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Integral Test for Convergence - AP Calculus BC

  1. Use the Integral Test to determine whether ∑n=1∞1(n+1)2\sum_{n=1}^{\infty}\frac{1}{(n+1)^2} converges.
    Answer: Converges, because ∫1∞1(x+1)2 dx\int_1^{\infty}\frac{1}{(x+1)^2}\,dx converges. This problem applies the Integral Test to determine series convergence. The function f(x) = 1/(x+1)² is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ 1/(x+1)² dx, we use the substitution u = x+1, du = dx, giving ∫₂^∞ 1/u² du = [-1/u]₂^∞ = 0 - (-1/2) = 1/2, a finite value. Since the integral converges, the series also converges. Choice C incorrectly claims the antiderivative is ln(x+1), which would be correct for 1/(x+1) but not for 1/(x+1)². Choice D has the wrong sign in the antiderivative evaluation. When applying the Integral Test, carefully compute antiderivatives of shifted functions using appropriate substitutions.
  2. A damping term is ∑n=1∞1n5/4\sum_{n=1}^{\infty} \frac{1}{n^{5/4}}. What does the integral test imply?
    Answer: Converges because ∫1∞x−5/4 dx\int_1^{\infty} x^{-5/4}\,dx converges. The integral test provides a fundamental approach in calculus to check series convergence with integrals. For ∑_{n=1}^∞ 1/n5/4n^{5/4}, f(x) = x−5/4x^{-5/4} is positive, continuous, and decreasing. The integral ∫_1^∞ x−5/4x^{-5/4} dx = [x−1/4x^{-1/4} / (-1/4)]_1^∞ converges since 5/4 > 1. Thus, the series converges. A distractor claims divergence because ∫ x−1/4x^{-1/4} dx diverges, but that's a different exponent. Confirm the p-value exceeds 1 in p-series for convergence in the integral test.
  3. Determine convergence of ∑n=1∞1n23\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n^2}} using the Integral Test.
    Answer: Diverges, because ∫1∞x−2/3 dx\int_1^{\infty}x^{-2/3}\,dx diverges. This problem requires using the Integral Test on the series ∑ 1/∛(n²) = ∑ n−2/3n^{-2/3}. The function f(x) = x−2/3x^{-2/3} is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ x−2/3x^{-2/3} dx, we use the power rule: ∫ x−2/3x^{-2/3} dx = x1/3x^{1/3}/(1/3) = 3x1/33x^{1/3} + C. Computing the improper integral: [3x1/33x^{1/3}]₁^∞ = ∞ - 3, which diverges to infinity. Since the integral diverges, the series also diverges. Choice C incorrectly suggests the antiderivative is 1/x2/3x^{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/npn^p, remember that convergence occurs when p > 1, and here p = 2/3 < 1.
  4. 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.​
    Answer: Converges, since ∫1∞1x3+1 dx\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx converges. 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.
  5. Use the Integral Test to decide whether ∑n=1∞1nn\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}} converges.
    Answer: Converges, because ∫1∞x−3/2 dx\int_1^{\infty}x^{-3/2}\,dx converges. This problem applies the Integral Test to ∑ 1/(n√n) = ∑ n−3/2n^{-3/2}. The function f(x) = x−3/2x^{-3/2} is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ x−3/2x^{-3/2} dx, we use the power rule: ∫ x−3/2x^{-3/2} dx = x−1/2x^{-1/2}/(-1/2) = -2x−1/22x^{-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/npn^p, convergence occurs when p > 1, and here p = 3/2 > 1, confirming convergence.
  6. Using the Integral Test, determine whether ∑n=1∞1n2+n\sum_{n=1}^{\infty}\frac{1}{n^2+n} converges.
    Answer: Converges, because ∫1∞1x2+x dx\int_1^{\infty}\frac{1}{x^2+x}\,dx converges. This problem tests the Integral Test for series convergence. The function f(x) = 1/(x²+x) = 1/(x(x+1)) is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ 1/(x²+x) dx, we use partial fractions: 1/(x(x+1)) = 1/x - 1/(x+1). Thus ∫ 1/(x²+x) dx = ln|x| - ln|x+1| = ln|x/(x+1)| + C. Computing the improper integral: [ln(x/(x+1))]₁^∞ = ln(1) - ln(1/2) = 0 - (-ln 2) = ln 2, a finite value. Since the integral converges, the series also converges. Choice C incorrectly suggests the antiderivative is ln(x²+x), which would require a factor of (2x+1) in the numerator. When encountering rational functions in the Integral Test, partial fractions decomposition is often the key technique.
  7. 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?
