AP Calculus BC Quiz: Integral Test For Convergence
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Integral Test For ConvergenceQuestion 1 of 20

A decay sequence is ∑n=1∞1(n2+1)3/4\sum_{n=1}^{\infty} \frac{1}{(n^2+1)^{3/4}}. Use the integral test to decide.

Converges because ∫1∞1(x2+1)3/4 dx\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}}\,dx converges.
Diverges because ∫1∞1(x2+1)3/4 dx\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}}\,dx converges.
Diverges because ∫1∞(x2+1)3/4 dx\int_1^{\infty} (x^2+1)^{3/4}\,dx diverges.
Converges because ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{\sqrt{x^2+1}}\,dx converges.
Diverges because ∫1∞1x3/2 dx\int_1^{\infty} \frac{1}{x^{3/2}}\,dx diverges.
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AP Calculus BC Quiz

AP Calculus BC Quiz: Integral Test For Convergence

Practice Integral Test For Convergence in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Integral Test For Convergence, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A decay sequence is ∑n=1∞1(n2+1)3/4\sum_{n=1}^{\infty} \frac{1}{(n^2+1)^{3/4}}. Use the integral test to decide.

  1. Converges because ∫1∞1(x2+1)3/4 dx\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}}\,dx converges. (correct answer)
  2. Diverges because ∫1∞1(x2+1)3/4 dx\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}}\,dx converges.
  3. Diverges because ∫1∞(x2+1)3/4 dx\int_1^{\infty} (x^2+1)^{3/4}\,dx diverges.
  4. Converges because ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{\sqrt{x^2+1}}\,dx converges.
  5. Diverges because ∫1∞1x3/2 dx\int_1^{\infty} \frac{1}{x^{3/2}}\,dx diverges.
Explanation: The integral test is a vital calculus method for testing series via improper integrals. For ∑_{n=1}^∞ 1/(n^2 + 1)^{3/4}, f(x) = 1/(x^2 + 1)^{3/4} is positive, continuous, and decreasing. Asymptotically, it resembles ∫ dx/x^{3/2}, with 3/2 > 1, so the integral converges. Thus, the series converges. A distractor claims divergence because ∫ dx/x^{3/2} diverges, but it actually converges. Focus on the dominant term for large x to apply the integral test effectively.

Question 2

A response curve uses ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}. What does the integral test conclude?

  1. Converges because ∫1∞1x dx\int_1^{\infty} \frac{1}{\sqrt{x}}\,dx converges.
  2. Diverges because ∫1∞1x dx\int_1^{\infty} \frac{1}{\sqrt{x}}\,dx diverges. (correct answer)
  3. Converges because ∫1∞x dx\int_1^{\infty} \sqrt{x}\,dx converges.
  4. Diverges because ∫1∞1x3/2 dx\int_1^{\infty} \frac{1}{x^{3/2}}\,dx diverges.
  5. Converges because ∫1∞1x3/2 dx\int_1^{\infty} \frac{1}{x^{3/2}}\,dx diverges.
Explanation: The integral test offers a key method in calculus to test series convergence via improper integrals. For ∑_{n=1}^∞ 1/√n, f(x) = 1/√x is positive, continuous, and decreasing for x ≥ 1. The integral ∫_1^∞ dx/√x = [2√x]_1^∞ diverges to infinity since the exponent 1/2 < 1. Thus, the series diverges. A distractor claims convergence based on ∫ dx/x^{3/2}, but that converges while the actual integral diverges. Compare exponents to 1 in p-series equivalents for effective integral test application.

Question 3

A signal decay is ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n\ln n}. Using the integral test, determine convergence.

