Ratio Test for Convergence: Practice Questions (Calculus BC 10.8)

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; all sums start at n = 1; “Σ from n = 1 to ∞ of aₙ” means a₁ + a₂ + a₃ + …; L always means lim (n → ∞) |aₙ₊₁/aₙ|.

Question 1 (multiple choice · foundation)

The ratio test is applied to Σ from n = 1 to ∞ of 7ⁿ/(n · 5ⁿ). Which statement is correct?

(A) L = 5/7, so the series converges.
(B) L = 7/5, so the series diverges.
(C) L = 1, so the test gives no conclusion.
(D) L = 7/5, so the series converges.

Answer and explanation

Answer: (B). aₙ₊₁/aₙ = [7ⁿ⁺¹/((n + 1) 5ⁿ⁺¹)] × [n · 5ⁿ/7ⁿ] = (7/5) · n/(n + 1). As n → ∞, n/(n + 1) → 1, so L = 7/5 > 1 and the series diverges.

(A) uses the upside-down ratio aₙ/aₙ₊₁, which gives 1/L and the opposite conclusion.
(C) looks only at the n/(n + 1) part and forgets the factor 7/5 from the powers.
(D) has the right L but reverses the rule: L > 1 means divergence.

Question 2 (multiple choice · core)

For which series does the ratio test give no conclusion?

(A) Σ 2ⁿ/n!
(B) Σ n/(n³ + 1)
(C) Σ n(0.9)ⁿ
(D) Σ 3ⁿ/n²

Answer and explanation

Answer: (B). The terms contain only powers of n, so the ratio [(n + 1)/((n + 1)³ + 1)] × [(n³ + 1)/n] tends to 1. The test is inconclusive. (The series converges by limit comparison with Σ 1/n², since the ratio of terms tends to 1.)

(A) ratio 2/(n + 1) → 0 < 1: converges.
(C) ratio 0.9(n + 1)/n → 0.9 < 1: converges.
(D) ratio 3n²/(n + 1)² → 3 > 1: diverges.

Question 3 (multiple choice · core)

Consider Σ from n = 1 to ∞ of (n + 2)!/(n! · 4ⁿ). Which statement is correct?

(A) L = 1/4, so the series converges.
(B) L = 4, so the series diverges.
(C) L = 1, so the test gives no conclusion.
(D) L = 1/4, but this only shows that the terms tend to 0; it says nothing about the series.

Answer and explanation

Answer: (A). First simplify: (n + 2)!/n! = (n + 2)(n + 1), so aₙ = (n + 1)(n + 2)/4ⁿ. Then aₙ₊₁/aₙ = [(n + 2)(n + 3)/4ⁿ⁺¹] × [4ⁿ/((n + 1)(n + 2))] = (n + 3)/[4(n + 1)] → 1/4. L < 1, so the series converges.

(B) flips the ratio (or the power of 4) and gets 1/L.
(C) cancels the factorials correctly but forgets the 4ⁿ in the denominator.
(D) confuses the ratio test with the nth term test. L < 1 proves that the series converges.

Question 4 (multiple choice · stretch)

Let k be a positive constant. For which values of k does the ratio test show that Σ from n = 1 to ∞ of n² kⁿ/3ⁿ converges?

(A) 0 < k < 3
(B) 0 < k ≤ 3
(C) 0 < k < 1/3
(D) k > 3

Answer and explanation

Answer: (A). aₙ₊₁/aₙ = (k/3) · (n + 1)²/n² → k/3. The test shows convergence when k/3 < 1, that is 0 < k < 3.

(B) includes k = 3, where L = 1 and the ratio test gives no conclusion. (At k = 3 the terms are n², which do not tend to 0, so the series diverges by the nth term test.)
(C) comes from writing the condition as 3k < 1 instead of k/3 < 1, that is, putting the 3 on the wrong side.
(D) is where L > 1, so the series diverges.

Question 5 (calculation · core)

Use the ratio test to determine whether Σ from n = 1 to ∞ of (−2)ⁿ (n + 1)/3ⁿ⁺¹ converges or diverges.

Worked solution

Terms: aₙ = (−2)ⁿ (n + 1)/3ⁿ⁺¹. The first terms are −4/9, 4/9, −32/81, 80/243, so the signs alternate. Keep the absolute value bars.
Ratio: |aₙ₊₁/aₙ| = |(−2)ⁿ⁺¹ (n + 2)/3ⁿ⁺²| × |3ⁿ⁺¹/((−2)ⁿ (n + 1))| = (2/3) · (n + 2)/(n + 1).
Limit: (n + 2)/(n + 1) → 1, so L = 2/3.
Conclude: L = 2/3 < 1, so the series converges by the ratio test (and Σ |aₙ| converges as well).

Suggested mark points (4): 1 for a correct ratio with the absolute value; 1 for simplifying the powers of −2 and 3 correctly; 1 for L = 2/3; 1 for the conclusion comparing L with 1 and naming the ratio test.

