Series & Sequences · Convergence Tests

The p-Series Test

Σ 1/nᵖ: the single most important benchmark in all of series theory. Understand exactly when it converges, why it diverges for p ≤ 1, and how to use it as a comparison anchor.

12Worked Examples
10Quiz Questions
3Interactive Diagrams
Share this page

§ 01What Is a p-Series?

A p-series is any series of the form Σ 1/nᵖ. A single number p determines everything: whether the series converges or diverges, and how quickly the partial sums grow.

p-Series — Definition & Theorem
Σn=1∞ 1⁄nᵖ = 1 + 1⁄2ᵖ + 1⁄3ᵖ + 1⁄4ᵖ + ··· Converges  if  p > 1     Diverges  if  p ≤ 1

p can be any real number, including fractions, decimals, or irrational numbers.

The p-series is not merely a theoretical curio, it is the primary comparison target for dozens of other series. Whenever you see a series whose terms behave like a power of n in the denominator, you are essentially looking at a p-series in disguise, and the comparison tests unlock the answer immediately.

The p-Value Spectrum

p ≤ 1   DIVERGES
p > 1   CONVERGES
p → −∞ (diverges fast) p = 1 (harmonic) p → +∞ (converges fast)
The boundary case p = 1 is the harmonic series Σ 1/n = 1 + 1/2 + 1/3 + 1/4 + ··· diverges, even though its terms go to zero. This is one of the most counterintuitive results in all of mathematics — see §03 for the proof and intuition.

§ 02Proof via the Integral Test

The p-series theorem follows directly from the Integral Test, which connects the convergence of Σ f(n) to the convergence of ∫ f(x)dx. The function f(x) = 1/xᵖ is continuous, positive, and decreasing for x ≥ 1, so the test applies.

The Key Integral
∫1∞ 1⁄xᵖ dx = ∫1∞ x−p dx

Case 1: p ≠ 1

Using the power rule for integration:

Evaluation (p ≠ 1)
= [x1−p⁄1−p]1∞ If p > 1:   1−p < 0, so x1−p → 0 as x → ∞  ⟹  integral = 1⁄p−1 If p < 1:   1−p > 0, so x1−p → ∞ as x → ∞  ⟹  integral diverges

The integral converges iff p > 1, giving the exact value 1/(p−1).

Case 2: p = 1 (The Harmonic Series)

Harmonic integral
∫1∞ 1⁄x dx = [ln x]1∞ = ∞

The logarithm grows without bound, the integral diverges, confirming Σ 1/n diverges.

What the Integral Test actually says The Integral Test says Σ f(n) and ∫f(x)dx share the same convergence/divergence (when f is positive, continuous, and decreasing). Crucially, the integral value is NOT the series sum — it only determines behaviour. For example, Σ 1/n² converges to π²/6 ≈ 1.6449, but ∫₁^∞ 1/x² dx = 1.

§ 03The Harmonic Series — Deep Dive

The harmonic series Σ 1/n is the most famous divergent series in mathematics. Its divergence is slow, surprising, and has profound consequences.

Even though the terms 1/n → 0, the series diverges. In fact, the partial sums Hₙ = 1 + 1/2 + 1/3 + ··· + 1/n, called harmonic numbers, grow like ln(n). Specifically, Hₙ ≈ ln(n) + γ where γ ≈ 0.5772 is the Euler-Mascheroni constant.

Cauchy's Condensation Proof (Elementary)

Group the terms in blocks whose sums each exceed 1/2:

Grouping argument
1 + 1⁄2 + (1⁄3 + 1⁄4) + (1⁄5 + ··· + 1⁄8) + ··· > 1 + 1⁄2 + (1⁄4 + 1⁄4) + (1⁄8 + ··· + 1⁄8) + ··· = 1 + 1⁄2 + 1⁄2 + 1⁄2 + ··· = ∞

Each block of 2ᵏ terms contributes more than 1/2. There are infinitely many blocks, so the sum grows without bound.

How long until the harmonic series exceeds 10? You need roughly e¹⁰ ≈ 22,026 terms. To exceed 100, you'd need about e¹⁰⁰ ≈ 2.7 × 10⁴³ terms — far more than the number of atoms in the observable universe. Divergence can be spectacularly slow.
Interactive — partial sums of Σ 1/nᵖ for varying p

Gold curve: partial sums Sₙ. Teal dashed: convergence limit (when p>1). Rust dashed: ln(n) growth reference (harmonic).

