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Unit 10.6 – Comparison Tests for Convergence BC ONLY

AP® Calculus BC | Comparing Series to Known Series

Why This Matters: Comparison tests are your go-to tools when series don't fit nice formulas! The idea is brilliant: compare your unknown series to a known series (like p-series or geometric). If you can show your series is smaller than a convergent series OR larger than a divergent series, you're done! Two versions: Direct Comparison Test (DCT) and Limit Comparison Test (LCT).

🎯 Direct Comparison Test (DCT)

Direct Comparison Test

THE TEST:

Suppose \(0 \leq a_n \leq b_n\) for all \(n \geq N\) (some integer N). Then:

1. If \(\sum b_n\) CONVERGES:
\[ \text{Then } \sum a_n \text{ also CONVERGES} \]

Smaller than convergent → converges

2. If \(\sum a_n\) DIVERGES:
\[ \text{Then } \sum b_n \text{ also DIVERGES} \]

Larger than divergent → diverges

📝 Key Insight: Think "squeeze" logic! If you're smaller than something that converges, you must converge. If you're bigger than something that diverges, you must diverge.

⭐ Limit Comparison Test (LCT)

Limit Comparison Test

THE TEST:

Let \(a_n > 0\) and \(b_n > 0\) for all \(n\). If

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L \]

where \(L\) is finite and positive (\(0 < L < \infty\)), then:

\[ \sum a_n \text{ and } \sum b_n \text{ BOTH converge or BOTH diverge} \]

Special Cases:

  • If \(L = 0\) and \(\sum b_n\) converges: Then \(\sum a_n\) converges
  • If \(L = \infty\) and \(\sum b_n\) diverges: Then \(\sum a_n\) diverges

📝 When to Use: LCT is MUCH easier than DCT when the inequality is hard to verify! Just compute a limit instead.

🤔 DCT vs. LCT: Which to Use?

Comparison Test Selection
Test When to Use Advantages Disadvantages
DCT Inequality obvious Direct, simple logic Must prove inequality
LCT Inequality unclear Just need a limit! More calculation

💡 Pro Tip: On AP® exams, LCT is often easier because you don't have to justify inequalities—just compute a limit!

🎨 Choosing the Right Comparison Series

Strategy for Choosing \(b_n\)

General Strategy:

  1. Identify dominant terms in numerator and denominator
  2. Drop less significant terms (smaller powers, constants)
  3. Compare to p-series or geometric series
  4. Verify your comparison makes sense
Common Patterns:
  • \(\frac{n^2 + 3n + 1}{n^4 + 5}\): Compare to \(\frac{n^2}{n^4} = \frac{1}{n^2}\)
  • \(\frac{1}{n^2 + n}\): Compare to \(\frac{1}{n^2}\) (convergent)
  • \(\frac{n}{n^2 + 1}\): Compare to \(\frac{1}{n}\) (divergent)
  • \(\frac{2^n}{3^n + 1}\): Compare to \(\left(\frac{2}{3}\right)^n\) (geometric)

📖 Comprehensive Worked Examples

Example 1: Direct Comparison Test (Converges)

Problem: Does \(\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}\) converge?

Solution using DCT:

Step 1: Find comparison series

For \(n \geq 1\): \(n^2 + 1 > n^2\), so \(\frac{1}{n^2 + 1} < \frac{1}{n^2}\)

Compare to \(\sum \frac{1}{n^2}\) (p-series with p=2, converges)

Step 2: Apply DCT

\[ 0 < \frac{1}{n^2+1} < \frac{1}{n^2} \]

Since \(\sum \frac{1}{n^2}\) converges and \(\frac{1}{n^2+1} < \frac{1}{n^2}\),

by DCT, \(\sum \frac{1}{n^2+1}\) CONVERGES

Example 2: Direct Comparison Test (Diverges)

Problem: Does \(\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}\) converge?

