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Σ
\(\frac{1}{n^p}\)

Harmonic Series
and \( p \)-Series

Master the convergence behavior of \( \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p} \) and understand why the value of \( p \) determines everything.

AP Calculus BC

What Is an Infinite Series?

An infinite series is the sum of infinitely many terms from a sequence. We write it as:

The central question in series analysis: does this sum approach a finite number (converge) or grow without bound (diverge)?

The Harmonic Series

The most famous divergent series in mathematics:

Even though the terms \( \frac{1}{n} \to 0 \), they shrink too slowly. The partial sums grow without bound: after 10 terms \( \approx 2.93 \), after 1000 terms \( \approx 7.49 \), after \( 10^6 \) terms \( \approx 14.4 \).

Partial Sums: Converging vs. Diverging

0 1 2 3 4 n (number of terms) Partial Sum 1 6 11 16
\(\frac{\pi^2}{6}\)
\(\frac{1}{n}\) (diverges)
\(\frac{1}{n^2}\) (converges)

The \( p \)-Series Definition

\( p = 1 \)
The harmonic series: \( \sum \frac{1}{n} \). Diverges.
\( p = 2 \)
Basel problem: \( \sum \frac{1}{n^2} = \frac{\pi^2}{6} \). Converges.
\( p = 3 \)
Apéry's constant: \( \sum \frac{1}{n^3} \approx 1.202 \). Converges.

The \( p \)-series is the generalization of the harmonic series. The exponent \( p \) controls how fast terms shrink, which determines convergence.

The \( p \)-Series Test

This is one of the most important convergence tests in calculus. Memorize it: convergence requires \( p \) strictly greater than 1.

✓ Converges
\( p = 1.01, \; p = 1.5, \; p = 2, \; p = 3, \; p = 100 \)
✕ Diverges
\( p = 1, \; p = 0.99, \; p = 0.5, \; p = 0, \; p = -1 \)

Why? The Integral Test Connection

The \( p \)-series test is proven via the integral test. The series \( \sum \frac{1}{n^p} \) converges if and only if the integral converges:

1
When \( p > 1 \): The antiderivative is \( \frac{x^{1-p}}{1-p} \), which approaches 0 as \( x \to \infty \). Integral is finite.
2
When \( p = 1 \): The integral becomes \( \int_1^{\infty} \frac{1}{x}\, dx = \ln x \big|_1^{\infty} = \infty \). Diverges.
3
When \( p < 1 \): The exponent \( 1 - p > 0 \), so \( x^{1-p} \to \infty \). Integral diverges.

\( p \)-Series vs. Geometric Series

\( p \)-Series Geometric Series
Form \( \sum \frac{1}{n^p} \) \( \sum ar^n \)
Variable is in Base (\( n \)) Exponent (\( r^n \))
Convergence \( p > 1 \) \( |r| < 1 \)
Sum formula Usually no closed form \( \frac{a}{1 - r} \)
Decay rate Polynomial (slow) Exponential (fast)

How to Apply the Test

1
Identify the form: Can you write the general term as \( \frac{1}{n^p} \) (or \( \frac{C}{n^p} \) with a constant)?
2
Extract \( p \): Rewrite radicals and fractions using exponents. For example, \( \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} \), so \( p = \frac{1}{2} \).
3
Compare to 1: If \( p > 1 \) → converges. If \( p \leq 1 \) → diverges.
4
State your conclusion with the test name: "By the \( p \)-series test with \( p = \ldots \), the series converges/diverges."

Example 1: Direct Recognition

Determine whether the series converges or diverges:

1
Identify: This is a \( p \)-series with \( p = 4 \).
2
Compare: Since \( 4 > 1 \), the test tells us it converges.
3
Conclusion: By the \( p \)-series test, \( \sum \frac{1}{n^4} \) converges. (In fact, it equals \( \frac{\pi^4}{90} \).)

Example 2: Hidden Exponent

Determine whether the series converges or diverges:

1
Rewrite: \( \frac{1}{\sqrt[3]{n^2}} = \frac{1}{n^{2/3}} \). So this is a \( p \)-series with \( p = \frac{2}{3} \).
2
Compare: Since \( \frac{2}{3} < 1 \), the series diverges.
3
Conclusion: By the \( p \)-series test with \( p = \frac{2}{3} \leq 1 \), the series diverges.

Example 3: Borderline Case

Determine whether the series converges or diverges:

1
Rewrite: \( \frac{3}{n\sqrt{n}} = \frac{3}{n \cdot n^{1/2}} = \frac{3}{n^{3/2}} = 3 \cdot \frac{1}{n^{3/2}} \).
2
Identify: Constant multiple of a \( p \)-series with \( p = \frac{3}{2} \).
3
Compare: Since \( \frac{3}{2} > 1 \), the series converges.

The Spectrum of \( p \)-Values

0
\(\frac{1}{2}\)
 1
\(\frac{3}{2}\)
 2 DIVERGES CONVERGES
\(p = 1\): Harmonic (diverges)
The boundary is strict: \(p = 1\) diverges

Why Does \( p = 1 \) Diverge?

Nicole Oresme's proof (c. 1350) groups terms to show each group sums to at least \( \frac{1}{2} \):

1
Group 1: \( \frac{1}{3} + \frac{1}{4} \geq \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)
2
Group 2: \( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \geq 4 \cdot \frac{1}{8} = \frac{1}{2} \)
3
Each doubling group adds at least \( \frac{1}{2} \), so the sum grows to \( \infty \).

Common Mistakes to Avoid

✕ Confusing \( p \)-series with geometric
\( \sum \frac{1}{n^2} \) is a \( p \)-series (\( n \) in base). \( \sum \frac{1}{2^n} \) is geometric (\( n \) in exponent). Different tests apply.
✕ Thinking \( a_n \to 0 \) means convergence
The harmonic series has \( \frac{1}{n} \to 0 \) yet diverges. This condition is necessary but not sufficient.
✕ Misidentifying \( p \)
In \( \sum \frac{1}{\sqrt{n^3}} \), students often say \( p = 3 \). Correct: \( \sqrt{n^3} = n^{3/2} \), so \( p = \frac{3}{2} \).

Summary Reference Table

Series \( p \) Value Result
\( \sum \frac{1}{n^3} \) \( p = 3 \) Converges
\( \sum \frac{1}{n^2} \) \( p = 2 \) Converges
\( \sum \frac{1}{n^{3/2}} \) \( p = \frac{3}{2} \) Converges
\( \sum \frac{1}{n} \) \( p = 1 \) Diverges
\( \sum \frac{1}{\sqrt{n}} \) \( p = \frac{1}{2} \) Diverges
\( \sum \frac{1}{\sqrt[3]{n}} \) \( p = \frac{1}{3} \) Diverges

Question 1 of 5

Which of the following series converges?

Question 2 of 5

The series \( \displaystyle\sum_{n=1}^{\infty} \frac{7}{\sqrt[4]{n^3}} \) is a \( p \)-series. What is the value of \( p \)?

Question 3 of 5

Which statement about the harmonic series is correct?

Question 4 of 5

For what values of \( k \) does \( \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2k-1}} \) converge?

Question 5 of 5

Which test is most directly used to prove the \( p \)-series test?