  1. Converges because ∫2∞1xln⁡x dx\int_2^{\infty} \frac{1}{x\ln x}\,dx converges.
  2. Diverges because ∫2∞1xln⁡x dx\int_2^{\infty} \frac{1}{x\ln x}\,dx diverges. (correct answer)
  3. Converges because ∫2∞ln⁡xx dx\int_2^{\infty} \frac{\ln x}{x}\,dx converges.
  4. Diverges because ∫2∞1(ln⁡x) dx\int_2^{\infty} \frac{1}{(\ln x)}\,dx diverges.
  5. Converges because ∫2∞1x(ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx diverges.
Explanation: The integral test serves as an essential technique in calculus to evaluate series convergence via improper integrals. For ∑_{n=2}^∞ 1/(n ln n), use f(x) = 1/(x ln x), positive, continuous, and decreasing for x ≥ 2. The integral ∫2^∞ dx/(x ln x) substitutes u = ln x, du = dx/x, resulting in ∫{ln 2}^∞ du/u, which is ln u from ln 2 to ∞, diverging to infinity. Therefore, the series diverges because the integral diverges. A distractor might propose convergence based on ∫ dx/(x (ln x)^2), but that converges while the given integral does not. To apply the integral test reliably, confirm the function's monotonicity and compute the integral's limit carefully.

Question 4

A learning-rate schedule is ∑n=2∞1n(ln⁡n)(ln⁡ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)(\ln\ln n)^2}. What does the integral test conclude?

  1. Converges because ∫2∞1x(ln⁡x)(ln⁡ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)(\ln\ln x)^2}\,dx converges. (correct answer)
  2. Diverges because ∫2∞1x(ln⁡x)(ln⁡ln⁡x)2 dx\int_2^{\infty} \frac{1}{x(\ln x)(\ln\ln x)^2}\,dx converges.
  3. Diverges because ∫2∞1xln⁡x dx\int_2^{\infty} \frac{1}{x\ln x}\,dx converges.
  4. Converges because ∫2∞1x(ln⁡x)(ln⁡ln⁡x) dx\int_2^{\infty} \frac{1}{x(\ln x)(\ln\ln x)}\,dx converges.
  5. Diverges because ∫2∞1(ln⁡ln⁡x)2 dx\int_2^{\infty} \frac{1}{(\ln\ln x)^2}\,dx diverges.
Explanation: The integral test is a key calculus method for determining series behavior via integrals. For ∑_{n=2}^∞ 1/(n ln n (ln ln n)^2), f(x) = 1/(x ln x (ln ln x)^2) is positive, continuous, and decreasing for large x. Substituting v = ln ln x gives ∫ dv/v^2, which converges to a finite value. Hence, the series converges. A distractor suggests divergence based on ∫ dx/(x ln x), but the additional terms ensure convergence. Use nested substitutions for multi-level logarithms in the integral test.

Question 5

A concentration model uses ∑n=1∞1n2 ⁣+ ⁣1\sum_{n=1}^{\infty} \frac{1}{n^2\!+\!1}. Using the integral test, determine convergence.

  1. Diverges because ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{x^2+1}\,dx diverges.
  2. Converges because ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{x^2+1}\,dx converges. (correct answer)
  3. Diverges because ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\,dx converges.
  4. Converges because ∫1∞(x2+1) dx\int_1^{\infty} (x^2+1)\,dx converges.
  5. Diverges because ∫1∞1x2 dx\int_1^{\infty} \frac{1}{x^2}\,dx diverges.
Explanation: The integral test is an important tool in calculus for evaluating series through integrals. For ∑_{n=1}^∞ 1/(n2n^2 + 1), f(x) = 1/(x2x^2 + 1) is positive, continuous, and decreasing. The integral ∫_1^∞ dx/(x2x^2 + 1) = arctan x to ∞ converges to π/4. Therefore, the series converges. A distractor suggests divergence because ∫ dx/x diverges, but that's irrelevant here. Match the integrand precisely to the series for correct integral test application.

Question 6

To estimate long-run signal loss, analyze ∑n=1∞1n3+1\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^3+1}} using the integral test for convergence.​

