What this is for. This diagnostic shows which topics of Unit 10, Infinite Sequences and Series, to revisit. The unit is BC only: it is not part of Calculus AB. There is one question per topic, and two for Topic 10.13. These are original Marlbridge practice questions, not past exam questions. They are not calibrated, and your result is not a predicted score.
Rules. No calculator; about 30 minutes. Answer everything before opening any answer, and in short answers name each test and check its conditions.
Question 1 (multiple choice · 10.1)
The series Σ aₙ (n ≥ 1) has nth partial sum Sₙ = 2 − 3/(n + 1). Find a₃ and describe the series.
- (A) a₃ = 5/4; the series converges to 2.
- (B) a₃ = 1/4; the series converges to 0.
- (C) a₃ = 1/4; the series converges to 2.
- (D) a₃ = 1/4; the series diverges because Sₙ increases.
Answer and explanation
Answer: (C). a₃ = S₃ − S₂ = 5/4 − 1 = 1/4, and lim Sₙ = 2.
- (A) gives S₃, not the third term.
- (B) uses lim aₙ = 0, not lim Sₙ.
- (D) Increasing partial sums can still have a limit.
If you missed this: Topic 10.1 study guide.
Question 2 (multiple choice · 10.2)
What is Σ from n = 2 to ∞ of 3(−2/5)ⁿ?
- (A) 15/7
- (B) −6/7
- (C) 4/5
- (D) 12/35
Answer and explanation
Answer: (D). r = −2/5 and the first term (n = 2) is 3(4/25) = 12/25. Sum = (12/25)/(1 + 2/5) = 12/35.
- (A) starts at n = 0.
- (B) starts at n = 1.
- (C) uses r = +2/5 in the denominator.
If you missed this: Topic 10.2 study guide.
Question 3 (multiple choice · 10.3)
For which series does the nth term test prove divergence?
- (A) Σ n/(n² + 3)
- (B) Σ arctan n
- (C) Σ (−1)ⁿ/√n
- (D) Σ 1/ln(n + 1)
Answer and explanation
Answer: (B). arctan n → π/2 ≠ 0, so Σ arctan n diverges.
- (A) and (D) diverge, but their terms tend to 0, so this test is silent.
- (C) converges; its terms also tend to 0.
If you missed this: Topic 10.3 study guide.
Question 4 (multiple choice · 10.4)
For x ≥ 2, f(x) = 1/(x(ln x)²) is positive, continuous and decreasing, and ∫ from 2 to ∞ of f(x) dx = 1/ln 2. What is true of Σ from n = 2 to ∞ of f(n)?
- (A) The series converges.
- (B) The series converges, and its sum is 1/ln 2.
- (C) The series diverges, because the integral is not 0.
- (D) The integral test cannot be used, because ln x is increasing.
Answer and explanation
Answer: (A). The conditions hold and the integral converges, so the series converges.
- (B) The test does not give the sum; the first three terms already exceed 1/ln 2.
- (C) The integral only has to converge, not be 0.
- (D) Only f has to decrease, and it does.
If you missed this: Topic 10.4 study guide.
Question 5 (multiple choice · 10.5)
Which series converges?
- (A) Σ 1/n^0.9
- (B) Σ √n/n
- (C) Σ ∛n/n^(3/2)
- (D) Σ 1/∛(n²)
Answer and explanation
Answer: (C). ∛n/n^(3/2) = 1/n^(3/2 − 1/3) = 1/n^(7/6), a p-series with p = 7/6 > 1.
- (A) p = 0.9 ≤ 1.
- (B) √n/n = 1/n^(1/2), so p = 1/2.
- (D) 1/n^(2/3), so p = 2/3.
If you missed this: Topic 10.5 study guide.
Question 6 (multiple choice · 10.6)
For the limit comparison test on Σ (n + 1)/√(n⁵ + 2), which is correct?
- (A) bₙ = 1/n; limit 0; the series diverges.
- (B) bₙ = 1/n^(3/2); limit 1; the series diverges.
- (C) bₙ = 1/n^(5/2); limit ∞; the series converges.
- (D) bₙ = 1/n^(3/2); limit 1; the series converges.
Answer and explanation
Answer: (D). The terms behave like n/n^(5/2) = 1/n^(3/2). lim aₙ/bₙ = 1, finite and positive, and Σ 1/n^(3/2) converges.
- (A) A limit of 0 with a divergent benchmark decides nothing.
- (B) Σ 1/n^(3/2) converges, so the series does too.