    Answer: Diverges because ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty} \frac{1}{x(\ln x)^{1/2}}\,dx diverges. The integral test provides an essential way in calculus to determine series convergence using integrals. For ∑_{n=2}^∞ 1/(n (lnn)1/2(ln n)^{1/2}), f(x) = 1/(x (lnx)1/2(ln x)^{1/2}) is positive, continuous, and decreasing for x ≥ 2. Substituting u = ln x gives ∫ du/u1/2u^{1/2} = [2u1/22u^{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.
  8. 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.
    Answer: Converges because ∫1∞1(x2+1)3/4 dx\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}}\,dx converges. The integral test is a vital calculus method for testing series via improper integrals. For ∑_{n=1}^∞ 1/(n2+1)3/4(n^2 + 1)^{3/4}, f(x) = 1/(x2+1)3/4(x^2 + 1)^{3/4} is positive, continuous, and decreasing. Asymptotically, it resembles ∫ dx/x3/2x^{3/2}, with 3/2 > 1, so the integral converges. Thus, the series converges. A distractor claims divergence because ∫ dx/x3/2x^{3/2} diverges, but it actually converges. Focus on the dominant term for large x to apply the integral test effectively.
  9. A resource model sums ∑n=1∞1n2+n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+n}}. Using the integral test, what occurs?
    Answer: Diverges because ∫1∞1x2+x dx\int_1^{\infty} \frac{1}{\sqrt{x^2+x}}\,dx diverges. 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.
  10. A workload estimate is ∑n=1∞1n+1\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}}; use the integral test to decide convergence.​​
    Answer: Diverges, since ∫1∞1x+1 dx\int_1^{\infty}\frac{1}{\sqrt{x+1}}\,dx diverges. This problem uses the integral test on a square root series. We evaluate ∫1∞1x+1 dx\int_1^{\infty}\frac{1}{\sqrt{x+1}}\,dx using substitution u=x+1u = x+1, so du=dxdu = dx. The integral becomes ∫2∞1u du=[2u]2∞=lim⁡b→∞2b−22=∞\int_2^{\infty}\frac{1}{\sqrt{u}}\,du = \left[2\sqrt{u}\right]_2^{\infty} = \lim_{b\to\infty}2\sqrt{b} - 2\sqrt{2} = \infty. Since the improper integral diverges to infinity, the series diverges by the integral test. Choice E incorrectly suggests convergence with antiderivative −2x+1-2\sqrt{x+1}, which has the wrong sign. Remember that ∑1np\sum\frac{1}{n^p} converges only when p>1p > 1; here we effectively have p=1/2<1p = 1/2 < 1, guaranteeing divergence.
  11. A population correction uses ∑n=1∞1(n+1)3/2\sum_{n=1}^{\infty}\frac{1}{(n+1)^{3/2}}. Use the integral test to classify the series.
    Answer: Converges, since ∫1∞1(x+1)3/2 dx\int_1^\infty \frac{1}{(x+1)^{3/2}}\,dx converges The integral test is a fundamental approach in AP Calculus BC for assessing series behavior through integrals. For ∑_{n=1}^∞ 1/(n+1)3/2(n+1)^{3/2}, f(x) = 1/(x+1)3/2(x+1)^{3/2} is positive, continuous, and decreasing for x ≥ 1. The integral ∫_1^∞ 1/(x+1)3/2(x+1)^{3/2} dx substitutes u = x+1 to get -2 u−1/2u^{-1/2} from 2 to ∞, which converges to 2/√2. Therefore, the series converges by the integral test. A tempting distractor might suggest divergence with a logarithmic antiderivative, but the power 3/2 > 1 ensures convergence like a p-series. 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.
  12. A population model yields ∑n=3∞1nln⁡n\sum_{n=3}^{\infty}\frac{1}{n\ln n}; apply the integral test to determine convergence.​
    Answer: Diverges, since ∫3∞1xln⁡x dx\int_3^{\infty}\frac{1}{x\ln x}\,dx diverges. This problem applies the integral test to a logarithmic series. We evaluate ∫3∞1xln⁡x dx\int_3^{\infty}\frac{1}{x\ln x}\,dx using the substitution u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x}dx. The integral transforms to ∫ln⁡3∞1u du=[ln⁡u]ln⁡3∞=lim⁡b→∞ln⁡b−ln⁡(ln⁡3)=∞\int_{\ln 3}^{\infty}\frac{1}{u}\,du = \left[\ln u\right]_{\ln 3}^{\infty} = \lim_{b\to\infty}\ln b - \ln(\ln 3) = \infty. Since the improper integral diverges to infinity, the series diverges by the integral test. Choice D correctly identifies both the divergence and the antiderivative ln⁡(ln⁡x)\ln(\ln x). The series ∑1nln⁡n\sum\frac{1}{n\ln n} is a classic example of a slowly diverging series; remember that 1xln⁡x\frac{1}{x\ln x} integrates to ln⁡(ln⁡x)\ln(\ln x), which grows without bound.