  1. Diverges, since ∫1∞1x3+1 dx\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx diverges.
  2. Converges, since ∫1∞1x3+1 dx\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx converges. (correct answer)
  3. Converges, since ∫1∞1x3+1 dx=2x3+1∣1∞\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx=\left.\frac{2}{\sqrt{x^3+1}}\right|_1^{\infty}.
  4. Diverges, since ∫1∞1x3+1 dx=x3+1∣1∞\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx=\left.\sqrt{x^3+1}\right|_1^{\infty}.
  5. Converges, since ∫1∞1x3+1 dx=ln⁡(x3+1)∣1∞\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx=\left.\ln(x^3+1)\right|_1^{\infty}.
Explanation: This problem requires using the integral test to determine convergence of the series. We need to evaluate ∫1∞1x3+1 dx\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx. For large xx, the integrand behaves like 1x3=1x3/2\frac{1}{\sqrt{x^3}} = \frac{1}{x^{3/2}}. Since ∫1∞1x3/2 dx=[−2x]1∞=0−(−2)=2\int_1^{\infty}\frac{1}{x^{3/2}}\,dx = \left[-\frac{2}{\sqrt{x}}\right]_1^{\infty} = 0 - (-2) = 2 converges, and our integrand is comparable, the original integral also converges. Therefore, by the integral test, the series converges. Choice D incorrectly claims the antiderivative is x3+1\sqrt{x^3+1}, which would incorrectly suggest divergence. The key strategy is to compare the integrand's behavior to known convergent or divergent integrals for large xx.

Question 7

A population model sums ∑n=1∞1n2+9\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+9}}. Using the integral test, determine whether it converges.

  1. Converges because ∫1∞1x2+9 dx\int_1^{\infty}\frac{1}{\sqrt{x^2+9}}\,dx converges.
  2. Diverges because ∫1∞1x2+9 dx\int_1^{\infty}\frac{1}{\sqrt{x^2+9}}\,dx diverges. (correct answer)
  3. Converges because ∫1∞1x2+9 dx=1x2+9∣1∞\int_1^{\infty}\frac{1}{\sqrt{x^2+9}}\,dx=\left.\frac{1}{\sqrt{x^2+9}}\right|_1^{\infty} is finite.
  4. Diverges because ∫1∞1x2+9 dx=ln⁡(x2+9)∣1∞\int_1^{\infty}\frac{1}{\sqrt{x^2+9}}\,dx=\left.\ln(\sqrt{x^2+9})\right|_1^{\infty} is infinite.
  5. Converges because ∫1∞1x2+9 dx\int_1^{\infty}\frac{1}{\sqrt{x^2+9}}\,dx is improper at x=1x=1.
Explanation: This problem tests the integral test for series convergence. For large xx, the integrand 1x2+9\frac{1}{\sqrt{x^2+9}} behaves like 1x2=1x\frac{1}{\sqrt{x^2}} = \frac{1}{x}. We know that ∫1∞1x dx=ln⁡x∣1∞=∞\int_1^{\infty}\frac{1}{x}\,dx = \ln x|_1^{\infty} = \infty, which diverges. By comparison, since 1x2+9∼1x\frac{1}{\sqrt{x^2+9}} \sim \frac{1}{x} for large xx, the integral ∫1∞1x2+9 dx\int_1^{\infty}\frac{1}{\sqrt{x^2+9}}\,dx also diverges. Choice A incorrectly claims convergence, possibly confusing this with cases where the denominator grows faster than linearly. For integrands of the form 1x2+c\frac{1}{\sqrt{x^2+c}}, the integral behaves like ∫1x dx\int\frac{1}{x}\,dx and thus diverges.

Question 8

Using the Integral Test, determine whether ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2} converges or diverges.

  1. Converges, because ∫2∞1x(ln⁡x)2 dx\int_2^{\infty}\frac{1}{x(\ln x)^2}\,dx converges. (correct answer)
  2. Diverges, because ∫2∞1x(ln⁡x)2 dx\int_2^{\infty}\frac{1}{x(\ln x)^2}\,dx diverges.
  3. Converges, because ∫2∞1x(ln⁡x)2 dx=1(ln⁡x)2∣2∞\int_2^{\infty}\frac{1}{x(\ln x)^2}\,dx=\left.\frac{1}{(\ln x)^2}\right|_2^{\infty} is finite.
  4. Diverges, because ∫2∞1x(ln⁡x)2 dx=ln⁡(ln⁡x)∣2∞\int_2^{\infty}\frac{1}{x(\ln x)^2}\,dx=\left.\ln(\ln x)\right|_2^{\infty} is infinite.
  5. Converges, because the Integral Test applies even though f(x)=1x(ln⁡x)2f(x)=\frac{1}{x(\ln x)^2} is not decreasing for x≥2x\ge2.
Explanation: This problem requires using the Integral Test to determine whether the series converges. The Integral Test states that if f(x) is positive, continuous, and decreasing for x ≥ 2, then the series and the improper integral have the same convergence behavior. For f(x) = 1/(x(ln x)²), we evaluate ∫₂^∞ 1/(x(ln x)²) dx using the substitution u = ln x, du = dx/x, which gives ∫{ln 2}^∞ 1/u² du = [-1/u]{ln 2}^∞ = 0 - (-1/ln 2) = 1/ln 2, a finite value. Since the integral converges, the series also converges. Choice C incorrectly evaluates the integral as 1/(ln x)² without proper integration, while choice D incorrectly computes the antiderivative. When applying the Integral Test, always verify that f(x) satisfies all conditions and carefully evaluate the improper integral using appropriate techniques.

