Calculus 2 Quiz: Integral Test
20 questions · exam conditions
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Integral TestQuestion 1 of 20

The series ∑n=2∞1n(ln⁡n)p\sum_{n=2}^\infty \frac{1}{n(\ln n)^p} is analyzed using the Integral Test. For which values of the constant pp does the corresponding improper integral converge?

p>1p > 1
p≥1p \ge 1
p<1p < 1
p>0p > 0
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Calculus 2 Quiz

Calculus 2 Quiz: Integral Test

Practice Integral Test in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Integral Test, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

The series ∑n=2∞1n(ln⁡n)p\sum_{n=2}^\infty \frac{1}{n(\ln n)^p} is analyzed using the Integral Test. For which values of the constant pp does the corresponding improper integral converge?

  1. p>1p > 1 (correct answer)
  2. p≥1p \ge 1
  3. p<1p < 1
  4. p>0p > 0
Explanation: To apply the Integral Test, we evaluate ∫2∞1x(ln⁡x)pdx\int_2^\infty \frac{1}{x(\ln x)^p} dx. We use the substitution u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x} dx. The limits of integration become ln⁡2\ln 2 and ∞\infty. The integral transforms to ∫ln⁡2∞1updu\int_{\ln 2}^\infty \frac{1}{u^p} du. This is a p-integral, which is known to converge if and only if the power p>1p > 1. Therefore, the original series converges for p>1p > 1.

Question 2

A function h(x)=ax+bcx3+dx2+ex+fh(x) = \frac{ax + b}{cx^3 + dx^2 + ex + f} where a,b,c,d,e,fa, b, c, d, e, f are positive constants with c>0c > 0, is being considered for the integral test. For large values of xx, the series ∑n=1∞h(n)\sum_{n=1}^{\infty} h(n) will have the same convergence behavior as which of the following?

  1. ∑n=1∞acn2\sum_{n=1}^{\infty} \frac{a}{cn^2} (correct answer)
  2. ∑n=1∞acn3\sum_{n=1}^{\infty} \frac{a}{cn^3}
  3. ∑n=1∞bfn\sum_{n=1}^{\infty} \frac{b}{fn}
  4. ∑n=1∞a+bc+d+e+f⋅1n2\sum_{n=1}^{\infty} \frac{a+b}{c+d+e+f} \cdot \frac{1}{n^2}
Explanation: For large xx, the highest degree terms dominate: h(x)≈axcx3=acx2h(x) \approx \frac{ax}{cx^3} = \frac{a}{cx^2}. By the limit comparison test or asymptotic analysis, ∑h(n)\sum h(n) behaves like ∑acn2\sum \frac{a}{cn^2}, which converges since it's a pp-series with p=2>1p = 2 > 1. Choice B gives the wrong power (n3n^3 instead of n2n^2). Choice C focuses on lower-order terms that don't dominate for large nn. Choice D incorrectly combines all coefficients rather than using the leading terms.

Question 3

Using the reasoning of the Integral Test (or the p-series test), for which value of kk does the series ∑n=1∞nkn3+1\sum_{n=1}^\infty \frac{n^k}{n^3+1} diverge?

  1. k=0k=0
  2. k=1k=1
  3. k=1.9k=1.9
  4. k=2k=2 (correct answer)
Explanation: For large nn, the term an=nkn3+1a_n = \frac{n^k}{n^3+1} behaves like nkn3=1n3−k\frac{n^k}{n^3} = \frac{1}{n^{3-k}}. This is a p-series with p=3−kp = 3-k. A p-series diverges if p≤1p \le 1. So, we need 3−k≤13-k \le 1, which means k≥2k \ge 2. Of the choices given, only k=2k=2 satisfies this condition. For k=2k=2, the integral ∫1∞x2x3+1dx\int_1^\infty \frac{x^2}{x^3+1} dx diverges, confirming the series diverges.

Question 4

What is the result of applying the Integral Test to the series ∑n=1∞1n(1+n)\sum_{n=1}^\infty \frac{1}{\sqrt{n}(1+\sqrt{n})}?

