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Unit 10 · Infinite Sequences & Series (BC) Flashcards Cheat Sheet Essentials Visual Review MC Practice FRQ Practice

AP Calculus AB/BC Unit 10 Visual Review

A topic-by-topic visual walkthrough of Unit 10 (BC): Infinite Sequences and Series — convergence tests, alternating series, Taylor and Maclaurin series, and power series.

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TOPIC 10.1 BC ONLY Convergent & Divergent Series Σ aₙ converges ⟺ the partial sums Sₙ approach a finite limit Partial sums Sₙ = a₁ + a₂ + … + aₙ If lim(n→∞) Sₙ = S (finite) → CONVERGES to the sum S. If the limit is ±∞ or DNE → DIVERGES. A series is the limit of its sequence of partial sums. Sequence vs. series A SEQUENCE is a list of terms aₙ. A SERIES is their sum Σaₙ. Convergence tests decide whether the sum is finite — Topics 10.2–10.9. Terms → 0 is necessary but not sufficient (10.3). Σ A series converges if its partial sums approach a finite limit. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.2 BC ONLY Geometric Series Σ a·rⁿ converges ⟺ |r| < 1 sum = a / (1 − r) (first term ÷ (1 − ratio)) Worked example Σ 6·(1/3)ⁿ, n=0…∞ a = 6, r = 1/3, |r| < 1 → converges: = 6 / (1 − 1/3) = 6/(2/3) = 9 Watch the starting index for "a" (the first term). The most useful series Geometric series have a KNOWN sum — the only common series you can sum exactly. |r| ≥ 1 → diverges (terms don't shrink to 0). Basis for power series (10.15) & repeating decimals. Geometric: converges iff |r|<1, sum = a/(1−r). The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.3 BC ONLY The nth Term Test for Divergence if lim(n→∞) aₙ ≠ 0 → Σ aₙ DIVERGES A quick first check If the terms don't approach 0, the sum can't settle → the series diverges. Σ n/(n+1): aₙ → 1 ≠ 0 → diverges immediately. Always try this test first. It can only prove DIVERGENCE If aₙ → 0, the test is INCONCLUSIVE — the series may still converge or diverge. Σ 1/n: aₙ → 0 but DIVERGES (the harmonic series — see 10.5). Terms → 0 is NECESSARY, not sufficient. If aₙ ↛ 0 the series diverges; if aₙ → 0 the test says nothing. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.4 BC ONLY The Integral Test Σ aₙ and ∫₁^∞ f(x) dx converge or diverge TOGETHER Conditions on f Let f(n) = aₙ, where f is: • POSITIVE • CONTINUOUS • DECREASING Then the series and the improper integral share the same fate. The integral's value is NOT the series' sum. Worked example Σ 1/n² ↔ ∫₁^∞ 1/x² dx The integral = 1 (converges), so the series CONVERGES. Best when f is easy to integrate. Confirms the p-series rule (10.5). ∫ Integral test: a positive, decreasing f — series & ∫f share the same fate. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.5 BC ONLY Harmonic & p-Series Σ 1/nᵖ converges ⟺ p > 1 The harmonic series (p = 1) Σ 1/n = 1 + 1/2 + 1/3 + … DIVERGES — even though its terms → 0. A famous, counter-intuitive result. p = 1 is exactly the borderline case. p-series rule p > 1 → CONVERGES. p ≤ 1 → DIVERGES. Σ1/n² (p=2) converges Σ1/√n (p=½) diverges A key comparison benchmark (10.6). Σ1/nᵖ converges iff p>1; the harmonic series (p=1) diverges. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.6 BC ONLY Comparison Tests Direct comparison Compare term-by-term to a known series: • 0 ≤ aₙ ≤ bₙ and Σbₙ converges → Σaₙ converges. • aₙ ≥ bₙ ≥ 0 and Σbₙ diverges → Σaₙ diverges. "Smaller than convergent → converges; bigger than divergent → diverges." 1/(n²+1) < 1/n² → converges Limit comparison Take the limit of the ratio: lim(n→∞) aₙ / bₙ = L If 0 < L < ∞ (finite & positive), both series do the SAME thing. Easier than direct comparison — just match the dominant terms. Σ(2n+1)/(n³+5): compare to Σ1/n² Choose bₙ from the leading powers. Compare to a known series directly, or via a finite limit of the ratio. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.7 BC ONLY The Alternating Series Test Σ (−1)ⁿ bₙ converges if BOTH hold: 1) bₙ decreasing 2) lim bₙ = 0 Signs must alternate The (−1)ⁿ flips each term's sign. Σ (−1)ⁿ⁺¹/n = 1 − ½ + ⅓ − … bₙ=1/n decreasing → 0, so it CONVERGES. (the alternating harmonic series) Test the bₙ (magnitudes) Check the POSITIVE part bₙ = |aₙ|: it must shrink monotonically to zero. Convergence here may be CONDITIONAL (see 10.9). Alternating series converge if bₙ decreases to 0. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.8 BC ONLY The Ratio Test L = lim(n→∞) | aₙ₊₁ / aₙ | Read the limit L L < 1 → CONVERGES (absolutely) L > 1 → DIVERGES L = 1 → INCONCLUSIVE Best for FACTORIALS and EXPONENTIALS in the terms — they simplify in the ratio. Worked example Σ 2ⁿ / n! |aₙ₊₁/aₙ| = 2/(n+1) → 0 = L < 1 So the series CONVERGES. Central to interval-of-convergence (10.13). Ratio test: L<1 converges, L>1 diverges, L=1 inconclusive. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.9 BC ONLY Absolute vs. Conditional Convergence Absolute convergence Σ|aₙ| CONVERGES. This is the stronger form — if the series of absolute values converges, so does Σaₙ. Σ(−1)ⁿ/n² : Σ1/n² conv. → absolute Rearranging terms won't change the sum. Conditional convergence Σaₙ converges BUT Σ|aₙ| diverges. The alternating signs are what save it. Σ(−1)ⁿ/n : converges, but Σ1/n diverges → conditional The alternating harmonic series is the classic case. Test the absolute-value series Check Σ|aₙ| first: if it converges → absolute. If not, but Σaₙ converges (e.g. by AST) → conditional. Σ|aₙ| converges → absolute; Σaₙ converges but Σ|aₙ| doesn't → conditional. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.10 BC ONLY Alternating Series Error Bound | S − Sₙ | ≤ b₍ₙ₊₁₎ (the next term's size) Error ≤ first omitted term For a CONVERGENT alternating series, the error from stopping at Sₙ is at most the absolute value of the NEXT term. Very easy to compute — just the size of b₍ₙ₊₁₎. Worked example Approx Σ(−1)ⁿ⁺¹/n with S₃ = 1 − ½ + ⅓: error ≤ b₄ = 1/4 = 0.25 To get error < 0.01, take enough terms so the next term is < 0.01. Only valid when the AST conditions hold. Alternating error ≤ the first omitted term |b₍ₙ₊₁₎|. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.11 BC ONLY Taylor Polynomial Approximations Pₙ(x) = Σ f⁽ᵏ⁾(a)/k! · (x − a)ᵏ, k = 0…n Match derivatives at a The Taylor polynomial agrees with f in value and first n derivatives at x = a. Centered at a = 0, it's a MACLAURIN polynomial. Higher degree → better fit near a. Worked example eˣ centered at 0: P₃(x) = 1 + x + x²/2 + x³/6 Since every derivative of eˣ is eˣ = 1 at 0. e^0.1 ≈ P₃(0.1) ≈ 1.10517 Divide the k-th derivative by k!. Pₙ(x)=Σf⁽ᵏ⁾(a)/k!·(x−a)ᵏ — matches f's derivatives at a. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.12 BC ONLY The Lagrange Error Bound | Rₙ(x) | ≤ ( max|f⁽ⁿ⁺¹⁾| / (n+1)! ) · |x − a|ⁿ⁺¹ Bounds the remainder Rₙ(x) = f(x) − Pₙ(x) is the error of the Taylor approximation. Use the MAXIMUM of the (n+1)-th derivative on the interval between a and x. Works for ANY Taylor series (not just alternating). The structure It looks like the NEXT term of the series, with the derivative replaced by its max. Smaller |x − a| and higher n → smaller error bound. Often you bound |f⁽ⁿ⁺¹⁾| by a convenient constant. Lagrange: error ≤ (max|f⁽ⁿ⁺¹⁾|/(n+1)!)·|x−a|ⁿ⁺¹. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.13 BC ONLY Radius & Interval of Convergence center a a − Ra + R radius R Use the ratio test on the power series Apply the ratio test to Σcₙ(x−a)ⁿ; require lim|aₙ₊₁/aₙ| < 1 and solve for x. This gives |x−a| < R. converges on (a−R, a+R); diverges outside Then CHECK THE ENDPOINTS x = a ± R separately (plug in — they may converge or not). R can be 0 (only at a), a finite number, or ∞ (all x). Ratio test gives radius R; then check the two endpoints for the full interval. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.14 BC ONLY Taylor & Maclaurin Series to Know // memorize these Maclaurin series (centered at 0) eˣ = 1 + x + x²/2! + x³/3! + … sin x = x − x³/3! + x⁵/5! − … cos x = 1 − x²/2! + x⁴/4! − … 1/(1−x) = 1 + x + x² + x³ + … (|x|<1) Manipulate the known series Substitute, multiply, differentiate, or integrate a known series instead of computing derivatives. e.g. e^(x²): replace x with x². Memorize the series for eˣ, sin x, cos x, 1/(1−x) and manipulate them. The Review Hub · AP Calculus AB/BC Unit 10 TOPIC 10.15 BC ONLY Representing Functions as Power Series Start from the geometric series 1/(1−x) = Σ xⁿ (|x| < 1) Substitute to build new series: 1/(1+x²) = Σ(−1)ⁿx²ⁿ (replace x with −x²). The new interval follows from the substitution. Differentiate / integrate term by term Integrate 1/(1+x²) = Σ(−1)ⁿx²ⁿ: arctan x = Σ(−1)ⁿx²ⁿ⁺¹/(2n+1) A power series can be differentiated and integrated WITHIN its interval. The radius R is unchanged (endpoints may differ). Approximate hard integrals Write a function as a series, then integrate term by term to evaluate integrals with no elementary antiderivative (e.g. ∫e^(−x²)dx). Build power series by substituting, differentiating, or integrating known ones. The Review Hub · AP Calculus AB/BC Unit 10
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How to use the visual review

Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I pick the right test or build this series from memory?"

Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.

This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 10.