Question 9

Use the Integral Test to decide whether ∑n=1∞1nn\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}} converges.

  1. Diverges, because ∫1∞x−3/2 dx\int_1^{\infty}x^{-3/2}\,dx diverges.
  2. Converges, because ∫1∞x−3/2 dx\int_1^{\infty}x^{-3/2}\,dx converges. (correct answer)
  3. Converges, because ∫1∞x−3/2 dx=ln⁡x∣1∞\int_1^{\infty}x^{-3/2}\,dx=\left.\ln x\right|_1^{\infty} is finite.
  4. Diverges, because ∫1∞x−3/2 dx=−2x1/2∣1∞\int_1^{\infty}x^{-3/2}\,dx=\left.-\frac{2}{x^{1/2}}\right|_1^{\infty} is infinite.
  5. Converges, because f(x)=x−3/2f(x)=x^{-3/2} is not required to be continuous for the Integral Test.
Explanation: This problem applies the Integral Test to ∑ 1/(n√n) = ∑ n^(-3/2). The function f(x) = x^(-3/2) is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ x^(-3/2) dx, we use the power rule: ∫ x^(-3/2) dx = x^(-1/2)/(-1/2) = -2x^(-1/2) + C = -2/√x + C. Computing the improper integral: [-2/√x]₁^∞ = 0 - (-2) = 2, a finite value. Since the integral converges, the series also converges. Choice D has the correct antiderivative but incorrectly claims divergence, while choice C suggests an incorrect antiderivative of ln x. For p-series ∑ 1/n^p, convergence occurs when p > 1, and here p = 3/2 > 1, confirming convergence.

Question 10

A resource model uses ∑n=1∞1(2n+1)2\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}; determine convergence with the integral test.​

  1. Diverges, since ∫1∞1(2x+1)2 dx\int_1^{\infty}\frac{1}{(2x+1)^2}\,dx diverges.
  2. Converges, since ∫1∞1(2x+1)2 dx\int_1^{\infty}\frac{1}{(2x+1)^2}\,dx converges. (correct answer)
  3. Diverges, since ∫1∞1(2x+1)2 dx=ln⁡(2x+1)∣1∞\int_1^{\infty}\frac{1}{(2x+1)^2}\,dx=\left.\ln(2x+1)\right|_1^{\infty}.
  4. Converges, since ∫1∞1(2x+1)2 dx=12x+1∣1∞\int_1^{\infty}\frac{1}{(2x+1)^2}\,dx=\left.\frac{1}{2x+1}\right|_1^{\infty}.
  5. Converges, since ∫1∞1(2x+1)2 dx=−1(2x+1)2∣1∞\int_1^{\infty}\frac{1}{(2x+1)^2}\,dx=\left.-\frac{1}{(2x+1)^2}\right|_1^{\infty}.
Explanation: This problem applies the integral test to a quadratic denominator series. We evaluate ∫1∞1(2x+1)2 dx\int_1^{\infty}\frac{1}{(2x+1)^2}\,dx using substitution u=2x+1u = 2x+1, so du=2dxdu = 2dx. The integral becomes 12∫3∞1u2 du=12[−1u]3∞=12(0−(−13))=16\frac{1}{2}\int_3^{\infty}\frac{1}{u^2}\,du = \frac{1}{2}\left[-\frac{1}{u}\right]_3^{\infty} = \frac{1}{2}(0 - (-\frac{1}{3})) = \frac{1}{6}. Since the improper integral converges to a finite value, the series converges by the integral test. Choice D shows an incorrect antiderivative missing the factor of −12-\frac{1}{2} from the chain rule. For integrals involving (ax+b)n(ax+b)^n with n>1n > 1, always remember to include the factor 1a\frac{1}{a} when substituting.