  13. In a decay model, test ∑n=1∞1(n+5)3/2\sum_{n=1}^{\infty}\frac{1}{(n+5)^{3/2}} for convergence using the integral test.​
    Answer: Converges, since ∫1∞1(x+5)3/2 dx\int_1^{\infty}\frac{1}{(x+5)^{3/2}}\,dx converges. This problem tests convergence using the integral test on a shifted power function. We evaluate ∫1∞1(x+5)3/2 dx\int_1^{\infty}\frac{1}{(x+5)^{3/2}}\,dx using the substitution u=x+5u = x+5, so du=dxdu = dx. The integral becomes ∫6∞1u3/2 du=[−2u]6∞=0−(−26)=26\int_6^{\infty}\frac{1}{u^{3/2}}\,du = \left[-\frac{2}{\sqrt{u}}\right]_6^{\infty} = 0 - (-\frac{2}{\sqrt{6}}) = \frac{2}{\sqrt{6}}. Since this improper integral converges to a finite value, the series converges by the integral test. Choice C incorrectly computes the antiderivative as 2x+52\sqrt{x+5} (wrong sign and power), which would suggest divergence. For integrals of the form 1(x+a)p\frac{1}{(x+a)^p} with p>1p > 1, the series converges; memorizing this pattern helps quickly identify convergent series.
  14. For an algorithm, total error is ∑n=1∞1n3+1\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^3+1}}. Use the integral test to decide convergence.
    Answer: Converges because ∫1∞1x3+1 dx\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx converges. This problem requires applying the integral test to determine series convergence. For large xx, the integrand 1x3+1\frac{1}{\sqrt{x^3+1}} behaves like 1x3=1x3/2\frac{1}{\sqrt{x^3}} = \frac{1}{x^{3/2}}. We know that ∫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, which converges. By the limit comparison test with the integral test, since ∫1∞1x3+1 dx\int_1^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx behaves like the convergent integral ∫1∞1x3/2 dx\int_1^{\infty}\frac{1}{x^{3/2}}\,dx, it also converges. Choice C incorrectly attempts to evaluate the antiderivative as 2x3+1\frac{2}{\sqrt{x^3+1}}, which is not correct. When the integrand contains expressions like xn+c\sqrt{x^n + c} in the denominator, compare with the simpler power function xn/2x^{n/2} for large xx behavior.
  15. A file-transfer protocol retries with delays ∑n=2∞1n(ln⁡n)1/2\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^{1/2}}. Use the integral test to decide convergence.
    Answer: Diverges because ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty}\frac{1}{x(\ln x)^{1/2}}\,dx diverges. This problem tests convergence using the integral test with a logarithmic integrand. To evaluate ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty}\frac{1}{x(\ln x)^{1/2}}\,dx, we substitute u=ln⁡xu = \ln x, giving du=1xdxdu = \frac{1}{x}dx. The integral becomes ∫ln⁡2∞1u1/2 du=[2u1/2]ln⁡2∞=[2(ln⁡x)1/2]2∞\int_{\ln 2}^{\infty}\frac{1}{u^{1/2}}\,du = \left[2u^{1/2}\right]_{\ln 2}^{\infty} = \left[2(\ln x)^{1/2}\right]_2^{\infty}. As x→∞x \to \infty, (ln⁡x)1/2→∞(\ln x)^{1/2} \to \infty, so the integral diverges. By the integral test, the series diverges. Choice A incorrectly claims convergence, confusing this with the case where the exponent on ln⁡x\ln x is greater than 1. The key principle: ∫2∞1x(ln⁡x)p dx\int_2^{\infty}\frac{1}{x(\ln x)^p}\,dx converges if and only if p>1p > 1.
  16. A decay budget is ∑n=1∞1n ln⁡(n+1)\sum_{n=1}^{\infty} \frac{1}{n\,\sqrt{\ln(n+1)}}. Using the integral test, what occurs?
    Answer: Diverges because ∫1∞1xln⁡(x+1) dx\int_1^{\infty} \frac{1}{x\sqrt{\ln(x+1)}}\,dx diverges. The integral test is a fundamental method in calculus for analyzing series through integrals. For ∑_{n=1}^∞ 1/(n √(ln(n+1))), f(x) = 1/(x √(ln(x+1))) is positive, continuous, and decreasing for large x. Substituting u = √(ln(x+1)) leads to ∫ du, diverging linearly. Hence, the series diverges. A distractor proposes convergence based on ∫ dx/(x (ln(x+1))3/2(ln(x+1))^{3/2}), but that converges while the given does not. Use substitutions to simplify square roots of logs in the integral test.
