# Chapter 26: Taylor and Power Series: How to Solve Problems > Radius and intervals of convergence, and how to bend known series — by substituting, differentiating, integrating and combining — into derivatives, integrals and limits that l'Hôpital cannot touch. **Canonical:** https://calculus.kitzo.work/chapter/power-series **Level:** Advanced **Estimated time:** 50 minutes **Exercises:** 8 (auto-marked; answers stay server-side) ## Topics - Radius of convergence - Interval of convergence - Substituting into series - Differentiating & integrating series - Combining series - Series for limits **Why this matters** Infinite series are the secret engine under your calculator: it does not evaluate e^0.3 directly, it adds up a power series. Learning to build, bend and evaluate power series turns impossible problems into arithmetic — and it is the final boss of first-year calculus. ## 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). **Worked example** **Example 1 — Radius of Σ xⁿ/n!.** 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 = ∞. **Example 2 — Interval of convergence of Σ (x − 1)ⁿ/(n·2ⁿ).** 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. **Watch out** The ratio test is useless exactly at the endpoints (L = 1) — always test them by hand. Also, differentiating or integrating a series keeps R but can flip endpoint behavior (harmonic vs alternating harmonic). ## 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. **Chapter formula sheet** Σₙ₌₀^∞ xⁿ = 1/(1 − x), |x| < 1 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. ## Practice exercises 1. The radius of convergence of Σₙ₌₀^∞ xⁿ/n! is: - A. 1 - B. e - C. ∞ - D. 0 2. Find the radius of convergence of Σₙ₌₁^∞ xⁿ/n² . 3. The interval of convergence of Σₙ₌₁^∞ xⁿ/n is [−1, 1). At x = 1, the series … (one word). 4. The interval of convergence of Σₙ₌₁^∞ (x − 1)ⁿ/(n·2ⁿ) is: - A. (−1, 3) - B. [−1, 3) - C. (−1, 3] - D. [−1, 3] 5. Using eˣ = Σ xⁿ/n! , the coefficient of x³ in the series is 1/6 ≈ 0.1667. Confirm as a decimal rounded to 3 places. 6. Does the Maclaurin series of e^(x²) converge for every real x? (one word) 7. Which series equals 1/(1 + x²) for |x| < 1? - A. Σ x²ⁿ - B. Σ (−1)ⁿ x²ⁿ - C. Σ (−1)ⁿ xⁿ - D. Σ xⁿ 8. Use series to evaluate lim x→0 (sin x − x)/x³ . --- Practice interactively: https://calculus.kitzo.work/chapter/power-series Citation: CalculusKitzo (https://calculus.kitzo.work/). Independent study notes inspired by the standard calculus syllabus; not affiliated with Princeton University Press. See https://calculus.kitzo.work/disclaimer