Question 11

A cooling model uses ∑n=1∞1(n+1)2/3\sum_{n=1}^{\infty} \frac{1}{(n+1)^{2/3}}. Using the integral test, what happens?

  1. Converges because ∫1∞1(x+1)2/3 dx\int_1^{\infty} \frac{1}{(x+1)^{2/3}}\,dx converges.
  2. Diverges because ∫1∞1(x+1)2/3 dx\int_1^{\infty} \frac{1}{(x+1)^{2/3}}\,dx diverges. (correct answer)
  3. Converges because ∫1∞(x+1)2/3 dx\int_1^{\infty} (x+1)^{2/3}\,dx converges.
  4. Diverges because ∫1∞1(x+1)5/3 dx\int_1^{\infty} \frac{1}{(x+1)^{5/3}}\,dx diverges.
  5. Converges because ∫1∞1x+1 dx\int_1^{\infty} \frac{1}{\sqrt{x+1}}\,dx converges.
Explanation: The integral test provides a vital approach in calculus to ascertain series convergence using improper integrals. For ∑_{n=1}^∞ 1/(n+1)^{2/3}, take f(x) = 1/(x+1)^{2/3}, positive, continuous, and decreasing for x ≥ 1. The integral ∫_1^∞ dx/(x+1)^{2/3} is similar to ∫ x^{-2/3} dx, where the exponent 2/3 < 1, causing divergence. Consequently, the series diverges due to the integral's divergence. A distractor suggests convergence based on ∫ dx/√(x+1), but that diverges and isn't the correct integral. Always check the exponent in p-integrals when applying the integral test for power-like functions.

Question 12

A transport model uses ∑n=1∞1(n+4)3/2\sum_{n=1}^{\infty} \frac{1}{(n+4)^{3/2}}. Using the integral test, determine convergence.

  1. Diverges because ∫1∞1(x+4)3/2 dx\int_1^{\infty} \frac{1}{(x+4)^{3/2}}\,dx diverges.
  2. Converges because ∫1∞1(x+4)3/2 dx\int_1^{\infty} \frac{1}{(x+4)^{3/2}}\,dx converges. (correct answer)
  3. Diverges because ∫1∞(x+4)3/2 dx\int_1^{\infty} (x+4)^{3/2}\,dx diverges.
  4. Converges because ∫1∞1x+4 dx\int_1^{\infty} \frac{1}{\sqrt{x+4}}\,dx converges.
  5. Diverges because ∫1∞1x+4 dx\int_1^{\infty} \frac{1}{x+4}\,dx converges.
Explanation: The integral test serves as a core tool in calculus for series convergence determination via integrals. For ∑_{n=1}^∞ 1/(n+4)^{3/2}, f(x) = 1/(x+4)^{3/2} is positive, continuous, and decreasing. Like ∫ dx/x^{3/2}, with 3/2 > 1, it converges. Therefore, the series converges. A distractor claims divergence because ∫ dx/√(x+4) diverges, but that's a different power. Adjust shifts like +4 by focusing on large x behavior in the integral test.

Question 13

A queue delay model uses ∑n=2∞1n(ln⁡n)1/2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{1/2}}. What does the integral test say?