  17. A signal-processing sum is ∑n=1∞1n2n\sum_{n=1}^{\infty}\frac{1}{n^2\sqrt{n}}. Use the integral test to determine convergence.
    Answer: Converges because ∫1∞1x2x dx\int_1^{\infty}\frac{1}{x^2\sqrt{x}}\,dx converges. This problem applies the integral test to a series with mixed powers. The integrand 1x2x=1x5/2\frac{1}{x^2\sqrt{x}} = \frac{1}{x^{5/2}} is a simple power function. We evaluate ∫1∞1x5/2 dx=[−23x−3/2]1∞=0−(−23)=23\int_1^{\infty}\frac{1}{x^{5/2}}\,dx = \left[-\frac{2}{3}x^{-3/2}\right]_1^{\infty} = 0 - (-\frac{2}{3}) = \frac{2}{3}. Since the integral converges to a finite value, the series converges by the integral test. Choice C incorrectly suggests using logarithms in the antiderivative, which is unnecessary for power functions. For integrals of the form ∫1∞1xp dx\int_1^{\infty}\frac{1}{x^p}\,dx, convergence occurs if and only if p>1p > 1; here p=5/2>1p = 5/2 > 1.
  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?
    Answer: Converges, since ∫2∞1x(ln⁡x)2 dx\int_2^\infty \frac{1}{x(\ln x)^2}\,dx converges 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 (lnn)2(ln n)^2), we consider the function f(x) = 1/(x (lnx)2(ln x)^2), which is positive, continuous, and decreasing for x ≥ 2. The improper integral ∫_2^∞ 1/(x (lnx)2(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.
  19. A decay process is approximated by ∑n=2∞1n(ln⁡n)3\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^3}. Using the integral test, determine convergence.
    Answer: Converges because ∫2∞1x(ln⁡x)3 dx\int_2^{\infty}\frac{1}{x(\ln x)^3}\,dx converges. This problem tests convergence using the integral test with a higher power of logarithm. To evaluate ∫2∞1x(ln⁡x)3 dx\int_2^{\infty}\frac{1}{x(\ln x)^3}\,dx, we substitute u=ln⁡xu = \ln x, giving du=1xdxdu = \frac{1}{x}dx. The integral becomes ∫ln⁡2∞1u3 du=[−12u2]ln⁡2∞=0−(−12(ln⁡2)2)=12(ln⁡2)2\int_{\ln 2}^{\infty}\frac{1}{u^3}\,du = \left[-\frac{1}{2u^2}\right]_{\ln 2}^{\infty} = 0 - (-\frac{1}{2(\ln 2)^2}) = \frac{1}{2(\ln 2)^2}. Since the integral converges to a finite value, the series converges by the integral test. Choice C incorrectly evaluates the antiderivative as ln⁡(ln⁡x)\ln(\ln x), which would be correct for ∫1xln⁡x dx\int\frac{1}{x\ln x}\,dx but not for the cube power. Remember: ∫2∞1x(ln⁡x)p dx\int_2^{\infty}\frac{1}{x(\ln x)^p}\,dx converges if and only if p>1p > 1.
  20. For a damping sequence, use the integral test to classify ∑n=2∞1n(ln⁡n)1/2\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^{1/2}}.​
    Answer: Diverges, since ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty}\frac{1}{x(\ln x)^{1/2}}\,dx diverges. This problem tests convergence of a logarithmic power series using the integral test. We evaluate ∫2∞1x(ln⁡x)1/2 dx\int_2^{\infty}\frac{1}{x(\ln x)^{1/2}}\,dx with substitution u=ln⁡xu = \ln x, giving du=1xdxdu = \frac{1}{x}dx. The integral becomes ∫ln⁡2∞1u1/2 du=[2u1/2]ln⁡2∞=lim⁡b→∞2b−2ln⁡2=∞\int_{\ln 2}^{\infty}\frac{1}{u^{1/2}}\,du = \left[2u^{1/2}\right]_{\ln 2}^{\infty} = \lim_{b\to\infty}2\sqrt{b} - 2\sqrt{\ln 2} = \infty. Since the improper integral diverges, the series diverges by the integral test. Choice E incorrectly suggests convergence with antiderivative −2ln⁡x-2\sqrt{\ln x}, which has the wrong sign. The key insight is that 1x(ln⁡x)p\frac{1}{x(\ln x)^p} converges only when p>1p > 1; here p=1/2<1p = 1/2 < 1, ensuring divergence.

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