Set up comparison:

For \(n \geq 1\): \(n^2 + 1 < n^2 + n^2 = 2n^2\)

So: \(\frac{n}{n^2+1} > \frac{n}{2n^2} = \frac{1}{2n}\)

Apply DCT:

\[ \frac{n}{n^2+1} > \frac{1}{2n} \]

Since \(\sum \frac{1}{2n} = \frac{1}{2}\sum \frac{1}{n}\) diverges (harmonic),

and our series is LARGER, it DIVERGES by DCT

Example 3: Limit Comparison Test

Problem: Does \(\sum_{n=1}^{\infty} \frac{3n^2 + 5n + 1}{7n^4 + 2n^2 + 8}\) converge?

Choose comparison series:

Dominant terms: \(\frac{3n^2}{7n^4} = \frac{3}{7n^2}\)

Compare to \(b_n = \frac{1}{n^2}\) (p-series, p=2, converges)

Compute limit:

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(3n^2+5n+1)/(7n^4+2n^2+8)}{1/n^2} \]
\[ = \lim_{n \to \infty} \frac{n^2(3n^2+5n+1)}{7n^4+2n^2+8} = \lim_{n \to \infty} \frac{3n^4+5n^3+n^2}{7n^4+2n^2+8} = \frac{3}{7} \]

Conclude:

Since \(0 < \frac{3}{7} < \infty\) and \(\sum \frac{1}{n^2}\) converges,

by LCT, our series CONVERGES

Example 4: LCT with Exponentials

Problem: Does \(\sum_{n=1}^{\infty} \frac{5^n}{3^n + 1}\) converge?

Choose comparison:

Dominant: \(\frac{5^n}{3^n} = \left(\frac{5}{3}\right)^n\)

Compare to geometric \(b_n = \left(\frac{5}{3}\right)^n\) (r > 1, diverges)

LCT calculation:

\[ \lim_{n \to \infty} \frac{5^n/(3^n+1)}{(5/3)^n} = \lim_{n \to \infty} \frac{5^n \cdot 3^n}{(3^n+1) \cdot 5^n} = \lim_{n \to \infty} \frac{3^n}{3^n+1} = 1 \]

Since limit = 1 and geometric series diverges, our series DIVERGES

Example 5: LCT with Limit = 0

Problem: \(\sum_{n=1}^{\infty} \frac{1}{n^3 + n}\) compared to \(\sum \frac{1}{n^2}\)

Compute limit:

\[ \lim_{n \to \infty} \frac{1/(n^3+n)}{1/n^2} = \lim_{n \to \infty} \frac{n^2}{n^3+n} = \lim_{n \to \infty} \frac{1}{n+1/n} = 0 \]

Limit = 0, and \(\sum \frac{1}{n^2}\) converges

By LCT (special case), our series CONVERGES

💡 Essential Tips & Strategies

✅ Success Strategies:

  • LCT is usually easier: No inequality to prove, just compute limit
  • Keep dominant terms: Ignore lower-order terms and constants
  • Compare to p-series first: Most common comparison
  • For ratios of polynomials: Compare degrees (top/bottom)
  • For exponentials: Compare bases (geometric series)
  • Check limit ≠ 0 or ∞: For standard LCT to work
  • Show work clearly: State which test and comparison series
  • Verify convergence of comparison: Before concluding!

🔥 Quick Guide: What to Compare To:

  • Rational functions: \(\frac{1}{n^p}\) (p-series)
  • Exponentials \(a^n/b^n\): \(\left(\frac{a}{b}\right)^n\) (geometric)
  • Factorials: Usually diverges (very fast growth)
  • Logarithms: Grows slower than any power
  • When in doubt: Try LCT with simplified version