  1. Converges because ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty} \frac{1}{x(\ln x)^{1/2}}\,dx converges.
  2. Diverges because ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty} \frac{1}{x(\ln x)^{1/2}}\,dx diverges. (correct answer)
  3. Converges because ∫2∞1xln⁡x dx\int_2^{\infty} \frac{1}{x\ln x}\,dx converges.
  4. Diverges because ∫2∞1x(ln⁡x)3/2 dx\int_2^{\infty} \frac{1}{x(\ln x)^{3/2}}\,dx converges.
  5. Converges because ∫2∞ln⁡xx dx\int_2^{\infty} \frac{\ln x}{x}\,dx converges.
Explanation: The integral test provides an essential way in calculus to determine series convergence using integrals. For ∑{n=2}^∞ 1/(n (ln n)^{1/2}), f(x) = 1/(x (ln x)^{1/2}) is positive, continuous, and decreasing for x ≥ 2. Substituting u = ln x gives ∫ du/u^{1/2} = [2u^{1/2}]{ln 2}^∞, which diverges. Therefore, the series diverges. A distractor suggests convergence based on ∫ dx/(x ln x), but both diverge, though the given has a square root making it still diverge. Adjust exponents carefully in substitutions for logarithmic terms in the integral test.

Question 14

A resource model sums ∑n=1∞1n2+n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+n}}. Using the integral test, what occurs?

  1. Converges because ∫1∞1x2+x dx\int_1^{\infty} \frac{1}{\sqrt{x^2+x}}\,dx converges.
  2. Diverges because ∫1∞1x2+x dx\int_1^{\infty} \frac{1}{\sqrt{x^2+x}}\,dx diverges. (correct answer)
  3. Converges because ∫1∞x2+x dx\int_1^{\infty} \sqrt{x^2+x}\,dx converges.
  4. Diverges because ∫1∞1x2+x dx\int_1^{\infty} \frac{1}{x^2+x}\,dx converges.
  5. Converges because ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\,dx converges.
Explanation: The integral test offers an essential technique in calculus for series convergence testing with integrals. For ∑_{n=1}^∞ 1/√(n2n^2 + n), f(x) = 1/√(x2x^2 + x) is positive, continuous, and decreasing. Asymptotically like ∫ dx/x, which diverges. Therefore, the series diverges. A distractor claims convergence based on ∫ dx/x, but it diverges, matching the conclusion but not the choice. Analyze square root denominators by factoring for accurate integral test results.

Question 15

A cooling experiment uses ∑n=2∞1nln⁡n\sum_{n=2}^{\infty}\frac{1}{n\ln n}. Using the integral test, what occurs?

  1. Converges, since ∫2∞1xln⁡x dx\int_2^\infty \frac{1}{x\ln x}\,dx converges
  2. Diverges, since ∫2∞1xln⁡x dx\int_2^\infty \frac{1}{x\ln x}\,dx diverges (correct answer)
  3. Converges, since ∫2∞1xln⁡x dx=−1ln⁡x∣2∞\int_2^\infty \frac{1}{x\ln x}\,dx=\left.-\frac{1}{\ln x}\right|_2^\infty
  4. Diverges, since ∫2∞1xln⁡x dx=1ln⁡x∣2∞\int_2^\infty \frac{1}{x\ln x}\,dx=\left.\frac{1}{\ln x}\right|_2^\infty
  5. Converges, because 1nln⁡n\frac{1}{n\ln n} is not continuous at n=1n=1
Explanation: The integral test is an essential technique in AP Calculus BC for evaluating series convergence via improper integrals. For the series ∑_{n=2}^∞ 1/(n ln n), f(x) = 1/(x ln x) is positive, continuous, and decreasing for x ≥ 2. The integral ∫_2^∞ 1/(x ln x) dx substitutes u = ln x to get ln(ln x) from 2 to ∞, which diverges to infinity. Therefore, the series diverges by the integral test. A tempting distractor might claim convergence because of an incorrect antiderivative like -1/ln x, but the proper evaluation shows divergence. A transferable strategy for the integral test is to always verify the conditions of the integral test: the function must be positive, continuous, and decreasing for sufficiently large x.

Question 16

A signal decay is modeled by ∑n=1∞1n2+4\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+4}}. Use the integral test to decide convergence.

