Short on time? This page is the recap. For the derivation, the graph and full worked examples, use the full study guide. BC-only material. Prerequisites (the definition of convergence from Topic 10.1, limits at infinity 1.15, exponent and log laws) are on the Calculus BC roadmap.
Recap
- Geometric means a constant ratio r = (any term)/(the term before it).
- Standard form: Σ from n = 0 to ∞ of a·rⁿ = a + ar + ar² + …, first term a.
- Partial sum: Sₙ = a(1 − rⁿ)/(1 − r) for r ≠ 1 (from subtracting r·Sₙ from Sₙ).
- If |r| < 1, rⁿ → 0, so the series converges to a/(1 − r).
- If |r| ≥ 1 and a ≠ 0, the partial sums have no limit: the series diverges.
Key relationships
| Situation | What to do | Example |
|---|---|---|
| Standard form, |r| < 1 | Sum = a/(1 − r) | Σ from n = 0 of 6(−½)ⁿ = 6/(3/2) = 4 |
| Later start index | Use the actual first term | Σ from n = 2 of 8(−3/4)ⁿ = (9/2)/(7/4) = 18/7 |
| Messy exponents | Rewrite as (constant)·(ratio)ⁿ | 3ⁿ⁺¹/5ⁿ = 3(3/5)ⁿ |
| |r| ≥ 1 | Diverges; no formula | Σ 4ⁿ/3ⁿ⁺¹, r = 4/3 |
| First n terms only | Sₙ = a(1 − rⁿ)/(1 − r) | 40 + 36 + …: S₁₀ ≈ 260.529 |
| Ratio contains x | Solve |r| < 1; endpoints diverge | ((x − 1)/3)ⁿ: −2 < x < 4, sum 3/(4 − x) |
| Repeating decimal | a = first block, r = 10⁻ᵏ | 0.4545… = 0.45/0.99 = 5/11 |
Mistakes to avoid
- Wrong first term when the series starts at n = 1 or n = 2.
- Using a/(1 − r) without checking |r| < 1.
- Including x-values where |r| = 1.
- Sign slips with negative r: 1 − (−¾) = 7/4.
- Total versus partial: “in the first n” means Sₙ; “in total” means the full sum.
- Not reversing the inequality when dividing by ln r < 0.
Quick self-check
- Find Σ from n = 0 to ∞ of 7(2/5)ⁿ. (7/(3/5) = 35/3)
- Find Σ from n = 1 to ∞ of (−1)ⁿ · 5/2ⁿ. (First term −5/2, r = −½: (−5/2)/(3/2) = −5/3)
- Does Σ from n = 0 to ∞ of 3(5/4)ⁿ converge? (No: r = 5/4 and |r| > 1, so it diverges)
Next: practice questions.