§ 04Special Values & Named Series

Several p-series have exact closed-form sums, many involving π. These are known as the Basel problem and its extensions, solved using Fourier analysis and complex analysis.

pSeries NameSumStatusDecimal
1/2Σ 1/√n—Divergesgrows ≈ 2√N
1Harmonic Series—Divergesgrows ≈ ln N
2Basel Problem (Euler, 1734)π²/6Converges≈ 1.6449
3Apéry's Constantζ(3) ≈ 1.202Converges≈ 1.2021
4—π⁴/90Converges≈ 1.0823
6—π⁶/945Converges≈ 1.0173
∞—1 (only first term)Converges= 1.0000
The Riemann Zeta Function The function ζ(s) = Σ 1/nˢ is the famous Riemann zeta function, defined for s > 1 as a real series (and extended analytically to the whole complex plane). The Riemann Hypothesis — one of the Millennium Prize Problems — concerns the zeros of ζ(s) in the complex plane. The humble p-series is thus the starting point of one of the deepest unsolved problems in mathematics.
Diagram — term magnitudes aₙ = 1/nᵖ for different p values

Gold bars = 1/nᵖ¹. Teal bars = 1/nᵖ². Larger p means terms decay faster.

§ 05Worked Examples

Example 1 — Direct application

Does Σn=1∞ 1/n3 converge?

  1. This is a p-series with p = 3.
  2. Since p = 3 > 1, the series converges by the p-series test.
  3. The exact sum is ζ(3) ≈ 1.202 (Apéry's constant).
  4. Converges (p = 3 > 1)
Example 2 — Boundary case

Does Σn=1∞ 1/n0.99 converge?

  1. p = 0.99. Since 0.99 < 1, the series diverges.
  2. Notice: even though p is very close to 1, the test is strict. p ≤ 1 always means divergence.
  3. The series diverges even though its terms decrease faster than the harmonic series.
  4. Diverges (p = 0.99 ≤ 1)
Example 3 — Fractional p

Does Σ 1/n3/2 converge?

  1. p = 3/2 = 1.5 > 1 ⟹ converges.
  2. This is often written as Σ 1/n√n or Σ 1/√(n³). All equivalent.
  3. Converges (p = 3/2 > 1)
Example 4 — Algebraic manipulation first

Does Σn=1∞ n² / n⁵ converge?

  1. Simplify: n²/n⁵ = 1/n³.
  2. So the series is Σ 1/n³, a p-series with p = 3 > 1.
  3. Converges (p = 3)
Example 5 — Constant multiple

Does Σn=1∞ 7 / n4 converge? If so, what does it sum to?

  1. Factor the constant: Σ 7/n⁴ = 7 · Σ 1/n⁴.
  2. Σ 1/n⁴ is a p-series with p = 4 > 1 ⟹ converges to π⁴/90.
  3. Therefore Σ 7/n⁴ = 7 · π⁴/90 = 7π⁴/90 ≈ 7.577.
  4. Converges to 7π⁴/90
Example 6 — Rewrite as p-series

Does Σn=1∞ 1 / ∛(n⁴) converge?

  1. Rewrite: ∛(n⁴) = n^(4/3), so the series is Σ 1/n^(4/3).
  2. p = 4/3 > 1 ⟹ converges.
  3. Converges (p = 4/3)
Example 7 — Sum of p-series split

Does Σ (3/n² + 5/n³) converge? Find the sum.

  1. Split: Σ (3/n² + 5/n³) = 3 · Σ 1/n² + 5 · Σ 1/n³.
  2. Both are p-series with p = 2 and p = 3, both > 1, both convergent.
  3. Sum = 3 · (π²/6) + 5 · ζ(3) = π²/2 + 5ζ(3) ≈ 4.935 + 6.010 ≈ 10.945.
  4. Converges to π²/2 + 5ζ(3)
Example 8 — p-series as comparison benchmark

Use the p-series to determine whether Σ 1/(n² + 4n + 4) converges.

  1. Factor: n² + 4n + 4 = (n+2)², so Σ 1/(n+2)².
  2. This is a shifted p-series. For large n, 1/(n+2)² ≤ 1/n².
  3. Formally, apply LCT: lim [1/(n+2)²] / [1/n²] = lim n²/(n+2)² = 1. Since Σ 1/n² converges, so does this.
  4. Converges (behaves like p = 2)
Example 9 — Negative p

Does Σn=1∞ n² (i.e. p = −2) converge?