  1. Converges, since ∫1∞1x2+4 dx\int_1^\infty \frac{1}{\sqrt{x^2+4}}\,dx converges
  2. Diverges, since ∫1∞1x2+4 dx\int_1^\infty \frac{1}{\sqrt{x^2+4}}\,dx diverges (correct answer)
  3. Converges, since ∫1∞1x2+4 dx=xx2+4∣1∞\int_1^\infty \frac{1}{\sqrt{x^2+4}}\,dx=\left.\frac{x}{\sqrt{x^2+4}}\right|_1^\infty
  4. Diverges, since ∫1∞1x2+4 dx=arctan⁡(x2)∣1∞\int_1^\infty \frac{1}{\sqrt{x^2+4}}\,dx=\left.\arctan\left(\frac{x}{2}\right)\right|_1^\infty
  5. Converges, because 1n2+4\frac{1}{\sqrt{n^2+4}} is decreasing only for n>4n>4
Explanation: The integral test is a core concept in AP Calculus BC for testing series convergence using improper integrals. For ∑_{n=1}^∞ 1/√(n2n^2 + 4), f(x) = 1/√(x2x^2 + 4) is positive, continuous, and decreasing for x ≥ 1. Asymptotically, it behaves like 1/x, and ∫_1^∞ 1/x dx diverges logarithmically. Hence, the integral diverges, so the series diverges by the integral test. A tempting distractor might propose convergence with an arctan antiderivative, but that's for 1/(x2x^2 + 4), not the square root, leading to an error. A transferable strategy for the integral test is to always verify the conditions of the integral test: the function must be positive, continuous, and decreasing for sufficiently large x.

Question 17

A data-smoothing method uses ∑n=2∞1n(ln⁡n)3\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^3}. Apply the integral test for convergence.

  1. Diverges, since ∫2∞1x(ln⁡x)3 dx\int_2^\infty \frac{1}{x(\ln x)^3}\,dx diverges
  2. Converges, since ∫2∞1x(ln⁡x)3 dx\int_2^\infty \frac{1}{x(\ln x)^3}\,dx converges (correct answer)
  3. Diverges, since ∫2∞1x(ln⁡x)3 dx=ln⁡ln⁡x∣2∞\int_2^\infty \frac{1}{x(\ln x)^3}\,dx=\left.\ln\ln x\right|_2^\infty
  4. Converges, since ∫2∞1x(ln⁡x)3 dx=1ln⁡x∣2∞\int_2^\infty \frac{1}{x(\ln x)^3}\,dx=\left.\frac{1}{\ln x}\right|_2^\infty
  5. Diverges, because 1n(ln⁡n)3\frac{1}{n(\ln n)^3} is not positive for all n≥2n\ge2
Explanation: The integral test is a crucial method in AP Calculus BC for testing series via improper integrals. For ∑_{n=2}^∞ 1/(n (ln n)^3), f(x) = 1/(x (ln x)^3) is positive, continuous, and decreasing for x ≥ 2. Substituting u = ln x gives ∫ u^{-3} du = -1/(2 u2u^2), evaluating from 2 to ∞ to converge to 1/(2 (ln 2)^2). Hence, the series converges by the integral test. A tempting distractor might claim divergence with an incorrect antiderivative like ln ln x, but the proper substitution shows finite value. A transferable strategy for the integral test is to always verify the conditions of the integral test: the function must be positive, continuous, and decreasing for sufficiently large x.

Question 18

A pollutant model gives ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}. Using the integral test, what is the series' behavior?

  1. Diverges, since ∫2∞1x(ln⁡x)2 dx\int_2^\infty \frac{1}{x(\ln x)^2}\,dx diverges
  2. Converges, since ∫2∞1x(ln⁡x)2 dx\int_2^\infty \frac{1}{x(\ln x)^2}\,dx converges (correct answer)
  3. Diverges, since ∫2∞1x(ln⁡x)2 dx=1ln⁡x∣2∞\int_2^\infty \frac{1}{x(\ln x)^2}\,dx=\left.\frac{1}{\ln x}\right|_2^\infty
  4. Converges, since ∫2∞1x(ln⁡x)2 dx=ln⁡(ln⁡x)∣2∞\int_2^\infty \frac{1}{x(\ln x)^2}\,dx=\left.\ln(\ln x)\right|_2^\infty
  5. Diverges, because 1n(ln⁡n)2\frac{1}{n(\ln n)^2} is not decreasing for n≥2n\ge2
Explanation: The integral test is a fundamental tool in AP Calculus BC for assessing the convergence of series by relating them to improper integrals. For the series ∑_{n=2}^∞ 1/(n (ln n)^2), we consider the function f(x) = 1/(x (ln x)^2), which is positive, continuous, and decreasing for x ≥ 2. The improper integral ∫_2^∞ 1/(x (ln x)^2) dx is evaluated using the substitution u = ln x, yielding -1/ln x evaluated from 2 to ∞, which converges to 1/ln 2. Since the integral converges, the series also converges by the integral test. A tempting distractor might claim the series diverges because the term is not decreasing for all n ≥ 2, but it is decreasing for sufficiently large n, satisfying the test conditions. A transferable strategy for the integral test is to always check the three conditions—positive, continuous, and eventually decreasing—before evaluating the integral.