  1. The series converges because the corresponding integral converges to ln⁡(2)\ln(2).
  2. The series converges because the corresponding integral converges to 2.
  3. The series diverges because the corresponding integral diverges. (correct answer)
  4. The test is not applicable as the function is not decreasing.
Explanation: The integral is ∫1∞1x(1+x)dx\int_1^\infty \frac{1}{\sqrt{x}(1+\sqrt{x})} dx. Use substitution with u=1+xu = 1+\sqrt{x}, so du=12xdxdu = \frac{1}{2\sqrt{x}} dx, or 2du=1xdx2du = \frac{1}{\sqrt{x}} dx. The integral becomes ∫2∞2udu=[2ln⁡∣u∣]2∞\int_2^\infty \frac{2}{u} du = [2 \ln|u|]_2^\infty. This evaluates to ∞−2ln⁡(2)\infty - 2\ln(2), which diverges. Therefore, the series diverges by the Integral Test.

Question 5

Let p(x)p(x) be a polynomial of degree d>1d > 1 with a positive leading coefficient, and p(x)>0p(x) > 0 for x≥1x \ge 1. According to the Integral Test, for what values of kk does the series ∑n=1∞nkp(n)\sum_{n=1}^\infty \frac{n^k}{p(n)} converge?

  1. k−d<−1k - d < -1 (correct answer)
  2. k−d>1k - d > 1
  3. k−d≤−1k - d \le -1
  4. k>0k > 0
Explanation: For large xx, the function f(x)=xkp(x)f(x) = \frac{x^k}{p(x)} behaves like xkcxd=Cxk−d\frac{x^k}{c x^d} = C x^{k-d} for some constants c,C>0c, C > 0. We can apply the Integral Test by analyzing the convergence of ∫1∞xk−ddx\int_1^\infty x^{k-d} dx. This is a p-integral with power p=−(k−d)=d−kp = -(k-d) = d-k. The integral converges if this power is greater than 1, so d−k>1d-k > 1. Rearranging this gives k<d−1k < d-1, which is equivalent to k−d<−1k-d < -1.

Question 6

Let S=∑n=1∞1n3S = \sum_{n=1}^\infty \frac{1}{n^3}. According to the remainder estimate derived from the Integral Test, the error R5=S−∑n=151n3R_5 = S - \sum_{n=1}^5 \frac{1}{n^3} is bounded by which inequality?

  1. R5≤150R_5 \le \frac{1}{50}
  2. 172≤R5≤150\frac{1}{72} \le R_5 \le \frac{1}{50} (correct answer)
  3. R5=150R_5 = \frac{1}{50}
  4. 0≤R5≤1720 \le R_5 \le \frac{1}{72}
Explanation: The remainder estimate for the Integral Test states that ∫N+1∞f(x)dx≤RN≤∫N∞f(x)dx\int_{N+1}^\infty f(x) dx \le R_N \le \int_N^\infty f(x) dx. Here, N=5N=5 and f(x)=1/x3f(x) = 1/x^3. We compute the bounds: ∫5∞1x3dx=[−12x2]5∞=0−(−12(52))=150\int_5^\infty \frac{1}{x^3} dx = [-\frac{1}{2x^2}]_5^\infty = 0 - (-\frac{1}{2(5^2)}) = \frac{1}{50}. And ∫6∞1x3dx=[−12x2]6∞=0−(−12(62))=172\int_6^\infty \frac{1}{x^3} dx = [-\frac{1}{2x^2}]_6^\infty = 0 - (-\frac{1}{2(6^2)}) = \frac{1}{72}. Therefore, 172≤R5≤150\frac{1}{72} \le R_5 \le \frac{1}{50}.

Question 7

A student wants to use the Integral Test on the series ∑n=1∞nn2+4\sum_{n=1}^\infty \frac{n}{n^2+4}. They correctly identify that the corresponding function f(x)=xx2+4f(x) = \frac{x}{x^2+4} is positive and continuous for x≥1x \ge 1. However, they must also verify that f(x)f(x) is eventually decreasing. For what values of xx is f(x)f(x) decreasing?

