Alternating Series Test for Convergence: Practice Questions (Calculus BC 10.7)

Answer: (C). The series is alternating with aₙ = 1/√(2n + 7) > 0. As n increases, 2n + 7 increases, so aₙ decreases. Also lim 1/√(2n + 7) = 0. Both conditions hold, so the series converges.

• (A) is not alternating: (−1)²ⁿ = 1 for every n, so this is Σ 1/n, the harmonic series, which diverges.
• (B) has aₙ = (n + 4)/n → 1, not 0. The alternating series test cannot be used, and the nth term test shows the series diverges.
• (D) has aₙ = (1.1)ⁿ/n, which grows without bound (exponential beats linear), so the terms do not tend to 0. The series diverges by the nth term test.

Answer: (B). With f(x) = x/(x² + 10), f′(x) = (10 − x²)/(x² + 10)², which is negative for x > √10 ≈ 3.16. So aₙ₊₁ ≤ aₙ for n ≥ 4. (In fact a₃ = 3/19 > a₄ = 2/13 as well, so the fall starts at n = 3.) Also lim n/(n² + 10) = 0. Decreasing from some point on is enough, so the series converges.

• (A) treats an early rise (a₁ < a₂ < a₃) as a failure. Only eventual behaviour matters, and a failed condition would not prove divergence anyway.
• (C) is wrong because the terms do tend to 0; the nth term test then gives no conclusion.
• (D) is wrong: the test is about the alternating series only. Σ n/(n² + 10) diverges by limit comparison with Σ 1/n (ratio limit 1).

Answer: (C). The terms do tend to 0, but they are not decreasing even eventually: for every even n, aₙ = 1/n³ is smaller than aₙ₊₁ = 1/(2n + 2). One condition fails, so the test gives no conclusion. (The series in fact diverges: the positive terms ½(1 + ⅓ + ⅕ + …) grow without bound, while the negative terms form a convergent p-series with p = 3.)

• (A) checks only the limit condition. Both conditions are needed.
• (B) claims the test proves divergence. It never does.
• (D) is false: 1/(2n) → 0 and 1/n³ → 0, so aₙ → 0.

Answer: (B). Here aₙ = nᵏ/(n² + 1) > 0. Dividing top and bottom by n² gives aₙ = n^(k − 2)/(1 + 1/n²), which tends to 0 exactly when k < 2. For decreasing, let f(x) = xᵏ/(x² + 1). Then f′(x) = x^(k − 1)[(k − 2)x² + k]/(x² + 1)². When k < 2, the bracket (k − 2)x² + k is negative for all large x, so f decreases from some point on. Both conditions hold for every k < 2.

• (A) k < 1 is the condition for Σ nᵏ/(n² + 1) without signs to converge (limit comparison with a p-series, p = 2 − k > 1). It leaves out 1 ≤ k < 2, such as k = 1.
• (C) includes k = 2, where aₙ = n²/(n² + 1) → 1. The series then diverges by the nth term test.
• (D) includes k ≥ 2, where the terms do not tend to 0.

1. Rewrite the sign. cos(nπ) = (−1)ⁿ, so the series is Σ (−1)ⁿ aₙ with aₙ = 6/(n + √n) > 0. It is alternating: −3 + 6/(2 + √2) − 6/(3 + √3) + 1 − … (about −3, 1.757, −1.268, 1, …).
2. Decreasing. Both n and √n increase with n, so (n + 1) + √(n + 1) > n + √n. A larger denominator with the same numerator 6 gives a smaller fraction: 6/((n + 1) + √(n + 1)) < 6/(n + √n). So aₙ₊₁ < aₙ for all n ≥ 1.
3. Limit. As n → ∞, n + √n → ∞, so lim aₙ = 0.
4. Conclude. aₙ is positive, decreasing and tends to 0, so the series converges by the alternating series test.

Suggested mark points (4): 1 for rewriting cos(nπ) as (−1)ⁿ and identifying aₙ; 1 for showing aₙ decreases with a reason; 1 for lim aₙ = 0; 1 for the conclusion that names the alternating series test.

Common error: writing “the terms get smaller” with no reason. Listing four terms does not prove it.

(a) lim (4n − 1)/(5n + 2) = lim (4 − 1/n)/(5 + 2/n) = 4/5. So the terms (−1)ⁿ⁺¹ (4n − 1)/(5n + 2) swing between values near 4/5 and −4/5 and do not tend to 0. By the nth term test, the series diverges.

(b) The alternating series test only ever concludes convergence. Here its limit condition fails (aₙ → 4/5, not 0), so it gives no conclusion at all. It cannot be used to show divergence.

(c) aₙ = n e⁻ⁿ = n/eⁿ > 0. Let f(x) = x e⁻ˣ. Then f′(x) = e⁻ˣ − x e⁻ˣ = (1 − x) e⁻ˣ, which is negative for x > 1, so f decreases on [1, ∞) and aₙ₊₁ ≤ aₙ for n ≥ 1. (Terms: about 0.368, 0.271, 0.149, 0.073.) The limit lim n/eⁿ has the form ∞/∞; by L’Hospital’s Rule it equals lim 1/eⁿ = 0. So the series converges by the alternating series test.

Total: 5 points. Acceptable alternative in (c): show aₙ₊₁/aₙ = (n + 1)/(n e) ≤ 2/e < 1 for n ≥ 1.

(a) Let f(x) = (ln x)²/x for x ≥ 2. By the quotient rule, f′(x) = [2 ln x · (1/x) · x − (ln x)²]/x² = ln x (2 − ln x)/x². For x > e² ≈ 7.39, ln x > 2, so 2 − ln x < 0 while ln x > 0 and x² > 0. So f′(x) < 0 and f decreases on [e², ∞). Since 8 > e², aₙ₊₁ ≤ aₙ for n ≥ 8.

(b) lim (ln x)²/x has the form ∞/∞. L’Hospital’s Rule: lim [2 ln x · (1/x)]/1 = lim 2 ln x / x, again ∞/∞. Once more: lim (2/x)/1 = 0. So lim aₙ = 0.

(c) Yes. For n ≥ 8, aₙ is positive, decreasing and tends to 0, so the series converges by the alternating series test.

(d) The test only needs the terms to decrease from some point on. The early terms (a₂ ≈ 0.240, a₃ ≈ 0.402, a₄ ≈ 0.480, rising to a₇ ≈ 0.541) do not affect convergence, because removing finitely many terms does not change whether a series converges.

(e) No. For n ≥ 3, ln n > 1, so (ln n)² > 1 and (ln n)²/n > 1/n > 0. Σ 1/n diverges (harmonic series), so Σ (ln n)²/n diverges by the direct comparison test.

Total: 5 points. Acceptable alternative for (e): the integral test, since an antiderivative of (ln x)²/x is (ln x)³/3, which grows without bound, so ∫ from 2 to ∞ of (ln x)²/x dx diverges.