Integral Test for Convergence
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AP Calculus BC › Integral Test for Convergence
A cooling model uses $\sum_{n=1}^{\infty} \frac{1}{(n+1)^{2/3}}$. Using the integral test, what happens?
Diverges because $\int_1^{\infty} \frac{1}{(x+1)^{2/3}},dx$ diverges.
Converges because $\int_1^{\infty} \frac{1}{(x+1)^{2/3}},dx$ converges.
Converges because $\int_1^{\infty} \frac{1}{\sqrt{x+1}},dx$ converges.
Converges because $\int_1^{\infty} (x+1)^{2/3},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{(x+1)^{5/3}},dx$ diverges.
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.
A decay sequence is $\sum_{n=1}^{\infty} \frac{1}{(n^2+1)^{3/4}}$. Use the integral test to decide.
Converges because $\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}},dx$ converges.
Diverges because $\int_1^{\infty} (x^2+1)^{3/4},dx$ diverges.
Diverges because $\int_1^{\infty} \frac{1}{x^{3/2}},dx$ diverges.
Converges because $\int_1^{\infty} \frac{1}{\sqrt{x^2+1}},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{(x^2+1)^{3/4}},dx$ converges.
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.
A response curve uses $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$. What does the integral test conclude?
Converges because $\int_1^{\infty} \sqrt{x},dx$ converges.
Converges because $\int_1^{\infty} \frac{1}{\sqrt{x}},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{x^{3/2}},dx$ diverges.
Diverges because $\int_1^{\infty} \frac{1}{\sqrt{x}},dx$ diverges.
Converges because $\int_1^{\infty} \frac{1}{x^{3/2}},dx$ 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.
A signal decay is $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$. Using the integral test, determine convergence.
Converges because $\int_2^{\infty} \frac{1}{x\ln x},dx$ converges.
Diverges because $\int_2^{\infty} \frac{1}{x\ln x},dx$ diverges.
Diverges because $\int_2^{\infty} \frac{1}{(\ln x)},dx$ diverges.
Converges because $\int_2^{\infty} \frac{1}{x(\ln x)^2},dx$ diverges.
Converges because $\int_2^{\infty} \frac{\ln x}{x},dx$ converges.
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.
A queue delay model uses $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{1/2}}$. What does the integral test say?
Diverges because $\int_2^{\infty} \frac{1}{x(\ln x)^{1/2}},dx$ diverges.
Converges because $\int_2^{\infty} \frac{1}{x(\ln x)^{1/2}},dx$ converges.
Converges because $\int_2^{\infty} \frac{\ln x}{x},dx$ converges.
Diverges because $\int_2^{\infty} \frac{1}{x(\ln x)^{3/2}},dx$ converges.
Converges because $\int_2^{\infty} \frac{1}{x\ln x},dx$ 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.
A learning-rate schedule is $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)(\ln\ln n)^2}$. What does the integral test conclude?
Diverges because $\int_2^{\infty} \frac{1}{x\ln x},dx$ converges.
Diverges because $\int_2^{\infty} \frac{1}{x(\ln x)(\ln\ln x)^2},dx$ converges.
Converges because $\int_2^{\infty} \frac{1}{x(\ln x)(\ln\ln x)^2},dx$ converges.
Diverges because $\int_2^{\infty} \frac{1}{(\ln\ln x)^2},dx$ diverges.
Converges because $\int_2^{\infty} \frac{1}{x(\ln x)(\ln\ln x)},dx$ converges.
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.
A resource model sums $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+n}}$. Using the integral test, what occurs?
Converges because $\int_1^{\infty} \frac{1}{x},dx$ converges.
Converges because $\int_1^{\infty} \frac{1}{\sqrt{x^2+x}},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{\sqrt{x^2+x}},dx$ diverges.
Diverges because $\int_1^{\infty} \frac{1}{x^2+x},dx$ converges.
Converges because $\int_1^{\infty} \sqrt{x^2+x},dx$ converges.
Explanation
The integral test offers an essential technique in calculus for series convergence testing with integrals. For $∑_{n=1}$^∞ $1/√(n^2$ + n), f(x) = $1/√(x^2$ + 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.
A transport model uses $\sum_{n=1}^{\infty} \frac{1}{(n+4)^{3/2}}$. Using the integral test, determine convergence.
Converges because $\int_1^{\infty} \frac{1}{(x+4)^{3/2}},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{x+4},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{(x+4)^{3/2}},dx$ diverges.
Diverges because $\int_1^{\infty} (x+4)^{3/2},dx$ diverges.
Converges because $\int_1^{\infty} \frac{1}{\sqrt{x+4}},dx$ 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.
A concentration model uses $\sum_{n=1}^{\infty} \frac{1}{n^2!+!1}$. Using the integral test, determine convergence.
Diverges because $\int_1^{\infty} \frac{1}{x},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{x^2+1},dx$ diverges.
Converges because $\int_1^{\infty} (x^2+1),dx$ converges.
Converges because $\int_1^{\infty} \frac{1}{x^2+1},dx$ converges.
Diverges because $\int_1^{\infty} \frac{1}{x^2},dx$ diverges.
Explanation
The integral test is an important tool in calculus for evaluating series through integrals. For $∑_{n=1}$^∞ $1/(n^2$ + 1), f(x) = $1/(x^2$ + 1) is positive, continuous, and decreasing. The integral ∫_1^∞ $dx/(x^2$ + 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.
A cooling experiment uses $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$. Using the integral test, what occurs?
Diverges, since $\int_2^\infty \frac{1}{x\ln x},dx$ diverges
Converges, since $\int_2^\infty \frac{1}{x\ln x},dx=\left.-\frac{1}{\ln x}\right|_2^\infty$
Converges, since $\int_2^\infty \frac{1}{x\ln x},dx$ converges
Converges, because $\frac{1}{n\ln n}$ is not continuous at $n=1$
Diverges, since $\int_2^\infty \frac{1}{x\ln x},dx=\left.\frac{1}{\ln x}\right|_2^\infty$
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.