1 · Power series in general
A power series is an infinite polynomial: Σ aₙ(x − c)ⁿ = a₀ + a₁(x − c) + a₂(x − c)² + … centered at c. Some series converge for every x (like eˣ, sin x, cos x), some only in a band around the center, some only at c itself. This chapter answers two questions: where does it converge, and how do we use it.
2 · Radius of convergence — the ratio test
Apply the ratio test to the general term: L = lim |aₙ₊₁(x − c)ⁿ⁺¹ ÷ aₙ(x − c)ⁿ|. Since the x-part factors out, L = |x − c|·k for some constant k. Converge when L < 1, diverge when L > 1. The inequality |x − c| < 1/k carves out the radius of convergence R = 1/k (R = ∞ if k = 0, R = 0 if the ratio explodes).
Step 1: ratio of consecutive terms = xⁿ⁺¹/(n+1)! ÷ xⁿ/n! = x/(n + 1).
Step 2: limit as n → ∞ is L = 0 for every x.
Step 3: L < 1 always ⇒ the series converges for all x. Radius R = ∞.
Step 1: ratio → |x − 1|/2.
Step 2: converge when |x − 1|/2 < 1, so R = 2 and the open interval is (−1, 3).
Step 3: test endpoints separately. x = 3 gives Σ 1/n (harmonic — diverges). x = −1 gives Σ (−1)ⁿ/n (alternating harmonic — converges).
Interval of convergence: [−1, 3). Endpoints are NEVER settled by the ratio test — plug them in and test each one.
3 · Substituting into known series
The cheapest new series comes from plugging into an old one. From 1/(1 − x) = Σ xⁿ (|x| < 1) you get 1/(1 + x²) = Σ (−1)ⁿ x²ⁿ, valid for x² < 1, i.e. |x| < 1. Same trick: e^(x²) from the series of eˣ, or sin(2x) from sin x. Substitution can only shrink the region of validity, never grow it past where the identity holds.
4 · Differentiating and integrating term by term
Inside its interval of convergence a power series acts like a polynomial: differentiate or integrate term by term. The radius stays the same; the endpoints must be re-tested. From 1/(1 − x) = Σ xⁿ, differentiating gives 1/(1 − x)² = Σ (n + 1) xⁿ, and integrating gives −ln(1 − x) = Σ xⁿ⁺¹/(n + 1) = x + x²/2 + x³/3 + …
5 · Combining series — add, subtract, multiply, divide
- Add/subtract term by term where both converge — the sum converges at least on the overlap.
- Multiply (Cauchy product): (Σaₙxⁿ)(Σbₙxⁿ) = Σ cₙxⁿ with cₙ = a₀bₙ + a₁bₙ₋₁ + … + aₙb₀. Keep only as many first terms as you need.
- Divide by long division or undetermined coefficients: assume the quotient series, cross-multiply, match coefficients.
6 · Derivatives and limits via series
To find f′(0) or higher derivatives, skip the derivative: read off the coefficient. If f(x) = Σ aₙxⁿ then f⁽ᵏ⁾(0) = k! · aₖ. For limits, series often beat l'Hôpital: replace each function by its Maclaurin series and cancel leading terms. For limₓ→0 (sin x − x)/x³, since sin x = x − x³/6 + … the numerator is −x³/6 + … and the limit is −1/6 — no derivatives of any order required.
eˣ = Σ xⁿ/n! (all x) · sin x = Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)! (all x)
cos x = Σ (−1)ⁿ x²ⁿ/(2n)! (all x)
Radius: R = 1 / lim |aₙ₊₁/aₙ| · f⁽ᵏ⁾(0) = k! · aₖ
Term-by-term differentiate/integrate (endpoints re-tested) · cₙ = a₀bₙ + a₁bₙ₋₁ + … + aₙb₀
Common mistakes
- Forgetting to test the endpoints — the ratio test is silent there.
- Assuming a manipulated series keeps its interval; substitution rescales it.
- Integrating 1/(1 + x) and losing the constant of integration.
- Multiplying series and mislining the powers of x — cₙ = a₀bₙ + a₁bₙ₋₁ + …
- Reaching for l'Hôpital four times when three lines of Maclaurin suffice.