Question 19

A sensor's cumulative drift is ∑n=2∞ln⁡nn2\sum_{n=2}^{\infty}\frac{\ln n}{n^2}. Use the integral test to determine convergence.

  1. Diverges, since ∫2∞ln⁡xx2 dx\int_2^\infty \frac{\ln x}{x^2}\,dx diverges
  2. Converges, since ∫2∞ln⁡xx2 dx\int_2^\infty \frac{\ln x}{x^2}\,dx converges (correct answer)
  3. Diverges, since ∫2∞ln⁡xx2 dx=(ln⁡x)22∣2∞\int_2^\infty \frac{\ln x}{x^2}\,dx=\left.\frac{(\ln x)^2}{2}\right|_2^\infty
  4. Converges, since ∫2∞ln⁡xx2 dx=ln⁡x∣2∞\int_2^\infty \frac{\ln x}{x^2}\,dx=\left.\ln x\right|_2^\infty
  5. Diverges, because ln⁡nn2\frac{\ln n}{n^2} is not decreasing for all n≥2n\ge2
Explanation: The integral test is an important tool in AP Calculus BC for convergence analysis via improper integrals. For ∑_{n=2}^∞ (ln n)/n^2, f(x) = (ln x)/x^2 is positive, continuous, and eventually decreasing for x ≥ 2. Integration by parts yields the antiderivative -(ln x + 1)/x, evaluating from 2 to ∞ to converge to (ln 2 + 1)/2. Thus, the series converges by the integral test. A tempting distractor might claim divergence because the term is not decreasing for all n ≥ 2, but it decreases for n ≥ 3, which suffices. A transferable strategy for the integral test is to always verify the conditions of the integral test: the function must be positive, continuous, and decreasing for sufficiently large x.

Question 20

Determine convergence of ∑n=1∞1n23\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n^2}} using the Integral Test.

  1. Converges, because ∫1∞x−2/3 dx\int_1^{\infty}x^{-2/3}\,dx converges.
  2. Diverges, because ∫1∞x−2/3 dx\int_1^{\infty}x^{-2/3}\,dx diverges. (correct answer)
  3. Converges, because ∫1∞x−2/3 dx=1x2/3∣1∞\int_1^{\infty}x^{-2/3}\,dx=\left.\frac{1}{x^{2/3}}\right|_1^{\infty} is finite.
  4. Diverges, because ∫1∞x−2/3 dx=32x2/3∣1∞\int_1^{\infty}x^{-2/3}\,dx=\left.\frac{3}{2}x^{2/3}\right|_1^{\infty} is infinite.
  5. Converges, because f(x)=x−2/3f(x)=x^{-2/3} is not required to be decreasing for the Integral Test.
Explanation: This problem requires using the Integral Test on the series ∑ 1/∛(n²) = ∑ n^(-2/3). The function f(x) = x^(-2/3) is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ x^(-2/3) dx, we use the power rule: ∫ x^(-2/3) dx = x^(1/3)/(1/3) = 3x^(1/3) + C. Computing the improper integral: [3x^(1/3)]₁^∞ = ∞ - 3, which diverges to infinity. Since the integral diverges, the series also diverges. Choice C incorrectly suggests the antiderivative is 1/x^(2/3), confusing the integrand with its antiderivative, while choice D correctly computes the antiderivative but has the limits evaluation backwards. For p-series ∑ 1/n^p, remember that convergence occurs when p > 1, and here p = 2/3 < 1.