  1. x>0x > 0
  2. x>1x > 1
  3. x>2x > 2 (correct answer)
  4. x>4x > 4
Explanation: To find where f(x)f(x) is decreasing, we must find where its derivative, f′(x)f'(x), is negative. Using the quotient rule, f′(x)=(x2+4)(1)−x(2x)(x2+4)2=4−x2(x2+4)2f'(x) = \frac{(x^2+4)(1) - x(2x)}{(x^2+4)^2} = \frac{4-x^2}{(x^2+4)^2}. For f′(x)<0f'(x) < 0, the numerator must be negative, since the denominator is always positive. 4−x2<04 - x^2 < 0 implies x2>4x^2 > 4, which for positive xx means x>2x > 2. The Integral Test can be applied starting from N=3N=3.

Question 8

Let f(x)f(x) be a positive, continuous, and decreasing function for x≥1x \ge 1. If ∑n=1∞f(n)\sum_{n=1}^\infty f(n) converges to SS by the Integral Test, which of the following provides the tightest guaranteed bound for SS?

  1. S=∫1∞f(x)dxS = \int_1^\infty f(x) dx
  2. S<f(1)+∫1∞f(x)dxS < f(1) + \int_1^\infty f(x) dx (correct answer)
  3. S>f(1)+∫1∞f(x)dxS > f(1) + \int_1^\infty f(x) dx
  4. S<∫1∞f(x)dxS < \int_1^\infty f(x) dx
Explanation: The sum of the series can be visualized as the sum of areas of rectangles. For a decreasing function, the sum of right-hand rectangles is an underestimate of the integral, and the sum of left-hand rectangles is an overestimate. Specifically, ∑n=2∞f(n)<∫1∞f(x)dx\sum_{n=2}^\infty f(n) < \int_1^\infty f(x) dx. Adding the first term, S=f(1)+∑n=2∞f(n)<f(1)+∫1∞f(x)dxS = f(1) + \sum_{n=2}^\infty f(n) < f(1) + \int_1^\infty f(x) dx. This is a standard result from the geometric argument for the Integral Test.

Question 9

For which of the following series are all the conditions of the Integral Test (positive, continuous, eventually decreasing) satisfied?

  1. ∑n=1∞n!2n\sum_{n=1}^\infty \frac{n!}{2^n}
  2. ∑n=1∞ncos⁡(1/n)\sum_{n=1}^\infty n \cos(1/n)
  3. ∑n=1∞3−sin⁡(n)n\sum_{n=1}^\infty \frac{3-\sin(n)}{n}
  4. ∑n=2∞1nln⁡n\sum_{n=2}^\infty \frac{1}{n \sqrt{\ln n}} (correct answer)
Explanation: A: The factorial function is not defined for non-integers, so a continuous function f(x)f(x) cannot be formed. B: The terms ncos⁡(1/n)n \cos(1/n) do not approach 0, so the series diverges by the nth Term Test, and the corresponding function is not decreasing. C: The function f(x)=(3−sin⁡x)/xf(x) = (3-\sin x)/x is not guaranteed to be decreasing due to the oscillating numerator. D: The function f(x)=1xln⁡xf(x) = \frac{1}{x \sqrt{\ln x}} is positive, continuous, and decreasing for x≥2x \ge 2 because the denominator is a product of positive, increasing functions.

Question 10

Consider the series ∑n=1∞ln⁡nn\sum_{n=1}^\infty \frac{\ln n}{n}. Applying the Integral Test leads to the improper integral ∫1∞ln⁡xxdx\int_1^\infty \frac{\ln x}{x} dx. What is the correct evaluation of this integral and the conclusion about the series?

  1. The integral converges to 1, so the series converges.
  2. The integral converges to 0, so the series converges.
  3. The integral diverges, so the series diverges. (correct answer)
  4. The test is invalid because the first term is zero.
Explanation: To evaluate ∫1∞ln⁡xxdx\int_1^\infty \frac{\ln x}{x} dx, use the substitution u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x} dx. The integral becomes ∫0∞udu=[u22]0∞\int_0^\infty u du = [\frac{u^2}{2}]_0^\infty, which diverges to ∞\infty. Therefore, by the Integral Test, the series diverges. The fact that the first term is zero does not invalidate the test; convergence is determined by the tail of the series, so we can start the test at n=2n=2.