Common error: leaving out the bars and writing the ratio as −(2/3)(n + 2)/(n + 1), then concluding from “L = −2/3”.

Question 6 (constructed response · core)

(a) Use the ratio test to show that Σ from n = 1 to ∞ of n⁵/(1.2)ⁿ converges.

(b) Show that the ratio test gives no conclusion for Σ from n = 1 to ∞ of 1/(4n + 1).

Worked solution and suggested Marlbridge rubric

(a) aₙ₊₁/aₙ = [(n + 1)⁵/(1.2)ⁿ⁺¹] × [(1.2)ⁿ/n⁵] = (1/1.2) · ((n + 1)/n)⁵ → 1/1.2 = 5/6. Since 5/6 < 1, the series converges by the ratio test.

(b) aₙ₊₁/aₙ = (4n + 1)/(4n + 5) → 1. L = 1, so the ratio test gives no conclusion.

(c) Limit comparison with the harmonic series Σ 1/n: lim [1/(4n + 1)]/(1/n) = lim n/(4n + 1) = 1/4, a finite positive number. Σ 1/n diverges, so Σ 1/(4n + 1) diverges by the limit comparison test.

Point	What earns it
1	(a) Correct ratio, with the factor 1/1.2 separated from ((n + 1)/n)⁵
1	(a) L = 5/6 and the conclusion “converges by the ratio test”
1	(b) Ratio (4n + 1)/(4n + 5) → 1, so no conclusion
1	(c) Correct limit comparison with Σ 1/n, limit 1/4 (finite and positive)
1	(c) Diverges, naming the test and the harmonic series

Total: 5 points. Acceptable alternatives for (c): direct comparison, since 1/(4n + 1) ≥ 1/(5n) for n ≥ 1 and Σ 1/(5n) diverges; or the integral test.

Question 7 (constructed response · stretch)

A sequence of positive numbers is defined by a₁ = 2 and aₙ₊₁ = [(3n + 1)/(5n + 2)] · aₙ for n ≥ 1.

(a) Find a₂ and a₃. (b) Does Σ from n = 1 to ∞ of aₙ converge? Justify your answer. (c) A second sequence has b₁ = 1 and bₙ₊₁ = [(2n + 5)/(n + 4)] · bₙ. Does Σ bₙ converge? Justify your answer. (d) Two more sequences start at c₁ = d₁ = 1, with cₙ₊₁ = [n/(n + 1)] · cₙ and dₙ₊₁ = [n/(n + 1)]² · dₙ. Show that the ratio test gives no conclusion for either series, then find formulas for cₙ and dₙ and decide whether each series converges.

Worked solution and suggested Marlbridge rubric

(a) a₂ = (4/7) · 2 = 8/7. a₃ = (7/12) · (8/7) = 2/3.

(b) The recursive rule gives the ratio directly: aₙ₊₁/aₙ = (3n + 1)/(5n + 2) → 3/5. Since 3/5 < 1, Σ aₙ converges by the ratio test.

(c) bₙ₊₁/bₙ = (2n + 5)/(n + 4) → 2. Since 2 > 1, Σ bₙ diverges by the ratio test. (Each ratio is already above 1, so the terms 1, 7/5, 21/10, 33/10, … keep growing.)

(d) Both ratios, n/(n + 1) and [n/(n + 1)]², tend to 1, so the ratio test gives no conclusion for either series.

cₙ: c₂ = 1/2, c₃ = (2/3)(1/2) = 1/3, … The product telescopes: cₙ = (1/2)(2/3)(3/4)…((n − 1)/n) = 1/n. Σ 1/n is the harmonic series, so Σ cₙ diverges.
dₙ: squaring each factor gives dₙ = 1/n². This is a p-series with p = 2 > 1, so Σ dₙ converges.

So two series with L = 1 can behave in opposite ways, which is why L = 1 gives no conclusion.

Point	What earns it
1	(a) a₂ = 8/7 and a₃ = 2/3
1	(b) Ratio taken from the rule, L = 3/5, converges by the ratio test
1	(c) L = 2 > 1, diverges by the ratio test
1	(d) Both ratios → 1, so the ratio test is inconclusive
1	(d) cₙ = 1/n diverges and dₙ = 1/n² converges, with the tests named

Total: 5 points. In (d), formulas found from a pattern in the first few terms are accepted if they are checked against the rule.

How did you do?

Q1 or Q3 wrong: check which way up the ratio goes and how powers and factorials cancel; see “Algebra you will need” and Worked example 2 in the study guide.
Q2, Q6 or Q7(d) wrong: reread “When the ratio test fails: L = 1”.
Q4 wrong: revisit Worked example 1 and the three cases of the test.
Q5 wrong: look again at why the absolute value bars are there.

Then tick off the topic checklist.