Skip to content
Marlbridge
Free trial

Flashcards · AP® Calculus BC · Unit 10

Unit 10: Infinite Sequences and Series flashcards

102 cards on topics 10.1 to 10.15 (BC only). Pick a topic or study the whole unit.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 102 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

10.1 Defining Convergent and Divergent Infinite Series

What does it mean for an infinite series to converge?
The series Σ aₙ converges to S exactly when the partial sums Sₙ = a₁ + … + aₙ have lim (n → ∞) Sₙ = S. If lim Sₙ does not exist (infinite or never settling), the series diverges and has no sum.
Is the sum of a series the limit of its terms?
No. The sum is lim Sₙ, the limit of the partial sums, not lim aₙ.
Sₙ = 2n/(n + 1) for Σ (n = 1 to ∞) aₙ. Find a₁, a₃ and the sum.
a₁ = S₁ = 1. a₃ = S₃ − S₂ = 3/2 − 4/3 = 1/6. The sum is lim Sₙ = 2.
How do you sum a telescoping series such as Σ (n = 1 to ∞) (1/(n + 1) − 1/(n + 2))?
Write out the first and last few brackets so the middle cancels: Sₙ = ½ − 1/(n + 2). Then take the limit: the sum is ½.
If aₙ → 0, must Σ aₙ converge?
No. Σ (√(n + 1) − √n) has terms tending to 0, but Sₙ = √(n + 1) − 1 → ∞, so it diverges. Σ ln((n + 1)/n) is another example.
Does 1 − 1 + 1 − 1 + … have a sum?
No. Its partial sums are 1, 0, 1, 0, …, which never settle, so it diverges even though they are bounded.
Does changing the starting index of a series change whether it converges?
No, only the sum changes. Σ (n = 3 to ∞) 2/((n + 1)(n + 3)) converges, with sum 9/20 instead of 5/6 from n = 1.

10.2 Working with Geometric Series

When does a geometric series converge, and to what?
Σ (n = 0 to ∞) arⁿ converges to a/(1 − r) when |r| is less than 1, where a is the first term. If |r| ≥ 1 (and a ≠ 0), it diverges.
What is the partial sum of a geometric series?
Sₙ = a(1 − rⁿ)/(1 − r) for r ≠ 1, found by subtracting r·Sₙ from Sₙ. For 40 + 36 + … (r = 0.9), S₁₀ ≈ 260.529.
Find Σ (n = 0 to ∞) 7(2/5)ⁿ.
a = 7 and r = 2/5, so the sum is 7/(1 − 2/5) = 7/(3/5) = 35/3.
Find Σ (n = 1 to ∞) (−1)ⁿ·5/2ⁿ.
The first term (n = 1) is −5/2 and r = −½, so the sum is (−5/2)/(3/2) = −5/3.
Find Σ (n = 2 to ∞) 8(−3/4)ⁿ.
Use the actual first term: 8(9/16) = 9/2. With r = −3/4, the sum is (9/2)/(1 + 3/4) = (9/2)/(7/4) = 18/7.
For which x does Σ ((x − 1)/3)ⁿ converge, and to what?
|(x − 1)/3| less than 1 gives −2 to 4, open at both ends (geometric endpoints diverge). The sum is 1/(1 − (x − 1)/3) = 3/(4 − x).
How do you write 0.4545… as a fraction with a geometric series?
a = 0.45 and r = 0.01, so the sum is 0.45/0.99 = 5/11.

10.3 The nth Term Test for Divergence

What is the nth term test for divergence?
If lim (n → ∞) aₙ ≠ 0 (or does not exist), then Σ aₙ diverges. It works because a convergent series has aₙ = Sₙ − Sₙ₋₁ → S − S = 0.
What does the nth term test tell you when lim aₙ = 0?
Nothing. The test never proves convergence: you must use another method. Example: Σ 1/n has terms → 0 but diverges.
Does Σ (4n + 1)/(9n − 2) converge?
No. aₙ → 4/9 ≠ 0, so the series diverges by the nth term test. It does not "converge to 4/9".
Does Σ cos(1/n) converge?
No. aₙ → cos 0 = 1 ≠ 0, so it diverges by the nth term test.
What does the nth term test say about Σ (−1)ⁿ n/(3n + 2)?
The terms swing between about +1/3 and −1/3, so lim aₙ does not exist. It is not 0, so the series diverges.
How should an nth term test justification be written?
"lim (n → ∞) aₙ = … ≠ 0, so by the nth term test, Σ aₙ diverges." Give the limit value and name the test. Note (1 + k/n)ⁿ → eᵏ, not 1.