Question 11

When using the Integral Test to determine the convergence of the series ∑n=2∞ln⁡nn2\sum_{n=2}^\infty \frac{\ln n}{n^2}, one must evaluate the integral ∫2∞ln⁡xx2dx\int_2^\infty \frac{\ln x}{x^2} dx. What is the value of this integral?

  1. The integral diverges
  2. ln⁡2+12\frac{\ln 2 + 1}{2} (correct answer)
  3. ln⁡22\frac{\ln 2}{2}
  4. 11
Explanation: We use integration by parts with u=ln⁡xu = \ln x and dv=x−2dxdv = x^{-2} dx. Then du=1xdxdu = \frac{1}{x} dx and v=−x−1v = -x^{-1}. The integral is ∫ln⁡xx2dx=−ln⁡xx−∫−1x2dx=−ln⁡xx−1x\int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} dx = -\frac{\ln x}{x} - \frac{1}{x}. Evaluating the improper integral: [−ln⁡xx−1x]2∞=lim⁡b→∞(−ln⁡bb−1b)−(−ln⁡22−12)[-\frac{\ln x}{x} - \frac{1}{x}]_2^\infty = \lim_{b \to \infty}(-\frac{\ln b}{b} - \frac{1}{b}) - (-\frac{\ln 2}{2} - \frac{1}{2}). The limit is 0, so the value is ln⁡2+12\frac{\ln 2 + 1}{2}.

Question 12

You are given that for a function f(x)f(x) which is positive, continuous, and decreasing for x≥1x \ge 1, the value of ∫1∞f(x)dx=7\int_1^\infty f(x) dx = 7. What can be concluded about the sum SS of the series ∑n=1∞f(n)\sum_{n=1}^\infty f(n)?

  1. The series diverges.
  2. The series converges and S=7S = 7.
  3. The series converges and S>7S > 7. (correct answer)
  4. The series converges and S<7S < 7.
Explanation: Since the integral converges, the series must also converge by the Integral Test. The relationship between the sum and the integral can be seen by visualizing the sum as a left-hand Riemann sum for the integral. For a decreasing function, a left-hand Riemann sum overestimates the integral. Thus, S=∑n=1∞f(n)>∫1∞f(x)dx=7S = \sum_{n=1}^\infty f(n) > \int_1^\infty f(x) dx = 7.

Question 13

Let ∑n=1∞an\sum_{n=1}^\infty a_n be a series of positive terms where an=f(n)a_n = f(n) for a continuous and decreasing function f(x)f(x). Let g(x)=∫1xf(t)dtg(x) = \int_1^x f(t) dt. Which condition guarantees that the series ∑an\sum a_n converges?

  1. lim⁡x→∞f(x)=0\lim_{x \to \infty} f(x) = 0
  2. lim⁡x→∞g(x)\lim_{x \to \infty} g(x) exists and is finite. (correct answer)
  3. lim⁡x→∞g′(x)=0\lim_{x \to \infty} g'(x) = 0
  4. g(x)g(x) is an increasing function for x≥1x \ge 1.
Explanation: The Integral Test states that ∑an\sum a_n converges if and only if the improper integral ∫1∞f(t)dt\int_1^\infty f(t) dt converges. The value of this improper integral is, by definition, lim⁡x→∞∫1xf(t)dt=lim⁡x→∞g(x)\lim_{x \to \infty} \int_1^x f(t) dt = \lim_{x \to \infty} g(x). Thus, the series converges if this limit exists and is finite. The conditions in A and C are necessary for convergence but not sufficient (e.g., f(x)=1/xf(x)=1/x). The condition in D is always true for a positive function f(x)f(x) and provides no information about convergence.

Question 14

A student attempts to prove the convergence of ∑n=1∞2+cos⁡(n)n2\sum_{n=1}^\infty \frac{2+\cos(n)}{n^2} using the Integral Test with f(x)=2+cos⁡(x)x2f(x) = \frac{2+\cos(x)}{x^2}. Why is this application of the Integral Test invalid?

