Unit 10.5 – Harmonic Series and p-Series BC ONLY

AP® Calculus BC | The Most Important Series Types

Why This Matters: The harmonic series and p-series are THE benchmark series in calculus! The harmonic series is the most famous divergent series (despite terms going to zero), and p-series give us a simple, powerful test. These appear on virtually EVERY BC exam and are essential for comparison tests!

🎯 The Harmonic Series

The Harmonic Series

DEFINITION:

\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots \]

⚠️ FAMOUS FACT: The harmonic series DIVERGES even though \(\lim_{n \to \infty} \frac{1}{n} = 0\)!

This is the classic example showing that \(\lim a_n = 0\) does NOT guarantee convergence!

📜 Proof: Harmonic Series Diverges

Integral Test Proof

Using the Integral Test:

Consider \(f(x) = \frac{1}{x}\) which is continuous, positive, and decreasing for \(x \geq 1\)

\[ \int_1^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} \int_1^t \frac{1}{x} \, dx = \lim_{t \to \infty} [\ln x]_1^t \]

\[ = \lim_{t \to \infty} (\ln t - \ln 1) = \lim_{t \to \infty} \ln t = \infty \]

Since the integral diverges, the harmonic series DIVERGES.

Alternative: Grouping Proof (Cauchy's Method)

Clever grouping shows sum → ∞:

\[ 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots \]

Each group sums to more than \(\frac{1}{2}\):

\(\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)
\(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > 4 \cdot \frac{1}{8} = \frac{1}{2}\)

Sum of infinitely many groups each ≥ \(\frac{1}{2}\) → ∞, so series DIVERGES!

⭐ The p-Series

p-Series Definition and Test

DEFINITION:

A p-series is a series of the form:

\[ \sum_{n=1}^{\infty} \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots \]

where \(p\) is a positive constant

THE p-SERIES TEST (MEMORIZE!)

\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{CONVERGES} & \text{if } p > 1 \\ \text{DIVERGES} & \text{if } p \leq 1 \end{cases} \]

📝 Critical Boundary: \(p = 1\) is the dividing line! Greater than 1 → converges, less than or equal to 1 → diverges.

🔍 Important Special Cases

Case 1: p = 1 (Harmonic Series)

\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots \quad \text{DIVERGES} \]

Case 2: p = 2

\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \quad \text{CONVERGES} \]

(Actually converges to \(\frac{\pi^2}{6}\)!)

Case 3: p = 1/2

\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots \quad \text{DIVERGES} \]

Case 4: p = 3/2

\[ \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} = 1 + \frac{1}{2\sqrt{2}} + \frac{1}{3\sqrt{3}} + \cdots \quad \text{CONVERGES} \]

📊 Complete p-Series Reference

p-Series Convergence Table
p value	Series	Name	Converges?
\(p = 1/4\)	\(\sum \frac{1}{n^{1/4}}\)	—	NO
\(p = 1/2\)	\(\sum \frac{1}{\sqrt{n}}\)	—	NO
\(p = 0.99\)	\(\sum \frac{1}{n^{0.99}}\)	—	NO
\(p = 1\)	\(\sum \frac{1}{n}\)	Harmonic	NO
\(p = 1.01\)	\(\sum \frac{1}{n^{1.01}}\)	—	YES
\(p = 3/2\)	\(\sum \frac{1}{n^{3/2}}\)	—	YES
\(p = 2\)	\(\sum \frac{1}{n^2}\)	Basel problem	YES
\(p = 3\)	\(\sum \frac{1}{n^3}\)	—	YES

📖 Comprehensive Worked Examples

Example 1: Identifying p-Series

Problem: Does \(\sum_{n=1}^{\infty} \frac{1}{n^{5/2}}\) converge?

Solution:

This is a p-series with \(p = \frac{5}{2} = 2.5\)

Since \(p = 2.5 > 1\), the series CONVERGES.

Example 2: Rewriting to Identify p

Problem: Does \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n^2}}\) converge?

Rewrite:

\[ \frac{1}{\sqrt[3]{n^2}} = \frac{1}{n^{2/3}} \]

This is p-series with \(p = \frac{2}{3}\)

Since \(p = \frac{2}{3} < 1\), the series DIVERGES.

Example 3: Not Quite p-Series

Problem: What about \(\sum_{n=1}^{\infty} \frac{n}{n^3}\)?

Simplify first:

\[ \frac{n}{n^3} = \frac{1}{n^2} \]

NOW it's p-series with \(p = 2\)

Since \(p = 2 > 1\), the series CONVERGES.

Example 4: Similar But NOT p-Series

Problem: Is \(\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}\) a p-series?

Analysis:

This is NOT a p-series because of the "+1" in denominator.

However, for large n: \(\frac{n}{n^2+1} \approx \frac{n}{n^2} = \frac{1}{n}\)

This behaves like harmonic series, so likely diverges.

(Would need integral test or comparison test to prove)

Example 5: Boundary Case

Problem: Does \(\sum_{n=1}^{\infty} \frac{1}{n^{1.0001}}\) converge?

Solution:

\(p = 1.0001 > 1\) (barely!)

Series CONVERGES, but very slowly.

Shows how sharp the boundary at \(p = 1\) is!