10.4 Integral Test for Convergence

What are the conditions and conclusion of the integral test?
If f is positive, continuous and decreasing for x ≥ N with aₙ = f(n), then Σ (n = N to ∞) aₙ and ∫ (N to ∞) f(x) dx both converge or both diverge.
Does the integral test give the sum of the series?
No. Σ 3/(n + 2)² converges because ∫ (1 to ∞) 3/(x + 2)² dx = 1 is finite, but the sum is not 1.
Use the integral test on Σ 1/(2n + 5).
f(x) = 1/(2x + 5) is positive, continuous and decreasing. ∫ (1 to b) f dx = ½ ln((2b + 5)/7) → ∞ as b → ∞, so the series diverges.
Why can you not use the integral test on Σ (−1)ⁿ/n²?
The terms are not all positive. The test needs f positive (and continuous and decreasing).
Why does a finite integral make the series converge?
Width-1 rectangles of height aₙ drawn below the curve show a₂ + … + aₙ ≤ ∫ (1 to ∞) f, so the partial sums are bounded above and increasing.
What mistakes are common with the integral test?
Claiming the sum equals the integral; omitting the conditions or not showing why f is decreasing; writing an antiderivative with no limit as b → ∞; arguing "f → 0 so the integral converges".

10.5 Harmonic Series and p-Series

When does a p-series Σ 1/nᵖ converge?
It converges if p > 1 and diverges if p ≤ 1. The boundary case p = 1 diverges.
What is the harmonic series and why is it important?
Σ 1/n, the p-series with p = 1. It diverges even though 1/n → 0, so terms tending to 0 never prove convergence.
Does Σ n²/n^(7/2) converge?
Simplify first: the term is 1/n^(3/2), so p = 3/2 > 1 and it converges.
Does Σ 1/(n^0.5·n^0.4) converge?
No. The term is 1/n^0.9, so p = 0.9 ≤ 1 and it diverges.
Does Σ (1/n³ + 1/n) converge?
No. Convergent plus divergent is divergent. (Divergent plus divergent gives no conclusion.)
Does the alternating harmonic series 1 − 1/2 + 1/3 − … converge?
Yes, by the alternating series test, although Σ 1/n with the same term sizes diverges.
Is Σ 1/3ⁿ a p-series?
No. The variable is in the exponent, so it is geometric (r = 1/3). In a p-series, n is in the base and p is constant.

10.6 Comparison Tests for Convergence

What does the direct comparison test say?
For positive terms: if 0 ≤ aₙ ≤ bₙ eventually and Σ bₙ converges, Σ aₙ converges. If aₙ ≥ bₙ ≥ 0 and Σ bₙ diverges, Σ aₙ diverges. Smaller than divergent or bigger than convergent gives no conclusion.
What does the limit comparison test say?
If aₙ, bₙ > 0 and lim aₙ/bₙ = L with L positive and finite, then Σ aₙ and Σ bₙ both converge or both diverge.
Does Σ 1/(n³ + n) converge?
Yes. 0 ≤ aₙ ≤ 1/n³ for n ≥ 1, and Σ 1/n³ converges (p = 3), so it converges by direct comparison.
Does Σ (n + 1)/(n² + 1) converge?
No. Limit comparison with 1/n gives L = lim (n² + n)/(n² + 1) = 1, and Σ 1/n diverges.
Does Σ 1/(5ⁿ − 1) converge?
Yes. Direct comparison with (1/5)ⁿ fails because aₙ > (1/5)ⁿ, but limit comparison gives L = 1 and the geometric series with r = 1/5 converges.
How do you choose a benchmark series for comparison?
Guess from the dominant terms, usually a p-series or geometric series. Useful bounds: |sin n| ≤ 1, 0 ≤ cos²n ≤ 1, arctan n between 0 and π/2.
What mistakes are common with comparison tests?
A useless direction ("smaller than Σ 1/n, so it diverges"); no positivity check; using lim aₙ instead of lim aₙ/bₙ; accepting L = 0 or ∞; not saying why the benchmark converges.