  1. Σ n² = Σ 1/n^(−2), so p = −2 ≤ 1.
  2. More directly: the nth-term test, aₙ = n² → ∞ ≠ 0, so the series diverges immediately.
  3. Diverges (terms → ∞)
Example 10 — The integral test value vs series sum

For Σ 1/n², compute ∫₁∞ 1/x² dx. Explain why this is not the series sum.

  1. ∫₁∞ x⁻² dx = [−x⁻¹]₁∞ = 0 − (−1) = 1.
  2. The integral test uses this value only to confirm convergence. The integral equals 1, but the series sums to π²/6 ≈ 1.6449.
  3. The Integral Test gives: 1 = ∫₁∞ 1/x² dx ≤ Σ 1/n² ≤ 1 + ∫₁∞ 1/x² dx = 2. This bounds the sum but does not give the exact value.
  4. The exact value π²/6 requires Fourier analysis (Basel problem proof).
Example 11 — p-series in disguise (shifting)

Does Σn=2∞ 1/(n−1)² converge?

  1. Substitute m = n − 1 (when n = 2, m = 1; as n → ∞, m → ∞):
  2. The series becomes Σm=1∞ 1/m², a standard p-series with p = 2.
  3. p = 2 > 1 ⟹ converges to π²/6.
  4. Converges (index shift to p = 2 series)
Example 12 — Finding p from given convergence info

For what values of k does Σ 1/n^(2k−1) converge?

  1. This is a p-series with p = 2k − 1.
  2. Converges iff p > 1 iff 2k − 1 > 1 iff 2k > 2 iff k > 1.
  3. The series converges for k > 1
  4. At k = 1: p = 1 (harmonic), diverges. At k = 1.5: p = 2, converges to π²/6.

§ 06Common Mistakes

Mistake 1 — Thinking p = 1 converges because terms go to zero The harmonic series Σ 1/n has terms that go to zero, yet it diverges. The nth-term test only tells you that terms going to zero is a necessary condition for convergence — it is not sufficient. The p-series test gives the definitive answer.
Mistake 2 — Using the integral value as the series sum ∫₁∞ 1/xᵖ dx = 1/(p−1) tells you the series converges, not what it converges to. Σ 1/n² = π²/6, not 1/(2−1) = 1. The integral provides bounds, not the exact sum.
Mistake 3 — Misidentifying p from a rewritten series Σ 1/√n has p = 1/2 (not p = 2). Σ √n = Σ n^(1/2) has p = −1/2. Always express the series as Σ 1/nᵖ = Σ n^(−p) before reading off p.
Mistake 4 — Shifted p-series — p still determines convergence Σ 1/(n+5)² converges for the same reason as Σ 1/n² — the shift affects only finitely many terms and the asymptotic behaviour is identical. Use LCT with bₙ = 1/n² to confirm formally.

§ 07Related Pages

Series
Series Convergence
Partial sums, nth-term test, definition of convergence
Series
Geometric Series
The other essential convergence benchmark alongside p-series
Series
Comparison Tests
p-series is the most common comparison target
Series
Ratio & Root Tests
Best for factorials and exponentials
Integration
Improper Integrals
The key tool in the Integral Test proof
Series
Alternating Series Test
The alternating harmonic series Σ(−1)ⁿ/n converges to ln 2
Series
Taylor Series
p-series appear in Taylor coefficients and radius of convergence

§ 08Quiz — Test Your Understanding

10 questions on the p-series test, harmonic series, integral connections, and algebraic manipulations.

Score: 0 / 0
Question 1 of 10

The p-series Σ 1/nᵖ converges if and only if:

Question 2 of 10

What is the sum of Σn=1∞ 1/n²?

Question 3 of 10

Does Σ 1/√n converge?

Question 4 of 10

What value of p is the boundary between convergence and divergence?

Question 5 of 10

Which of these is also a p-series (after simplification)?

Question 6 of 10

The harmonic series Σ 1/n diverges because:

Question 7 of 10

For what values of k does Σ n^k converge?

Question 8 of 10

∫₁∞ 1/x³ dx = ?

Question 9 of 10

What is the p-value of Σ 1/n^(2/3)?

Question 10 of 10

Σ 1/n converges to ln(2). True or False?

0/10

Cookie Settings