These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. BC-only material. Assumptions: no calculator; angles in radians; every series starts at n = 1 unless stated. Notation: “∑ aₙ” means the series ∑ from n = 1 to ∞ of aₙ, and Sₙ = a₁ + a₂ + … + aₙ is its nth partial sum.
Question 1 (multiple choice · foundation)
For which series does the nth term test prove divergence?
- (A) ∑ (6n + 1)/(3n − 1)
- (B) ∑ 4/(n² + 1)
- (C) ∑ (0.6)ⁿ
- (D) ∑ 5/√n
Answer and explanation
Answer: (A). The highest powers match, so (6n + 1)/(3n − 1) → 6/3 = 2. Since 2 ≠ 0, the series diverges by the nth term test.
- (B) 4/(n² + 1) → 0, so the test gives no conclusion. (The series actually converges, but this test cannot show that.)
- (C) (0.6)ⁿ → 0, so the test gives no conclusion. It is a convergent geometric series with |r| = 0.6 < 1.
- (D) 5/√n → 0, so the test gives no conclusion. (This series diverges, but you need a later test, from Topic 10.5, to show it.)
Question 2 (multiple choice · foundation)
A series ∑ aₙ has terms with lim (n → ∞) of aₙ = 0. What can you conclude from the nth term test alone?
- (A) The series converges.
- (B) The series diverges.
- (C) The series converges, and its sum is 0.
- (D) Nothing: the test gives no conclusion.
Answer and explanation
Answer: (D). The nth term test can only show divergence, and only when the terms do not approach 0. When aₙ → 0, the series might converge (for example ∑ (0.5)ⁿ) or diverge (for example the block series 1 + ½ + ½ + ⅓ + ⅓ + ⅓ + … in Worked example 3 of the study guide).
- (A) treats “terms approach 0” as enough for convergence. It is necessary, not sufficient.
- (B) reverses the test: divergence follows when the limit is not 0.
- (C) confuses the limit of the terms with the sum. The sum is the limit of the partial sums; ∑ (0.5)ⁿ, for instance, has sum 1.
Question 3 (multiple choice · core)
Consider ∑ n sin(3/n). Which statement is correct?
- (A) lim aₙ = 0, so the nth term test is inconclusive.
- (B) lim aₙ = 3, so the series diverges by the nth term test.
- (C) lim aₙ = 1, so the series diverges by the nth term test.
- (D) lim aₙ = 3, so the series converges to 3.
Answer and explanation
Answer: (B). Write aₙ = sin(3/n)/(1/n), a 0/0 form. With a continuous x and L’Hospital’s Rule: [cos(3/x) · (−3/x²)] / (−1/x²) = 3 cos(3/x) → 3. So lim aₙ = 3 ≠ 0 and the series diverges.
- (A) uses sin(3/n) → 0 and forgets the factor n, which grows. The product is an ∞ · 0 form, not 0.
- (C) uses the limit sin u / u → 1 but forgets the chain-rule factor 3 (u = 3/n, so n sin(3/n) = 3 · sin u / u).
- (D) finds the right limit but confuses the limit of the terms with the sum. Terms approaching 3 make the series diverge.
Question 4 (multiple choice · stretch)
A series ∑ aₙ is known to converge, and aₙ ≠ −2 for every n. What is lim (n → ∞) of (6aₙ + 1)/(aₙ + 2)?
- (A) 1/2
- (B) 6
- (C) 0
- (D) It cannot be found without knowing the sum of the series.
Answer and explanation
Answer: (A). If a series converges, its terms approach 0 (this is the fact behind the nth term test). So aₙ → 0, and the expression is continuous at 0: (6 · 0 + 1)/(0 + 2) = 1/2.
- (B) treats aₙ as if it grows without bound and takes the ratio of the coefficients, 6/1. For a convergent series the terms approach 0.
- (C) assumes every part of the expression approaches 0, forgetting the constants 1 and 2.
- (D) confuses the limit of the terms with the sum. Only lim aₙ = 0 is needed, and that follows from convergence alone.
Question 5 (calculation · core)
For each series, find lim aₙ and state what the nth term test shows.
(a) ∑ (2n³ − n)/(5n³ + 4n²) (b) ∑ (−1)ⁿ n²/(n² + 1) (c) ∑ 6n/(n² + 5) (d) ∑ [ln(5n + 1) − ln n]
Worked solution
(a) Divide top and bottom by n³: (2 − 1/n²)/(5 + 4/n) → 2/5 ≠ 0. Diverges by the nth term test.
(b) The size n²/(n² + 1) → 1, and the sign alternates. Even terms → +1 and odd terms → −1, so lim aₙ does not exist. It is not 0, so the series diverges by the nth term test.
(c) The bottom has the higher power: 6n/(n² + 5) = (6/n)/(1 + 5/n²) → 0. The nth term test is inconclusive.
(d) Combine: ln(5n + 1) − ln n = ln((5n + 1)/n) = ln(5 + 1/n) → ln 5 ≠ 0 (about 1.609). Diverges by the nth term test.