10.7 Alternating Series Test for Convergence

What is the alternating series test?
For Σ (−1)ⁿaₙ or Σ (−1)ⁿ⁺¹aₙ with aₙ > 0: if aₙ₊₁ ≤ aₙ from some n on and lim aₙ = 0, the series converges.
Why does the alternating series test work?
Each partial sum overshoots the last by less than before, so the zigzag of partial sums closes onto a single limit.
Does Σ (−1)ⁿ/n^(1/4) converge?
Yes. 1/n^(1/4) is positive, decreasing and tends to 0, so it converges by the alternating series test.
Does Σ (−1)ⁿ(n + 1)/n converge?
No. aₙ → 1, so it diverges by the nth term test. The alternating series test can never show divergence.
Is Σ cos(nπ)/(n² + 1) alternating, and does it converge?
Yes and yes: cos(nπ) = (−1)ⁿ, and 1/(n² + 1) decreases to 0.
What mistakes are common with the alternating series test?
Checking only one condition; claiming "diverges by the alternating series test"; proving "decreasing" from a few terms instead of an inequality or derivative; missing an early rise (√n/(n + 5)); no conclusion sentence.

10.8 Ratio Test for Convergence

What is the ratio test?
Find L = lim (n → ∞) |aₙ₊₁/aₙ|. If L is less than 1, Σ aₙ converges absolutely. If L > 1 or L = ∞, it diverges. If L = 1, there is no conclusion.
Why is L = 1 inconclusive in the ratio test?
Σ 1/n diverges and Σ 1/n² converges, yet both give L = 1. Use a p-series or comparison test instead.
Use the ratio test on Σ 4ⁿ/n!.
|aₙ₊₁/aₙ| = 4ⁿ⁺¹n!/((n + 1)!4ⁿ) = 4/(n + 1) → 0, which is less than 1, so the series converges.
Use the ratio test on Σ 2ⁿ/n².
|aₙ₊₁/aₙ| = 2n²/(n + 1)² → 2 > 1, so the series diverges.
Which factorial simplifications are needed for the ratio test?
(n + 1)!/n! = n + 1 and (2n + 2)!/(2n)! = (2n + 2)(2n + 1). Note (n + 1)! = (n + 1)·n!, not n! + 1.
When is the ratio test the right choice?
When terms contain n!, (2n)! or powers like 3ⁿ. For terms with only powers of n, the ratio test gives L = 1, so use a p-series or comparison.
Use the ratio test on Σ n³/2ⁿ.
|aₙ₊₁/aₙ| = (n + 1)³/(2n³) → 1/2, which is less than 1, so the series converges.

10.9 Determining Absolute or Conditional Convergence

What is the difference between absolute and conditional convergence?
Absolutely convergent: Σ |aₙ| converges (then Σ aₙ converges too). Conditionally convergent: Σ aₙ converges but Σ |aₙ| diverges, relying on cancellation.
Classify Σ (−1)ⁿ⁺¹/nᵖ for different p.
p > 1: absolutely convergent. p > 0 and p ≤ 1: conditionally convergent (alternating series test). p ≤ 0: divergent, because the terms do not approach 0.
Classify Σ (−1)ⁿ⁺¹/n^(1/3).
Conditionally convergent: Σ 1/n^(1/3) diverges (p = 1/3 ≤ 1), but the alternating series test shows Σ aₙ converges.
Classify Σ (−1)ⁿ2ⁿ/n.
Divergent: the ratio of sizes 2n/(n + 1) → 2 > 1, so the terms do not approach 0.
If Σ aₙ converges, must Σ |aₙ| converge?
No. That reverses the theorem. The alternating harmonic series converges but Σ 1/n diverges.
Can you reorder the terms of a convergent series without changing its sum?
Yes if it is absolutely convergent. A conditionally convergent series can have its sum changed by reordering.
How do you handle a series with irregular signs, such as Σ sin n/n²?
The alternating series test does not apply. Test Σ |aₙ| with comparison: |sin n|/n² ≤ 1/n², which converges, so the series converges absolutely.

10.10 Alternating Series Error Bound

What is the alternating series error bound?
If Σ (−1)ⁿ⁺¹aₙ meets the alternating series test conditions, then |S − Sₙ| ≤ aₙ₊₁: the error is at most the size of the first term left out.
How do you tell whether an alternating partial sum is an over- or underestimate?
Look at the sign of the first omitted term. Positive: Sₙ is too small. Negative: Sₙ is too big. The sum S lies between Sₙ and Sₙ₊₁.
For Σ (−1)ⁿ⁺¹/n⁵, give a bound for |S − S₂|.
The first omitted term is a₃ = 1/3⁵ = 1/243 ≈ 0.00412.
How many terms of Σ (−1)ⁿ⁺¹/(4n) guarantee an error below 0.01?
Need 1/(4(n + 1)) less than 0.01, so n + 1 > 25. n + 1 = 26 works, so 25 terms.
An alternating series meeting the conditions has S₇ = 2.316 and 8th term −0.005. Give an interval for S.
2.311 ≤ S ≤ 2.316. The next term is negative, so S₇ is an overestimate.
What conditions must hold to use the alternating series error bound?
The series alternates, the term sizes decrease (at least from where you stop) and approach 0. Otherwise use another bound, such as Lagrange.
What mistakes are common with the alternating series error bound?
Using aₙ instead of aₙ₊₁; calling the bound the error ("at most"); no comparison with the tolerance; off-by-one counts when the series starts at n = 0; the wrong direction.

