interactive Lesson

Part 1: Partial Sums & Geometric Series

Infinite Series — Convergence Tests

Part 1 of 7 — The Integral Test

Review: Series Convergence

∑n=1∞an\sum_{n=1}^\infty a_n converges if and only if lim⁡N→∞SN\lim_{N \to \infty} S_N exists (finite).

We already know:

  • Geometric series: ∣r∣<1|r| < 1
  • pp-series: p>1p > 1
  • nnth term test: an↛0  ⟹  a_n \not\to 0 \implies divergence

Now we develop systematic convergence tests.

The Integral Test

If f(x)f(x) is continuous, positive, and decreasing for x≥1x \ge 1, and an=f(n)a_n = f(n), then:

∑n=1∞an and ∫1∞f(x) dx either both converge or both diverge.\boxed{\sum_{n=1}^\infty a_n \text{ and } \int_1^\infty f(x)\,dx \text{ either both converge or both diverge.}}

Key Fact: The integral test does NOT give the sum — only whether the series converges.

Example 1 — Proving pp-series

Show ∑1/n2\sum 1/n^2 converges using the integral test.

f(x)=1/x2f(x) = 1/x^2: continuous, positive, decreasing for x≥1x \ge 1. ✓

∫1∞1x2 dx=lim⁡b→∞[−1x]1b=0−(−1)=1\int_1^\infty \frac{1}{x^2}\,dx = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_1^b = 0 - (-1) = 1

Integral converges   ⟹  \implies series converges. ✓

Example 2 — Harmonic series diverges

f(x)=1/xf(x) = 1/x: continuous, positive, decreasing.

∫1∞1x dx=lim⁡b→∞ln⁡b=∞\int_1^\infty \frac{1}{x}\,dx = \lim_{b \to \infty} \ln b = \infty

Integral diverges   ⟹  \implies ∑1/n\sum 1/n diverges. ✓

Remainder Estimate

If ∑an\sum a_n converges and SS denotes its sum, then:

∫N+1∞f(x) dx≤S−SN≤∫N∞f(x) dx\int_{N+1}^\infty f(x)\,dx \le S - S_N \le \int_N^\infty f(x)\,dx

Practice Problems

Concept Checks

Computation

Summary

  • Integral test: compare ∑an\sum a_n with ∫f(x) dx\int f(x)\,dx (same convergence behavior)
  • Requires: continuous, positive, decreasing ff
  • Does NOT give the sum, only convergence/divergence
  • Useful for proving pp-series results and testing unfamiliar series

Next: Part 2 — Comparison and Limit Comparison tests.

Part 2: Telescoping Series & Divergence Test

Infinite Series — Comparison Tests

Part 2 of 7 — Direct & Limit Comparison Tests

Direct Comparison Test (DCT)

For 0≤an≤bn0 \le a_n \le b_n:

If...Then...
∑bn\sum b_n converges∑an\sum a_n converges
∑an\sum a_n diverges∑bn\sum b_n diverges

Intuition: Smaller than convergent   ⟹  \implies convergent. Bigger than divergent   ⟹  \implies divergent.

Limit Comparison Test (LCT)

If an,bn>0a_n, b_n > 0 and lim⁡n→∞anbn=L\lim_{n \to \infty} \frac{a_n}{b_n} = L where 0<L<∞0 < L < \infty, then:

∑an and ∑bn either both converge or both diverge.\boxed{\sum a_n \text{ and } \sum b_n \text{ either both converge or both diverge.}}

AP Tip: The LCT is the most versatile comparison test. Choose bnb_n to be a simpler series (pp-series or geometric) that behaves like ana_n.

Examples

Example 1 (DCT): ∑1n2+n\sum \frac{1}{n^2 + n}

1n2+n<1n2\frac{1}{n^2 + n} < \frac{1}{n^2} and ∑1/n2\sum 1/n^2 converges (p=2p = 2).

