Short on time? This page is the recap. For the proofs, the partial-sum graph and worked examples, use the full study guide. BC-only material. Prerequisites (partial sums 10.1, geometric series 10.2, nth term test 10.3, integral test 10.4) are on the Calculus BC roadmap.
Recap
- A p-series is ∑ 1/nᵖ with p a fixed number. The variable n is in the base.
- p > 1: converges. p ≤ 1: diverges. Proof: the integral test with ∫ from 1 to ∞ of x^(−p) dx (p > 0), and the nth term test for p ≤ 0.
- The harmonic series ∑ 1/n is the case p = 1. It diverges, although 1/n → 0.
- The alternating harmonic series 1 − 1/2 + 1/3 − … converges (proof in Topic 10.7). Same term sizes as ∑ 1/n, different signs, different outcome.
Key relationships
| Idea | Statement | Note |
|---|---|---|
| p-series rule | ∑ 1/nᵖ converges ⇔ p > 1 | Boundary p = 1 diverges |
| Integral behind it | ∫ from 1 to ∞ of x^(−p) dx = 1/(p − 1) for p > 1 | Not the sum of the series |
| Harmonic grouping | S(2ᵏ) ≥ 1 + k/2 | Partial sums are unbounded |
| Constant multiple | ∑ c·aₙ behaves like ∑ aₙ (c ≠ 0) | ∑ 1/(2n) diverges |
| Shifted start | Dropping finitely many terms keeps the behaviour | ∑ 1/(n + 5) diverges |
| Sums | conv + conv = conv; conv + div = div | div + div: no conclusion |
Assumptions
- p is a constant. If the power changes with n, it is not a p-series.
- Use the index laws: √n = n^(1/2), n·∛n = n^(4/3), nᵃ/nᵇ = n^(a−b).
Mistakes to avoid
- “Terms → 0, so it converges.” The harmonic series breaks this.
- Using the nth term test on ∑ 1/n. It is inconclusive there.
- p ≥ 1 or p > 0 instead of p > 1.
- Reading p before simplifying. n/n³ is 1/n², so p = 2.
- Calling ∑ 1/3ⁿ a p-series. It is geometric.
- Giving a decision without a reason. Name the series, state p, compare with 1.
Quick self-check
- Does ∑ n²/n^(7/2) converge? (Yes: the term is 1/n^(3/2), p = 3/2 > 1.)
- Does ∑ 1/(n^0.5 · n^0.4) converge? (No: p = 0.9 ≤ 1.)
- Does ∑ (1/n³ + 1/n) converge? (No: convergent plus divergent diverges.)
Next: practice questions.