🔄 Common Variations

Different Starting Index:

\[ \sum_{n=5}^{\infty} \frac{1}{n^p} \]

Starting index doesn't affect convergence! Still converges if \(p > 1\)

With Constant Multiple:

\[ \sum_{n=1}^{\infty} \frac{7}{n^p} = 7\sum_{n=1}^{\infty} \frac{1}{n^p} \]

Constant doesn't affect convergence! Still use p-series test

Alternating p-Series:

\[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^p} \]

Different test needed (alternating series test)

Can converge even when \(p \leq 1\)!

💡 Essential Tips & Strategies

✅ Success Strategies:

MEMORIZE: \(p > 1\) converges, \(p \leq 1\) diverges
Harmonic series diverges: Most important divergent series
Look for \(\frac{1}{n^p}\) form: Rewrite if needed
Simplify before identifying: Cancel common factors
Compare exponents: Is power of n in denominator > 1?
Starting index irrelevant: For convergence behavior
Constant multiples don't matter: For convergence
Be careful with negatives: Alternating series are different

🔥 Quick Recognition:

\(\sum \frac{1}{n}\): Diverges (harmonic)
\(\sum \frac{1}{n^2}\): Converges (p=2)
\(\sum \frac{1}{\sqrt{n}}\): Diverges (p=1/2)
\(\sum \frac{1}{n^{3/2}}\): Converges (p=3/2)
If denominator grows faster than n: Likely converges
If denominator grows like n or slower: Likely diverges

❌ Common Mistakes to Avoid

Mistake 1: Thinking harmonic series converges because terms → 0
Mistake 2: Confusing \(p > 1\) with \(p < 1\) for convergence
Mistake 3: Not simplifying before identifying p
Mistake 4: Forgetting \(p = 1\) is the boundary (diverges!)
Mistake 5: Applying p-series test to non-p-series
Mistake 6: Thinking \(\sum \frac{1}{n^2+1}\) is a p-series
Mistake 7: Not recognizing \(\frac{1}{\sqrt[k]{n^m}} = \frac{1}{n^{m/k}}\)
Mistake 8: Thinking constants affect convergence behavior
Mistake 9: Confusing with geometric series
Mistake 10: Not checking if series is actually in form \(\frac{1}{n^p}\)

📝 Practice Problems

Determine if each series converges or diverges:

\(\sum_{n=1}^{\infty} \frac{1}{n^{4/3}}\)
\(\sum_{n=1}^{\infty} \frac{1}{n^{0.9}}\)
\(\sum_{n=1}^{\infty} \frac{5}{n^3}\)
\(\sum_{n=1}^{\infty} \frac{1}{\sqrt[5]{n^3}}\)
\(\sum_{n=1}^{\infty} \frac{n^2}{n^4}\)
\(\sum_{n=10}^{\infty} \frac{1}{n}\)

Answers:

CONVERGES (p = 4/3 > 1)
DIVERGES (p = 0.9 < 1)
CONVERGES (constant doesn't matter, p = 3 > 1)
DIVERGES (p = 3/5 < 1)
CONVERGES (simplifies to \(\frac{1}{n^2}\), p = 2 > 1)
DIVERGES (still harmonic, starting index doesn't matter)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Identify as p-series: "This is a p-series with p = ..."
State p value clearly: Show what p is
Apply the test: "Since p > 1, series converges" or "Since p ≤ 1, series diverges"
Show simplification: If needed to identify p
For harmonic series: State "This is the harmonic series, which diverges"
Be explicit: Don't just say "converges"—say WHY
Check form: Make sure it's actually \(\frac{1}{n^p}\)

💯 Exam Strategy:

Check if series is in form \(\frac{c}{n^p}\) (constant over n to a power)
Simplify if necessary to identify p
State "p-series with p = [value]"
Apply rule: p > 1 → converges, p ≤ 1 → diverges
If p = 1, can also say "harmonic series, diverges"
Remember: this is one of the quickest tests!

⚡ Quick Reference Guide

p-SERIES ESSENTIALS

The Test:

\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{CONVERGES} & \text{if } p > 1 \\ \text{DIVERGES} & \text{if } p \leq 1 \end{cases} \]

Harmonic Series (p = 1):

\[ \sum_{n=1}^{\infty} \frac{1}{n} \quad \text{DIVERGES} \]

Key Examples:

\(\sum \frac{1}{n}\) → DIVERGES (p=1)
\(\sum \frac{1}{n^2}\) → CONVERGES (p=2)
\(\sum \frac{1}{\sqrt{n}}\) → DIVERGES (p=1/2)

Remember:

p > 1 is the magic condition!
Harmonic series (p=1) DIVERGES!
Must be in form \(\frac{1}{n^p}\)!

Master p-Series and Harmonic Series! The harmonic series \(\sum \frac{1}{n}\) is the most famous divergent series—diverges despite terms → 0, proving that \(\lim a_n = 0\) is necessary but NOT sufficient for convergence. p-Series \(\sum \frac{1}{n^p}\) converges if and only if \(p > 1\), diverges if \(p \leq 1\). Boundary case \(p=1\) is harmonic (diverges). Proven by integral test: \(\int_1^{\infty} \frac{1}{x^p}\,dx\) converges when \(p > 1\). Key examples: \(p=2\) converges to \(\pi^2/6\); \(p=1/2\) diverges; \(p=3/2\) converges. Must be in exact form \(\frac{1}{n^p}\)—if denominator has additions like \(n^2+1\), not a p-series. Constants don't affect convergence. Starting index irrelevant. These are benchmark series for comparison tests. Appears on EVERY BC exam—absolutely essential! 🎯✨