By DCT, ∑1n2+n\sum \frac{1}{n^2 + n} converges. ✓

Example 2 (LCT): ∑3n2+1n4−2n+7\sum \frac{3n^2 + 1}{n^4 - 2n + 7}

Compare with bn=1/n2b_n = 1/n^2 (dominant terms give 3n2/n4=3/n23n^2/n^4 = 3/n^2).

anbn=(3n2+1)n2n4−2n+7→3n4n4=3\frac{a_n}{b_n} = \frac{(3n^2+1)n^2}{n^4-2n+7} \to \frac{3n^4}{n^4} = 3

Since L=3∈(0,∞)L = 3 \in (0, \infty) and ∑1/n2\sum 1/n^2 converges, the given series converges by LCT. ✓

Example 3 (LCT): ∑1n−1\sum \frac{1}{\sqrt{n} - 1}

Compare with bn=1/nb_n = 1/\sqrt{n}: anbn=nn−1→1\frac{a_n}{b_n} = \frac{\sqrt{n}}{\sqrt{n}-1} \to 1.

∑1/n\sum 1/\sqrt{n} diverges (p=1/2p = 1/2), so ∑1n−1\sum \frac{1}{\sqrt{n}-1} diverges. ✓

Practice

Test Selection

LCT Practice

Summary

  • DCT: Bound ana_n above by convergent or below by divergent
  • LCT: Compare an/bn→L∈(0,∞)a_n/b_n \to L \in (0, \infty) — same behavior
  • Choose bnb_n by identifying dominant terms
  • Both tests require positive terms

Next: Part 3 — The Ratio and Root Tests.

Part 3: Integral Test & p-Series

Infinite Series — Ratio & Root Tests

Part 3 of 7 — The Ratio and Root Tests

The Ratio Test

Let L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|. Then:

LL valueConclusion
L<1L < 1Converges absolutely
L>1L > 1 (or L=∞L = \infty)Diverges
L=1L = 1Inconclusive

The Root Test

Let L=lim⁡n→∞∣an∣nL = \lim_{n \to \infty} \sqrt[n]{|a_n|}. Same conclusions as ratio test.

Key Fact: The ratio test works best with factorials and exponentials. The root test works best with nnth powers. Both are inconclusive for pp-series.

Examples

Ratio Test: ∑n!3n\sum \frac{n!}{3^n}

an+1an=(n+1)!3n+1⋅3nn!=n+13→∞\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{3^{n+1}} \cdot \frac{3^n}{n!} = \frac{n+1}{3} \to \infty

L=∞>1L = \infty > 1: diverges.

Ratio Test: ∑2nn!\sum \frac{2^n}{n!}

an+1an=2n+1(n+1)!⋅n!2n=2n+1→0\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2}{n+1} \to 0

L=0<1L = 0 < 1: converges absolutely.

Root Test: ∑(n2n+1)n\sum \left(\frac{n}{2n+1}\right)^n

ann=n2n+1→12\sqrt[n]{a_n} = \frac{n}{2n+1} \to \frac{1}{2}

L=1/2<1L = 1/2 < 1: converges absolutely.

Practice Problems

Test Selection

Ratio Test Computation

Summary

  • Ratio test: L=lim⁡∣an+1/an∣L = \lim |a_{n+1}/a_n| — best for factorials and exponentials
  • Root test: L=lim⁡∣an∣nL = \lim \sqrt[n]{|a_n|} — best for nnth powers
  • L<1L < 1: converges; L>1L > 1: diverges; L=1L = 1: inconclusive
  • Both tests are inconclusive for pp-series (use comparison or integral test instead)

L<1  ⟹  convergesL>1  ⟹  divergesL=1  ⟹  inconclusive\boxed{L < 1 \implies \text{converges} \qquad L > 1 \implies \text{diverges} \qquad L = 1 \implies \text{inconclusive}}

Next: Part 4 — Absolute and conditional convergence.

Part 4: Comparison Tests

Infinite Series — Absolute & Conditional Convergence

Part 4 of 7 — Types of Convergence

Definitions

TypeDefinition
Absolutely convergent$\sum
Conditionally convergent∑an\sum a_n converges but $\sum
Divergent∑an\sum a_n does not converge

Key Theorem

Absolute convergence  ⟹  Convergence\boxed{\text{Absolute convergence} \implies \text{Convergence}}

But NOT vice versa!

