Integral Test for Convergence: Practice Questions (Calculus BC 10.4)

Answer: (D). The test says only that the series and the integral both converge or both diverge. Their values are usually different, so equality is not a condition (and is generally false).

• (A) is required. Without positive terms, partial sums can go down as well as up, and the rectangle comparison fails.
• (B) is required. The improper integral must make sense on [N, ∞).
• (C) is required. Decreasing is what keeps the rectangles on one side of the curve.

Answer: (B). f(x) = 6/(2x + 1)² is positive and continuous for x ≥ 1, and f′(x) = −24/(2x + 1)³ < 0, so it is decreasing. The integral converges, so the series converges by the integral test. The test does not give the sum. (In fact the sum is about 1.402. It must lie between the integral, 1, and a₁ + 1 = 2/3 + 1 ≈ 1.667.)

• (A) treats the integral’s value as the sum of the series.
• (C) confuses a convergent integral with a zero integral. A finite value, of any size, means convergence.
• (D) is false: the denominator grows as x grows, so the fraction decreases.

Answer: (B). The terms are positive (2 + sin n ≥ 1), but they are not decreasing: sin n wobbles, so the terms sometimes go up. For example, a₅ = (2 + sin 5)/25 ≈ 0.042 and a₆ = (2 + sin 6)/36 ≈ 0.048. Since f(x) = (2 + sin x)/x² is never decreasing on a whole interval [N, ∞), the condition fails. (A comparison test from Topic 10.6 handles this series.)

• (A) f(x) = 1/(x + 5)² is positive, continuous and decreasing: f′(x) = −2/(x + 5)³ < 0.
• (C) f(x) = e^(−x/3) is positive, continuous and decreasing: f′(x) = −⅓ e^(−x/3) < 0.
• (D) f(x) = x/(x² + 1) has f′(x) = (1 − x²)/(x² + 1)², which is ≤ 0 for x ≥ 1. The test applies (and shows divergence).

Answer: (B). Let f(x) = x(x² + 1)^(−k). Then f′(x) = (x² + 1)^(−k−1)(1 + (1 − 2k)x²), which is negative for x > 1 when k ≥ 1 (the bracket equals 2 − 2k ≤ 0 at x = 1 and decreases after that), so the integral test applies for every k ≥ 1.

• For k ≠ 1: ∫ x(x² + 1)^(−k) dx = (x² + 1)^(1−k)/(2(1 − k)). As x → ∞ this has a finite limit only when 1 − k < 0, that is, k > 1. (For k = 2, the integral from 1 to ∞ is 1/4.)
• For k = 1: ∫ x/(x² + 1) dx = ½ ln(x² + 1) → ∞. Diverges.

So the series converges exactly when k > 1. (For 1/2 < k < 1 the function is decreasing from some point on and the integral diverges; for k ≤ 1/2 the terms do not approach 0, so the nth term test shows divergence.)

• (A) includes values such as k = 3/4, where the integral grows like ∫ x^(−1/2) dx and diverges.
• (C) includes k = 1, where the integral is a logarithm and diverges.
• (D) leaves out values that work, such as k = 2 (integral 1/4) and k = 3/2.

1. Conditions. f(x) = 1/(x² + 2x + 5) = 1/((x + 1)² + 4). The denominator is positive, so f is positive and continuous. For x ≥ 1 the denominator increases (its derivative 2x + 2 > 0), so f is decreasing. ✓
2. Antiderivative. Complete the square and use ∫ 1/(u² + a²) du = (1/a) arctan(u/a) with u = x + 1, a = 2: ∫ 1/((x + 1)² + 4) dx = ½ arctan((x + 1)/2) + C.
3. Improper integral. ∫ from 1 to b = ½ arctan((b + 1)/2) − ½ arctan(1). As b → ∞, arctan((b + 1)/2) → π/2, so the integral converges to ½ · π/2 − ½ · π/4 = π/8 (about 0.393).
4. Conclusion. The integral converges, so ∑ 1/(n² + 2n + 5) converges by the integral test. (Its sum is about 0.460, not π/8.)

Suggested mark points (4): 1 for the three conditions with a reason for “decreasing”; 1 for the antiderivative ½ arctan((x + 1)/2); 1 for the limit π/8 with arctan → π/2; 1 for the conclusion naming the integral test.

Common error: writing the final answer as “the series converges to π/8”.

(a) 6/(3n + 1) → 0. The nth term test can only show divergence when the limit is not 0, so here it is inconclusive.

(b) For x ≥ 1, 3x + 1 > 0, so f(x) > 0, and f is continuous there. f′(x) = −18/(3x + 1)² < 0, so f is decreasing.

(c) ∫ 6/(3x + 1) dx = 2 ln(3x + 1) + C (substitution u = 3x + 1). So ∫ from 1 to b of f(x) dx = 2 ln(3b + 1) − 2 ln 4 = 2 ln((3b + 1)/4). As b → ∞, this → ∞. The integral diverges. (At b = 100 it is already about 8.64, and it keeps growing.)

(d) f is positive, continuous and decreasing for x ≥ 1, and ∫ from 1 to ∞ of f(x) dx diverges, so ∑ 6/(3n + 1) diverges by the integral test.

Total: 5 points. A common error in (c) is the antiderivative 6 ln(3x + 1), which misses the factor 1/3. It still diverges, but the antiderivative point is lost.

(a) f′(x) = e^(−x) − x e^(−x) = e^(−x)(1 − x). For x > 1, 1 − x < 0, so f′(x) < 0 and f is decreasing for x ≥ 1. (f′(1) = 0 at a single point does not stop f being decreasing on [1, ∞).)

(b) Parts with u = x (du = dx) and dv = e^(−x) dx (v = −e^(−x)): ∫ x e^(−x) dx = −x e^(−x) + ∫ e^(−x) dx = −x e^(−x) − e^(−x) + C. ∫ from 1 to b = (−b e^(−b) − e^(−b)) − (−e⁻¹ − e⁻¹) = 2e⁻¹ − b e^(−b) − e^(−b). As b → ∞, e^(−b) → 0 and b e^(−b) = b/eᵇ → 0 (L’Hospital’s Rule: 1/eᵇ → 0). So the integral converges to 2/e.

(c) f is positive (x > 0, e^(−x) > 0), continuous and decreasing for x ≥ 1, and ∫ from 1 to ∞ of f(x) dx converges, so ∑ n e^(−n) converges by the integral test.

(d) S₃ = e⁻¹ + 2e⁻² + 3e⁻³ ≈ 0.368 + 0.271 + 0.149 = 0.788. But 2/e ≈ 0.736. All the remaining terms are positive, so the full sum is greater than S₃ and therefore greater than 2/e. The sum cannot be 2/e. (Its actual value is about 0.921.)

Total: 5 points. Acceptable alternative for (d): use the rectangle picture. Rectangles of height aₙ on [n, n + 1] lie above the curve (strictly, since f is strictly decreasing for x > 1), so the sum is greater than the integral 2/e. Any valid numerical comparison with S₂ or S₃ earns the point if it explains why later terms cannot bring the total back down to 2/e.