10.11 Finding Taylor Polynomial Approximations of Functions

What is the Taylor polynomial of degree n for f about x = a?
Pₙ(x) = Σ (k = 0 to n) f⁽ᵏ⁾(a)/k!·(x − a)ᵏ. It matches f and its first n derivatives at a. With a = 0 it is a Maclaurin polynomial.
Why does each Taylor coefficient divide by n!?
Differentiating (x − a)ⁿ n times gives n!, so cₙ = f⁽ⁿ⁾(a)/n! makes the nth derivative of Pₙ match f⁽ⁿ⁾(a).
Find the second-degree Maclaurin polynomial for e^(3x).
f(0) = 1, f′(0) = 3, f″(0) = 9, so P₂(x) = 1 + 3x + (9/2)x².
Find the second-degree Taylor polynomial for sin x about x = π/2.
f(π/2) = 1, f′(π/2) = cos(π/2) = 0, f″(π/2) = −1, so P₂(x) = 1 − ½(x − π/2)².
If f⁽⁴⁾(1) = 48, what is the coefficient of (x − 1)⁴?
48/4! = 48/24 = 2. Working backwards, f⁽ᵏ⁾(a) = k! × (coefficient of (x − a)ᵏ).
What are the low-degree Maclaurin polynomials for cos x and sin x?
cos x: P₄(x) = 1 − x²/2 + x⁴/24. sin x: P₃(x) = x − x³/6, which equals P₄ because the x⁴ coefficient is 0.
What mistakes are common with Taylor polynomials?
Leaving out the factorial; using xᵏ when the centre is a ≠ 0, or reading (x + 3) as centre 3; evaluating derivatives at the estimate point instead of the centre; trusting estimates far from a.

10.12 Lagrange Error Bound

What is the Lagrange error bound?
If |f⁽ⁿ⁺¹⁾(z)| ≤ M for all z between a and x, then |f(x) − Pₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!. It is the next Taylor term with the derivative replaced by its worst case.
How do you choose M in the Lagrange error bound?
M is the largest size of f⁽ⁿ⁺¹⁾ between a and x, or any bigger number. Check both ends of the interval. For sin and cos, M = 1; for eᶻ, round up (for example 3).
P₂ is used about x = a, |f‴| ≤ 4 between a and x, and x − a = 0.3. What is the Lagrange bound?
4(0.3)³/3! = 4(0.027)/6 = 0.018.
P₂(x) = 1 + x + x²/2 estimates e^0.2. Find a Lagrange bound and justify M.
f‴(z) = eᶻ ≤ e^0.2, which is less than 3 for z in [0, 0.2], so take M = 3. The bound is 3(0.2)³/6 = 0.004.
Is the Lagrange error bound the actual error?
No. It is a guarantee: the true value lies in [Pₙ(x) − bound, Pₙ(x) + bound], and the actual error may be smaller.
What mistakes are common with the Lagrange error bound?
Using f⁽ⁿ⁾ or n! instead of f⁽ⁿ⁺¹⁾ and (n + 1)!; taking M at the centre when the derivative is bigger at the other end; no comparison sentence; using the alternating bound when terms do not alternate.

10.13 Radius and Interval of Convergence of Power Series

What are the possible convergence sets of a power series Σ aₙ(x − c)ⁿ?
It converges only at c (R = 0), for all x (R = ∞), or for |x − c| less than R and diverges for |x − c| > R. It always converges at x = c.
How do you find the interval of convergence of a power series?
Use the ratio test to get |x − c| less than R, then test each endpoint c ± R separately with another test (the ratio test gives L = 1 there).
Find the interval of convergence of Σ (n = 1 to ∞) (x − 1)ⁿ/n.
The ratio test gives |x − 1| less than 1. At x = 2: Σ 1/n diverges. At x = 0: Σ (−1)ⁿ/n converges. The interval is [0, 2).
What is the radius of convergence of Σ (2x − 5)ⁿ/n²?
|2x − 5| less than 1 means |x − 5/2| less than 1/2, so R = 1/2. Both endpoints give convergent series (p = 2), so the interval is [2, 3].
f(x) = Σ (n = 0 to ∞) (x − 2)ⁿ/(n + 1) for |x − 2| less than 1. Find f‴(2).
f⁽ⁿ⁾(c) = n!·aₙ, and a₃ = 1/4, so f‴(2) = 3! × 1/4 = 3/2.
Does differentiating or integrating a power series change its interval of convergence?
The radius stays the same, but endpoints may change. Σ xⁿ/n² converges on [−1, 1]; its derivative converges on [−1, 1).
What mistakes are common with intervals of convergence?
Calling the interval length the radius (length is 2R); reading R from |2x − 5| without factoring; centre sign errors ((x + 1)ⁿ is centred at −1); skipping endpoints.