The Classic Example

∑n=1∞(−1)n+1n=1−12+13−14+⋯=ln⁡2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2

This converges, but ∑1/n\sum 1/n diverges, so it converges conditionally.

AP Tip: When a problem asks "does the series converge absolutely, conditionally, or diverge?" — test ∑∣an∣\sum |a_n| first. If it converges, you're done (absolute). If not, check ∑an\sum a_n separately.

Strategy for Classification

Step 1: Test ∑∣an∣\sum |a_n| for convergence.

  • If ∑∣an∣\sum |a_n| converges → absolutely convergent ✓

Step 2: If ∑∣an∣\sum |a_n| diverges, test ∑an\sum a_n (typically with AST).

  • If ∑an\sum a_n converges → conditionally convergent
  • If ∑an\sum a_n diverges → divergent

Why Conditional Convergence Matters

Conditionally convergent series have surprising properties:

  • Riemann Rearrangement Theorem: By rearranging terms, you can make the series sum to ANY value (or diverge). This is why absolute convergence is "safer."
  • On the AP exam, conditionally convergent series typically appear in the context of interval of convergence endpoints.

Practice Problems

Classification Practice

Quick Classification

Summary

  • Absolute convergence: ∑∣an∣\sum |a_n| converges
  • Conditional convergence: ∑an\sum a_n converges but ∑∣an∣\sum |a_n| diverges
  • Absolute convergence   ⟹  \implies convergence (not vice versa)
  • Test ∑∣an∣\sum |a_n| first, then ∑an\sum a_n if needed

Next: Part 5 — Choosing the right test (decision flowchart).

Part 5: Ratio & Root Tests

Infinite Series — Choosing the Right Test

Part 5 of 7 — Convergence Test Strategy

Decision Flowchart for ∑an\sum a_n

StepAsk YourselfAction
1Does an→0a_n \to 0?If NO → diverges (Divergence Test)
2Geometric or telescoping?Identify and use closed form
3Is it a pp-series ∑1/np\sum 1/n^p?Converges iff p>1p > 1
4Alternating sign?Try AST
5Contains n!n!, nnn^n, or ana^n?Try Ratio Test
6Contains nn-th powers (f(n))n(f(n))^n?Try Root Test
7Similar to pp-series or geometric?Try Comparison (DCT/LCT)
8Positive, decreasing, integratable?Try Integral Test

No single test works for all series — practice builds intuition.\boxed{\text{No single test works for all series — practice builds intuition.}}

AP Tip: The exam frequently asks "which test is appropriate?" or "justify your answer using [a specific test]." Know the hypotheses of each test cold.

Test Selection Examples

Example 1: ∑n=1∞3nn!\sum_{n=1}^\infty \frac{3^n}{n!}

  • Contains n!n! → Ratio Test: an+1an=3n+1→0<1\frac{a_{n+1}}{a_n} = \frac{3}{n+1} \to 0 < 1 → converges ✓

Example 2: ∑n=1∞n23n2+1\sum_{n=1}^\infty \frac{n^2}{3n^2 + 1}

  • an→1/3≠0a_n \to 1/3 \neq 0 → Divergence Test → diverges ✓

Example 3: ∑n=2∞1nln⁡n\sum_{n=2}^\infty \frac{1}{n \ln n}

  • Positive, decreasing, integratable → Integral Test: ∫2∞dxxln⁡x=ln⁡(ln⁡x)∣2∞=∞\int_2^\infty \frac{dx}{x\ln x} = \ln(\ln x)\big|_2^\infty = \infty → diverges ✓

Example 4: ∑n=1∞(−1)nnn+1\sum_{n=1}^\infty \frac{(-1)^n n}{n+1}

  • ∣an∣=n/(n+1)→1≠0|a_n| = n/(n+1) \to 1 \neq 0 → Divergence Test → diverges ✓ (not alternating — fails hypothesis)