- (C) A limit of ∞ with a convergent benchmark decides nothing.
If you missed this: Topic 10.6 study guide.
Question 7 (multiple choice · 10.7)
Which statement about Σ from n = 2 to ∞ of (−1)ⁿ/ln n is true?
- (A) It diverges, because Σ 1/ln n diverges.
- (B) It converges by the alternating series test.
- (C) The alternating series test cannot be used, because 1/ln n decreases too slowly.
- (D) It converges absolutely, by comparison with Σ 1/n.
Answer and explanation
Answer: (B). 1/ln n is positive, decreasing and tends to 0, and the signs alternate.
- (A) Σ |aₙ| diverging does not make the series diverge.
- (C) The test has no speed condition.
- (D) 1/ln n > 1/n, so Σ |aₙ| diverges by direct comparison.
If you missed this: Topic 10.7 study guide.
Question 8 (multiple choice · 10.8)
The ratio test is applied to Σ 3ⁿ n!/(n + 1)ⁿ. Which statement is correct?
- (A) L = 3/e, so the series diverges.
- (B) L = 3/e, so the series converges.
- (C) L = 3, so the series diverges.
- (D) L = 1, so the test gives no conclusion.
Answer and explanation
Answer: (A). aₙ₊₁/aₙ = 3(n + 1)(n + 1)ⁿ/(n + 2)ⁿ⁺¹ = 3((n + 1)/(n + 2))ⁿ⁺¹. With m = n + 1, this is 3/(1 + 1/m)ᵐ → 3/e. As e < 3, L > 1.
- (B) treats 3/e as less than 1.
- (C) loses the factor ((n + 1)/(n + 2))ⁿ⁺¹.
- (D) treats that 1^∞ form as 1.
If you missed this: Topic 10.8 study guide.
Question 9 (short answer · 10.9)
Classify Σ from n = 1 to ∞ of (−1)ⁿ⁺¹ n/(n² + 9) as absolutely convergent, conditionally convergent or divergent. Justify.
Worked answer
Absolute values. lim [n/(n² + 9)] ÷ (1/n) = lim n²/(n² + 9) = 1, and Σ 1/n diverges, so by limit comparison Σ |aₙ| diverges.
The series itself. aₙ = n/(n² + 9) > 0 and aₙ → 0. With f(x) = x/(x² + 9), f′(x) = (9 − x²)/(x² + 9)² < 0 for x > 3, so the terms decrease for n ≥ 3. By the alternating series test, the series converges.
So the series is conditionally convergent.
If you missed this: Topic 10.9 study guide.
Question 10 (multiple choice · 10.10)
S = 1 − 1/2 + 1/6 − 1/24 + … = Σ (−1)ⁿ⁺¹/n! (n ≥ 1), and S₄ = 0.625. Which interval for S does the alternating series error bound give?
- (A) 0.625 − 1/120 < S < 0.625
- (B) 0.625 − 1/24 < S < 0.625
- (C) 0.625 < S < 0.625 + 1/120
- (D) 0.625 < S < 0.625 + 1/24
Answer and explanation
Answer: (C). The test conditions hold, so |S − S₄| < a₅ = 1/120. The first omitted term, +1/120, is positive, so S₄ is an underestimate.
- (A) gets the direction wrong.
- (B) and (D) use the last term kept, not the first term left out. (D) is true, but it is not the bound.
If you missed this: Topic 10.10 study guide.
Question 11 (multiple choice · 10.11)
What is the second-degree Taylor polynomial for f(x) = x ln x about x = 1?
- (A) (x − 1) + (x − 1)²
- (B) (x − 1) − ½(x − 1)²
- (C) x + ½x²
- (D) (x − 1) + ½(x − 1)²
Answer and explanation
Answer: (D). f(1) = 0, f′(x) = ln x + 1 gives f′(1) = 1, and f″(x) = 1/x gives f″(1) = 1; divide by 2!.
- (A) forgets to divide by 2!.
- (B) has a sign error in f″.
- (C) is centred at 0, not 1.
If you missed this: Topic 10.11 study guide.
Question 12 (short answer · 10.12)
Let f(x) = e^(−x).
(a) Find P₃(x), the third-degree Maclaurin polynomial for f, and P₃(0.3). (b) Use the Lagrange error bound to bound |f(0.3) − P₃(0.3)|. (c) Hence show that e^(−0.3) > 0.74.
