Determining Absolute or Conditional Convergence: Practice Questions (Calculus BC 10.9)

Answer: (B). Absolute convergence implies convergence. The proof uses 0 ≤ aₙ + |aₙ| ≤ 2|aₙ| and the comparison test.

• (A) is the converse, and it is false: Σ (−1)ⁿ⁺¹/n converges but Σ 1/n diverges.
• (C) is false for the same series: Σ |aₙ| = Σ 1/n diverges, yet Σ aₙ converges (conditionally).
• (D) is false: terms approaching 0 is necessary for convergence but not enough. For aₙ = 1/n, aₙ → 0 but Σ |aₙ| = Σ 1/n diverges.

Answer: (B). Σ |aₙ| = Σ 1/nᵖ is a p-series. It diverges when p ≤ 1. For p > 0 the terms 1/nᵖ decrease to 0, so the alternating series test gives convergence of Σ aₙ. Both happen exactly when 0 < p ≤ 1.

• (A) gives the values where the series is absolutely convergent, because Σ 1/nᵖ converges for p > 1.
• (C) includes p = 1 (conditional, correct) but also p > 1, where the convergence is absolute, not conditional.
• (D) is the set where Σ aₙ converges at all. It mixes the conditional values with the absolute ones.

Answer: (B).

• I diverges. n/(n + 1) → 1, so the terms alternate between values near 1 and −1 and do not approach 0 (nth term test).
• II is absolutely convergent. Ratio test on |aₙ|: [5ⁿ⁺¹/(n + 1)!] × [n!/5ⁿ] = 5/(n + 1) → 0 < 1.
• III is conditionally convergent. Limit comparison of 1/(√n + 1) with 1/√n gives limit 1, and Σ 1/√n diverges (p = ½), so Σ |aₙ| diverges. The terms 1/(√n + 1) decrease to 0, so the alternating series test shows Σ aₙ converges.

Distractors: (A) treats series I as convergent because it alternates; alternating signs cannot rescue terms that do not approach 0. (C) treats III as absolutely convergent, but III only converges because of cancellation. (D) makes both errors.

Answer: (A). Σ |aₙ| converges (to 7), so Σ aₙ is absolutely convergent. Any rearrangement of an absolutely convergent series converges to the same value, 4.

• (B) confuses the sum of the series with the sum of its absolute values. Reordering does not turn negative terms positive.
• (C) and (D) describe what can happen to a conditionally convergent series. This one converges absolutely, so neither can happen.

(a) Use the integral test with f(x) = 1/(x ln x) on [2, ∞). f is positive and continuous there, and it is decreasing because x ln x is increasing for x ≥ 2 (both factors are positive and increasing). ∫ from 2 to b of 1/(x ln x) dx = [ln(ln x)] from 2 to b = ln(ln b) − ln(ln 2) (substitution u = ln x). As b → ∞, ln(ln b) → ∞, so the improper integral diverges. By the integral test, Σ 1/(n ln n) diverges.

(b) Alternating series test with bₙ = 1/(n ln n) > 0 for n ≥ 2:

• the signs alternate because of (−1)ⁿ;
• bₙ is decreasing, since n ln n increases (b₂ ≈ 0.721, b₃ ≈ 0.303, b₄ ≈ 0.180);
• bₙ → 0, since n ln n → ∞. So Σ (−1)ⁿ/(n ln n) converges.

(c) The series converges, but the series of absolute values diverges, so it is conditionally convergent.

Total: 4 points. Note for (a): direct comparison with the harmonic series does not work here, because 1/(n ln n) is smaller than 1/n (for n ≥ 3, where ln n > 1), and being smaller than a divergent series proves nothing. An answer that only says “smaller than 1/n, so diverges” earns no credit for (a).

(a) cos(π) = −1, cos(2π) = 1, cos(3π) = −1, and so on: cos(πn) is −1 for odd n and 1 for even n, which is (−1)ⁿ. The terms are a₁ = −2, a₂ = 3/8, a₃ = −4/27, a₄ = 5/64.

(b) |aₙ| = (n + 1)/n³. Use the limit comparison test with bₙ = 1/n²: lim as n → ∞ of [(n + 1)/n³] ÷ (1/n²) = lim of (n + 1)/n = 1, which is finite and positive. Σ 1/n² is a p-series with p = 2 > 1, so it converges. Therefore Σ |aₙ| converges and the series is absolutely convergent.

(Direct comparison also works: n + 1 ≤ 2n for n ≥ 1, so (n + 1)/n³ ≤ 2/n², and Σ 2/n² converges.)

(c) The series converges absolutely (part (b)). If a series converges absolutely, then any series formed by regrouping or rearranging its terms converges to the same value. So the grouped series has the same sum.

Total: 5 points. A (c) answer that only says “brackets do not change a sum” earns no credit: that is true for finite sums, but for an infinite series the reason is absolute convergence.

(a) Let t = 1/n. As n → ∞, t → 0⁺, and n sin(1/n) = sin(t)/t. Since lim as t → 0 of sin(t)/t = 1, lim of n sin(1/n) = 1.

(b) For n ≥ 1, 0 < 1/n ≤ 1 < π/2, so sin(1/n) > 0 and |aₙ| = sin(1/n). Use the limit comparison test with bₙ = 1/n: lim of sin(1/n) ÷ (1/n) = lim of n sin(1/n) = 1 (part (a)), which is finite and positive. Σ 1/n is the harmonic series, which diverges. Therefore Σ |aₙ| diverges.

(c) Let bₙ = sin(1/n) > 0. The signs alternate because of (−1)ⁿ.

• Decreasing: as n increases, 1/n decreases, and sin is increasing on (0, π/2), so sin(1/n) decreases (b₁ ≈ 0.841, b₂ ≈ 0.479, b₃ ≈ 0.327).
• Limit 0: 1/n → 0 and sin is continuous, so sin(1/n) → sin 0 = 0.

By the alternating series test, Σ aₙ converges. With (b), the series is conditionally convergent.

(d) The student has tested only the series of absolute values. Its divergence shows the series is not absolutely convergent, but a series can still converge through cancellation. Here the alternating series test shows it does converge.

Total: 5 points. Acceptable alternative for (a): L’Hospital’s rule on sin(t)/t, or on sin(1/x) ÷ (1/x) as x → ∞. Acceptable alternative for (b): direct comparison using sin x ≥ 2x/π for 0 ≤ x ≤ π/2, which gives sin(1/n) ≥ (2/π)(1/n).