10.14 Finding Taylor or Maclaurin Series for a Function

What is the Taylor series of f about x = c?
Σ (n = 0 to ∞) f⁽ⁿ⁾(c)(x − c)ⁿ/n! = f(c) + f′(c)(x − c) + f″(c)(x − c)²/2! + …. With c = 0 it is the Maclaurin series.
What are the Maclaurin series for eˣ, sin x and cos x?
eˣ = Σ xⁿ/n! = 1 + x + x²/2! + …; sin x = x − x³/3! + x⁵/5! − … (odd powers); cos x = 1 − x²/2! + x⁴/4! − … (even powers). All converge for every x.
The Taylor series for f about x = 2 has (x − 2)⁴ coefficient 1/3. Find f⁽⁴⁾(2).
f⁽⁴⁾(2) = 4! × 1/3 = 8.
What exactly is 1 − 1/3! + 1/5! − 1/7! + …?
sin 1: the Maclaurin series for sin x at x = 1. Similarly Σ 3ⁿ/n! = e³.
What do the first coefficients of a Taylor series about c tell you about f near c?
The sign of a₁ = f′(c) gives increasing or decreasing; the sign of a₂ = f″(c)/2 gives concavity. a₁ = −3, a₂ = 2 means decreasing and concave up.
How do you get the series for e^(−2x) from eˣ?
Substitute the whole expression: e^(−2x) = Σ (−2x)ⁿ/n!, valid for all x.
What mistakes are common with Taylor series?
Leaving out n!; writing xⁿ instead of (x − c)ⁿ; swapping the sin and cos patterns; using 1/(1 − x) = Σ xⁿ when |x| ≥ 1; treating a coefficient as the derivative itself.

10.15 Representing Functions as Power Series

How do you find a power series for a function such as 2/(5 − x)?
Rewrite in the form a/(1 − r): 2/(5 − x) = (2/5)·1/(1 − x/5) = Σ 2xⁿ/5ⁿ⁺¹, valid for |x/5| less than 1, so −5 to 5 (geometric endpoints diverge).
Write the series for 1/(1 − x³) and its interval.
Substitute x³ for x in 1/(1 − x): 1 + x³ + x⁶ + … = Σ x^(3n), valid for |x³| less than 1, so x between −1 and 1.
Find the first three nonzero terms of the series for (sin x)/x.
Divide the series for sin x by x: 1 − x²/6 + x⁴/120.
How do you get the series for ln(4 + x)?
Integrate the series for 1/(4 + x) = Σ (−1)ⁿxⁿ/4ⁿ⁺¹ term by term and add the constant ln 4: ln(4 + x) = ln 4 + Σ (−1)ⁿxⁿ⁺¹/((n + 1)4ⁿ⁺¹), valid for x greater than −4 and up to 4 (x = 4 is included).
Find the coefficient of x⁵ in the series for xe^(2x).
It is the x⁴ coefficient of e^(2x): 2⁴/4! = 16/24 = 2/3.
Use x/(1 − x)² = Σ nxⁿ to find Σ (n = 1 to ∞) n/3ⁿ.
Put x = 1/3: (1/3)/(2/3)² = (1/3)/(4/9) = 3/4.
What can you do with a power series inside its interval of convergence?
Differentiate and integrate term by term. Differentiating 1/(1 − x) gives 1/(1 − x)² = Σ (n = 1 to ∞) nxⁿ⁻¹. The radius is kept; endpoints must be retested.
What mistakes are common when building series from known ones?
No 1 in front (rewrite 5/(3 + x) as (5/3)·1/(1 + x/3)); substituting outside the powers (sin 3x has (3x)³ = 27x³); keeping the old interval; missing the constant after integrating.

Study Unit 10 further

AP® and Advanced Placement® are trademarks registered by the College Board, which is not affiliated with, and does not endorse, this website. Marlbridge is an independent tuition provider. It is not an authorized AP course provider, an AP school or an exam centre, and these resources are not official College Board materials. Reading them or doing simulations does not meet the laboratory requirement of an AP science course.