Worked answer
(a) The derivatives at 0 are 1, −1, 1, −1, so P₃(x) = 1 − x + x²/2 − x³/6, and P₃(0.3) = 1 − 0.3 + 0.045 − 0.0045 = 0.7405.
(b) f⁽⁴⁾(z) = e^(−z) ≤ 1 for 0 ≤ z ≤ 0.3, so take M = 1. The bound is 1 × 0.3⁴/4! = 0.0081/24 = 0.0003375.
(c) f(0.3) ≥ 0.7405 − 0.0003375 = 0.7401625 > 0.74.
If you missed this: Topic 10.12 study guide.
Question 13 (multiple choice · 10.13)
Find the interval of convergence of Σ (x + 2)ⁿ/(n · 3ⁿ), n ≥ 1.
- (A) (−5, 1]
- (B) [−5, 1)
- (C) [−1, 5)
- (D) (−5, 1)
Answer and explanation
Answer: (B). The ratio test gives |x + 2|/3 < 1: centre −2, R = 3. At x = 1 the series is Σ 1/n (diverges); at x = −5 it is Σ (−1)ⁿ/n (converges).
- (A) swaps the endpoint results.
- (C) uses centre +2.
- (D) never tests the endpoints.
If you missed this: Topic 10.13 study guide.
Question 14 (multiple choice · 10.13)
f(x) = Σ (x − 1)ⁿ/(n · 2ⁿ) (n ≥ 1) converges on [−1, 3). On what interval does the term-by-term series for f′(x) converge?
- (A) (−1, 3)
- (B) [−1, 3)
- (C) [−1, 3]
- (D) (−2, 2)
Answer and explanation
Answer: (A). f′(x) = Σ (x − 1)ⁿ⁻¹/2ⁿ. R stays 2. At x = 3 and x = −1 the terms are ±1/2, so both endpoints fail the nth term test.
- (B) assumes the endpoints never change.
- (C) gains an endpoint.
- (D) centres the interval at 0.
If you missed this: Topic 10.13 study guide.
Question 15 (multiple choice · 10.14)
What is Σ from n = 1 to ∞ of 2ⁿ/n!?
- (A) e²
- (B) e² − 1
- (C) e² − 2
- (D) 2e
Answer and explanation
Answer: (B). From the series for eˣ, Σ from n = 0 of 2ⁿ/n! = e². The given sum leaves out the n = 0 term, 1.
- (A) keeps the n = 0 term.
- (C) also removes the n = 1 term.
- (D) multiplies instead of using the series for eˣ.
If you missed this: Topic 10.14 study guide.
Question 16 (short answer · 10.15)
Let f(x) = x e^(−x²).
(a) Write the first four nonzero terms and the general term of the Maclaurin series for f. (b) Write ∫ from 0 to 1 of f(x) dx as a series. Use three terms to estimate it, and bound the error.
Worked answer
(a) Put u = −x² into eᵘ, then multiply by x: f(x) = x − x³ + x⁵/2 − x⁷/6 + … + (−1)ⁿx²ⁿ⁺¹/n! + …, for all x.
(b) ∫ from 0 to 1 = Σ (−1)ⁿ/(n!(2n + 2)) = 1/2 − 1/4 + 1/12 − 1/48 + … . Three terms give 1/3. The terms alternate, decrease and tend to 0, so the error is less than 1/48.
If you missed this: Topic 10.15 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 10.1 Convergence | 1 | Guide |
| 10.2 Geometric series | 2 | Guide |
| 10.3 nth term test | 3 | Guide |
| 10.4 Integral test | 4 | Guide |
| 10.5 p-series | 5 | Guide |
| 10.6 Comparison tests | 6 | Guide |
| 10.7 Alternating test | 7 | Guide |
| 10.8 Ratio test | 8 | Guide |
| 10.9 Absolute or conditional | 9 | Guide |
| 10.10 Alternating error bound | 10 | Guide |
| 10.11 Taylor polynomials | 11 | Guide |
| 10.12 Lagrange error bound | 12 | Guide |
| 10.13 Interval of convergence | 13, 14 | Guide |
| 10.14 Taylor series | 15 | Guide |
| 10.15 Building power series | 16 | Guide |
How to use your result
- Mark each topic secure, shaky (a slip) or gap (wrong).
- Fix gaps in Topics 10.1, 10.2 and 10.11 first: later topics build on them.
- In Questions 9, 12 and 16, check that you named each test and showed its conditions.
- For a gap, read the guide, then do the topic’s practice set.
- Then try the Unit 10 mixed review.