Short on time? This page is the recap. For the graph, the telescoping method and full worked examples, use the full study guide. BC-only material. Prerequisites (limits at infinity 1.15, partial fractions 6.12, laws of logarithms) are on the Calculus BC roadmap.
Recap
- Terms aₙ are the numbers being added. The series Σ aₙ is the instruction to add them all.
- The nth partial sum Sₙ = a₁ + a₂ + … + aₙ is the total of the first n terms.
- The series converges to S exactly when lim as n → ∞ of Sₙ = S. Then S is the sum, an exact number.
- If lim Sₙ does not exist (Sₙ → ±∞, or Sₙ never settles), the series diverges and has no sum.
Key relationships
| Idea | Statement | Example |
|---|---|---|
| Partial sum | Sₙ = Σ from k = 1 to n of aₖ | ½ + ¼ + ⅛: S₃ = ⅞ |
| Convergence | Σ aₙ = S ⇔ lim Sₙ = S | Sₙ = 3n/(n + 2) → 3 |
| Terms from Sₙ | a₁ = S₁; aₙ = Sₙ − Sₙ₋₁ (n ≥ 2) | Sₙ = 3n/(n + 2) gives a₄ = 1/5 |
| Telescoping | Split aₙ, cancel, find Sₙ, take the limit | Σ 2/((n + 1)(n + 3)) = 5/6 |
| Divergence with aₙ → 0 | Sₙ can still grow without bound | Σ ln((n + 1)/n): Sₙ = ln(n + 1) → ∞ |
| Bounded but divergent | Sₙ oscillates | 1 − 1 + 1 − …: Sₙ = 1, 0, 1, 0, … |
| Changing the start | Same convergence, different sum | Σ from n = 3 of 2/((n + 1)(n + 3)) = 9/20 |
Mistakes to avoid
- Using lim aₙ as the sum. The sum is lim Sₙ.
- “aₙ → 0, so it converges.” Counter-example: Σ ln((n + 1)/n).
- Trusting a table of partial sums as proof.
- Assigning a value to 1 − 1 + 1 − … or writing “the sum is ∞” as if it were a sum.
- Using S₀ for the first term. Use a₁ = S₁.
- Losing surviving pieces when telescoping: write out the first few and last few brackets.
Quick self-check
- Sₙ = 2n/(n + 1) for Σ from n = 1 of aₙ. Find a₁, a₃ and the sum. (a₁ = 1, a₃ = 3/2 − 4/3 = 1/6, sum 2)
- Find Σ from n = 1 to ∞ of (1/(n + 1) − 1/(n + 2)). (Sₙ = ½ − 1/(n + 2) → ½)
- Does Σ from n = 1 to ∞ of (√(n + 1) − √n) converge? (No: Sₙ = √(n + 1) − 1 → ∞, so it diverges, although the terms → 0)
Next: practice questions.