Common Pitfalls

MistakeCorrection
AST on $a_n
Ratio/Root gives L=1L = 1Test is inconclusive — try another
Comparison in wrong directionan≤bna_n \le b_n and ∑bn\sum b_n converges → ∑an\sum a_n converges. NOT the other way for convergence

Which Test? Practice

Test Strategy Application

Ratio Test Application

Key Takeaways

SituationGo-To Test
an↛0a_n \not\to 0Divergence Test
Factorials or exponentialsRatio Test
nn-th power structureRoot Test
Polynomial-like termsComparison / LCT
Positive, continuous, decreasingIntegral Test
Alternating signsAST

Key Fact: On the AP exam, you'll almost never need more than one test per series. The challenge is identifying which one.

Next: Part 6 — Problem-Solving Workshop.

Part 6: Practice Workshop

Infinite Series — Problem-Solving Workshop

Part 6 of 7 — Practice with All Tests

Work through these problems carefully. For each series, identify the appropriate convergence test, verify hypotheses, and state a conclusion.

Warm-Up: Test Identification

For each series, think about which test best applies before solving.

SeriesKey FeatureBest Test
∑n!/nn\sum n!/n^nBoth n!n! and nnn^nRatio Test
∑1/(nln⁡n)\sum 1/(n\sqrt{\ln n})Continuous, decreasing, positiveIntegral Test
∑(−1)n/(2n+1)\sum (-1)^n/(2n+1)Alternating, bn→0b_n \to 0AST
∑n/(n3+1)\sum n/(n^3+1)Behaves like 1/n21/n^2LCT with 1/n21/n^2

Workshop Problems

Classify Each Series

Computation Challenge

Workshop Takeaways

  • Always check an→0a_n \to 0 first (Divergence Test)
  • Factorials and exponentials → Ratio Test
  • Powers of nn → Comparison / pp-series
  • Alternating signs → AST (after verifying bnb_n is decreasing and →0\to 0)
  • For classification: test ∑∣an∣\sum |a_n| first, then ∑an\sum a_n

Next: Part 7 — Comprehensive Review.

Part 7: Final Assessment

Infinite Series — Comprehensive Review

Part 7 of 7 — Review All Convergence Tests

Quick Reference: All Tests

TestHypothesesConclusion
Divergencelim⁡an≠0\lim a_n \neq 0Diverges
Geometric∑arn\sum ar^nConverges iff $
pp-Series∑1/np\sum 1/n^pConverges iff p>1p > 1
Integralff positive, continuous, decreasing∑an\sum a_n and ∫f\int f converge/diverge together
DCT0≤an≤bn0 \le a_n \le b_n∑bn\sum b_n conv. ⇒∑an\Rightarrow \sum a_n conv.
LCTlim⁡an/bn=L>0\lim a_n/b_n = L > 0Both converge or both diverge
Ratio$L = \lima_{n+1}/a_n
Root$L = \lim \sqrt[n]{a_n
AST(−1)nbn(-1)^n b_n, bnb_n decreasing, bn→0b_n \to 0Converges

Absolute convergence  ⟹  Convergence  ⟹  an→0\boxed{\text{Absolute convergence} \implies \text{Convergence} \implies a_n \to 0}

AP Exam Note: You MUST state the test name and verify its hypotheses for full credit. A correct answer with no justification earns minimal credit.

Comprehensive MC Review

More Review Problems

Final Classification Drill

Final Computation

Infinite Series — Complete Summary

You've mastered:

  • Integral Test — connects series and improper integrals
  • Comparison Tests — DCT and LCT for bounding series
  • Ratio & Root Tests — best for factorials, exponentials, and nn-th powers
  • Absolute vs. Conditional Convergence — fundamental classification
  • Test Selection Strategy — choosing the right tool for each series

Key Fact: Series convergence is a major BC topic, typically appearing in both MC and FRQ sections. Expect 3-5 questions on the AP exam.

Up Next: Alternating Series — deep dive into the Alternating Series Test, error bounds, and applications.