Short on time? This page is the recap. For the number-line diagram and worked examples, use the full study guide. BC-only material. Prerequisites (geometric series 10.2, p-series 10.5, alternating series test 10.7, ratio test 10.8) are on the Calculus BC roadmap.
Recap
- A power series centred at c is Σ aₙ(x − c)ⁿ. It always converges at x = c (sum a₀).
- It converges only at c (R = 0), for all x (R = ∞), or for |x − c| < R and diverges for |x − c| > R.
- The endpoints c ± R must be tested one at a time.
- If R > 0, the series is the Taylor series of its sum f about c.
The method
- Form |uₙ₊₁/uₙ| using the whole term uₙ = aₙ(x − c)ⁿ.
- Take the limit as n → ∞. Set it < 1.
- Rewrite as |x − c| < R. Factor out any coefficient of x first.
- Substitute each endpoint and test the numerical series.
- Write the interval with the correct brackets.
Key relationships
| Idea | Fact | Example |
|---|---|---|
| Limit is 0 for all x | R = ∞ | Σ xⁿ/n! |
| Limit is ∞ for x ≠ c | R = 0 | Σ n! xⁿ |
| Coefficient on x | |2x − 5| < 1 means |x − 5/2| < 1/2, so R = 1/2 | Interval [2, 3] for Σ (2x − 5)ⁿ/n² |
| Coefficients and derivatives | f⁽ⁿ⁾(c) = n! · aₙ | a₄ = 1/6 gives f⁽⁴⁾(c) = 4 |
| Term-by-term calculus | Same R; endpoints may change | Σ xⁿ/n² on [−1, 1]; derivative on [−1, 1) |
Endpoint tests
- Σ 1/nᵖ: p-series (converges if p > 1).
- Alternating with terms decreasing to 0: alternating series test.
- Terms not tending to 0: nth term test (diverges).
- Never the ratio test: its limit is 1 at an endpoint.
Mistakes to avoid
- Calling the interval length the radius (length is 2R).
- Reading R from |2x − 5| < 1 without factoring.
- Wrong centre sign: (x + 1)ⁿ is centred at −1.
- Skipping the endpoints, or testing them with the ratio test.
- Leaving out a conditionally convergent endpoint.
- Assuming a differentiated series keeps the same endpoints.
Quick self-check
- Radius of Σ xⁿ/4ⁿ? (4)
- Interval of Σ from n = 1 to ∞ of (x − 1)ⁿ/n? ([0, 2): x = 2 gives Σ 1/n, which diverges; x = 0 gives Σ (−1)ⁿ/n, which converges)
- If f(x) = Σ from n = 0 to ∞ of (x − 2)ⁿ/(n + 1) for |x − 2| < 1, what is f‴(2)? (3! · 1/4 = 3/2)
Next: practice questions.