❌ Common Mistakes to Avoid

  • Mistake 1: Wrong direction of inequality for DCT
  • Mistake 2: Not stating which test you're using
  • Mistake 3: Comparing to unknown/wrong series
  • Mistake 4: Not checking if comparison series converges/diverges
  • Mistake 5: Limit = 0 or ∞ without using special cases
  • Mistake 6: Forgetting to state conclusion clearly
  • Mistake 7: Using DCT when inequality is wrong direction
  • Mistake 8: Not simplifying before taking limit in LCT
  • Mistake 9: Comparing terms instead of series
  • Mistake 10: Not verifying 0 < L < ∞ for standard LCT

📝 Practice Problems

Determine convergence using comparison tests:

  1. \(\sum_{n=1}^{\infty} \frac{1}{2^n + n}\)
  2. \(\sum_{n=1}^{\infty} \frac{n+1}{n^3 + 2}\)
  3. \(\sum_{n=1}^{\infty} \frac{3n^2 + 1}{n^3 + 5n}\)
  4. \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + n}}\)
  5. \(\sum_{n=1}^{\infty} \frac{2^n + 1}{3^n}\)

Answers:

  1. CONVERGES (compare to \(\frac{1}{2^n}\), geometric)
  2. CONVERGES (LCT with \(\frac{1}{n^2}\))
  3. DIVERGES (LCT with \(\frac{1}{n}\))
  4. CONVERGES (compare to \(\frac{1}{n^{3/2}}\))
  5. CONVERGES (LCT with \(\left(\frac{2}{3}\right)^n\))

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Name the test: "By Direct Comparison Test..." or "By Limit Comparison Test..."
  • State comparison series: "Compare to \(\sum \frac{1}{n^2}\)"
  • Verify comparison converges/diverges: "which is p-series with p=2, converges"
  • For DCT: Show inequality: "\(a_n < b_n\) because..."
  • For LCT: Show limit calculation: Full work with proper notation
  • State the limit value: "\(\lim \frac{a_n}{b_n} = L\)"
  • Apply test correctly: Correct logic direction
  • Clear conclusion: "Therefore, series converges/diverges"

💯 Exam Strategy:

  1. Identify dominant terms (highest powers)
  2. Decide: p-series or geometric comparison?
  3. Choose LCT unless inequality is obvious
  4. State test name and comparison series
  5. Verify comparison series behavior
  6. Execute test (inequality or limit)
  7. Apply test logic correctly
  8. State conclusion with "because..."

⚡ Quick Reference Guide

COMPARISON TESTS ESSENTIALS

Direct Comparison Test (DCT):

If \(0 \leq a_n \leq b_n\):

  • \(\sum b_n\) converges → \(\sum a_n\) converges
  • \(\sum a_n\) diverges → \(\sum b_n\) diverges

Limit Comparison Test (LCT):

If \(\lim_{n\to\infty} \frac{a_n}{b_n} = L\) where \(0 < L < \infty\):

\(\sum a_n\) and \(\sum b_n\) both converge or both diverge

Choosing Comparison:

  • Rational functions → p-series
  • Exponentials → geometric
  • Keep dominant terms only

Remember:

  • LCT usually easier (just compute limit)!
  • State test name and comparison series!
  • Verify comparison series first!

Master Comparison Tests! Direct Comparison Test (DCT): if \(0 \leq a_n \leq b_n\), then \(\sum b_n\) converges → \(\sum a_n\) converges; or \(\sum a_n\) diverges → \(\sum b_n\) diverges. Must prove inequality holds. Limit Comparison Test (LCT): if \(\lim \frac{a_n}{b_n} = L\) with \(0 < L < \infty\), both series have same behavior. LCT often easier—just compute limit! Strategy: identify dominant terms, compare to p-series \(\frac{1}{n^p}\) or geometric \(r^n\). For rational functions: compare degrees. For exponentials: compare bases. Special LCT cases: \(L=0\) with convergent \(b_n\) → converges; \(L=\infty\) with divergent \(b_n\) → diverges. Always state test name, comparison series, verify comparison behavior, then conclude. These are THE most versatile convergence tests—work on nearly anything! Master choosing good comparison series—key skill for BC exams! 🎯✨