Suggested mark points (4): 1 for each part with the correct limit and the correct conclusion. A part with “converges” in (c) earns 0 for that part.
Common error in (d): treating ln(5n + 1) − ln n as “∞ − ∞ = 0”. Combine the logs first.
Question 6 (constructed response · core)
For a constant c, let aₙ = (c n² + 4n)/(5n² + 1).
(a) Let c = 3. Show that ∑ aₙ diverges, using the nth term test. (b) Find the value of c for which the nth term test gives no conclusion about ∑ aₙ. Show your reasoning. (c) For the value of c in (b), a student writes: “The terms approach 0, so by the nth term test ∑ aₙ converges.” Explain the error. (d) Let c = −5. What does the nth term test show? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) Divide top and bottom by n²: (3 + 4/n)/(5 + 1/n²) → 3/5. Since 3/5 ≠ 0, ∑ aₙ diverges by the nth term test.
(b) In general, (c + 4/n)/(5 + 1/n²) → c/5. The test gives no conclusion only when this limit is 0, so c = 0. Then aₙ = 4n/(5n² + 1) → 0.
(c) The nth term test can only prove divergence, and only when lim aₙ ≠ 0. When lim aₙ = 0 it gives no conclusion, so it cannot show that the series converges. (Terms approaching 0 are needed for convergence but are not enough on their own.)
(d) aₙ → −5/5 = −1. Since −1 ≠ 0, ∑ aₙ diverges by the nth term test. A negative limit counts: the test asks only whether the limit is 0.
| Point | What earns it |
|---|---|
| 1 | Limit 3/5 for c = 3, using the highest powers |
| 1 | Conclusion “diverges”, with the limit compared with 0 and the test named |
| 1 | General limit c/5 (or equivalent reasoning) leading to c = 0 |
| 1 | States that the nth term test is inconclusive when the limit is 0, so it cannot prove convergence |
| 1 | Limit −1 and conclusion “diverges” for c = −5 |
Total: 5 points. An answer of “diverges” with no limit stated earns at most 1 of the 2 points in part (a), and no point in part (d).
Question 7 (constructed response · stretch)
Two fictional savings plans add money to an account at the end of each month. In month n, Plan A adds dₙ = 50n/(n + 4) dollars and Plan B adds bₙ = 80(0.75)ⁿ⁻¹ dollars. Assume each plan runs forever.
(a) Find lim dₙ. Use the nth term test to decide whether the total ∑ dₙ is finite. Interpret your answer in context. (b) Find lim bₙ. What does the nth term test tell you about ∑ bₙ? (c) Decide whether ∑ bₙ converges. If it does, find the total amount Plan B ever adds. (d) A student says: “Plan A’s monthly amounts level off at 50 dollars, so its total must level off too.” Explain why this is wrong.
Worked solution and suggested Marlbridge rubric
(a) dₙ = 50n/(n + 4) = 50/(1 + 4/n) → 50. Since 50 ≠ 0, ∑ dₙ diverges by the nth term test. In context: Plan A adds nearly 50 dollars every month in the long run (d₁₀ = 250/7 ≈ 35.71, d₁₀₀ = 625/13 ≈ 48.08), so the total grows without bound.
(b) bₙ = 80(0.75)ⁿ⁻¹ → 0, because 0 < 0.75 < 1. The nth term test is inconclusive for ∑ bₙ.
(c) ∑ bₙ is geometric with first term 80 and ratio r = 0.75 (the terms are 80, 60, 45, …). Since |r| < 1 it converges, and the total is 80/(1 − 0.75) = 320 dollars.
(d) Levelling off of the monthly amounts (the terms) is not the same as levelling off of the total (the partial sums). Each month adds close to 50 dollars, so after N more months the total has risen by roughly 50N dollars. For the total to have a finite limit, the monthly amounts would have to approach 0.
| Point | What earns it |
|---|---|
| 1 | lim dₙ = 50 with a correct reason |
| 1 | Concludes ∑ dₙ diverges by the nth term test, with an interpretation (total grows without bound) |
| 1 | lim bₙ = 0 and states the nth term test is inconclusive |
| 1 | Identifies a geometric series with a = 80 and r = 0.75 (between −1 and 1), and finds the total 320 dollars |
| 1 | Explains the difference between terms and partial sums in (d) |
Total: 5 points. Acceptable alternative for (c): Sₙ = 320(1 − 0.75ⁿ) → 320. An answer to (c) of “converges because bₙ → 0” earns no point, even with 320 dollars, because the reason given is not valid.
How did you do?
- Q1, Q3 or Q5 wrong: practise finding limits; see Worked examples 1 and 2 in the study guide.
- Q2 or Q6(c) wrong: reread “Why the terms of a convergent series must go to 0” and Worked example 3.
- Q4 wrong: revisit “Using the fact forwards”.
- Q6(a), (b) or (d) wrong: practise dominant terms again (Worked example 1) and check you compare the limit with 0, not with 1.
- Q7 wrong: compare terms and totals again, and review geometric series from Topic 10.2.
Then tick off the topic checklist.