  1. The function f(x)f(x) is not continuous for x≥1x \ge 1.
  2. The function f(x)f(x) is not positive for x≥1x \ge 1.
  3. The integral ∫1∞f(x)dx\int_1^\infty f(x) dx diverges.
  4. The function f(x)f(x) is not eventually decreasing. (correct answer)
Explanation: To apply the Integral Test, the function f(x)f(x) must be positive, continuous, and eventually decreasing. The function f(x)=2+cos⁡(x)x2f(x) = \frac{2+\cos(x)}{x^2} is positive and continuous for x≥1x \ge 1. However, due to the oscillating cos⁡(x)\cos(x) term in the numerator, the function is not monotonically decreasing for all xx greater than some NN. The derivative contains an oscillating term that prevents it from staying negative. Therefore, a key condition of the Integral Test is not met.

Question 15

If the Integral Test is successfully applied to a series ∑n=1∞an\sum_{n=1}^\infty a_n and shows convergence, which of the following statements is not necessarily true?

  1. The series ∑n=100∞an\sum_{n=100}^\infty a_n converges.
  2. The value of the series' sum is equal to the value of the corresponding integral. (correct answer)
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0.
  4. The sequence of partial sums SN=∑n=1NanS_N = \sum_{n=1}^N a_n is bounded.
Explanation: The Integral Test determines whether a series converges or diverges, but it does not find the sum. The value of the sum is generally not equal to the value of the integral. If a series converges, its tail (A) must also converge, its terms must approach zero (C), and its sequence of partial sums must be bounded and convergent (D).

Question 16

Consider the series ∑n=2∞1n(ln⁡n)p\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} where pp is a positive constant. For which values of pp does the integral test guarantee convergence of this series?

  1. p>1p > 1 (correct answer)
  2. p≥1p \geq 1
  3. p>0p > 0
  4. p≥2p \geq 2
Explanation: To apply the integral test, we evaluate ∫2∞1x(ln⁡x)pdx\int_2^{\infty} \frac{1}{x(\ln x)^p} dx. Using substitution u=ln⁡xu = \ln x, du=1xdxdu = \frac{1}{x}dx, this becomes ∫ln⁡2∞1updu\int_{\ln 2}^{\infty} \frac{1}{u^p} du. This integral converges when p>1p > 1 and diverges when p≤1p \leq 1. Choice B is incorrect because when p=1p = 1, the integral diverges. Choice C is too broad as it includes values 0<p≤10 < p \leq 1 where the series diverges. Choice D is too restrictive as convergence occurs for all p>1p > 1, not just p≥2p \geq 2.

Question 17

A student claims that since ∫1∞sin⁡2xx2dx\int_1^{\infty} \frac{\sin^2 x}{x^2} dx converges, the series ∑n=1∞sin⁡2nn2\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} must also converge by the integral test. Which statement best explains the validity of this reasoning?

  1. The reasoning is correct because f(x)=sin⁡2xx2f(x) = \frac{\sin^2 x}{x^2} satisfies all conditions for the integral test
  2. The reasoning is correct because both the integral and series involve the same function values
  3. The reasoning is incorrect because the integral test requires sin⁡2x\sin^2 x to be constant
  4. The reasoning is incorrect because f(x)=sin⁡2xx2f(x) = \frac{\sin^2 x}{x^2} is not monotonic for x≥1x \geq 1 (correct answer)
Explanation: When you encounter a question about the integral test, remember that this test has specific conditions that must be satisfied - the function must be positive, continuous, and decreasing (monotonic) on the interval of interest. The integral test states that for a function f(x)f(x) that is positive, continuous, and decreasing for x≥Nx \geq N, the series ∑n=N∞f(n)\sum_{n=N}^{\infty} f(n) and the integral ∫N∞f(x)dx\int_N^{\infty} f(x) dx either both converge or both diverge. The critical issue here is that f(x)=sin⁡2xx2f(x) = \frac{\sin^2 x}{x^2} is not monotonically decreasing for x≥1x \geq 1. While 1x2\frac{1}{x^2} decreases, the sin⁡2x\sin^2 x factor oscillates between 0 and 1, causing the entire function to oscillate rather than decrease consistently. This violates the monotonicity requirement, making the integral test inapplicable. Choice A is wrong because f(x)f(x) fails the monotonicity condition. Choice B misses the point entirely - having the same function values doesn't guarantee the integral test applies if the conditions aren't met. Choice C incorrectly suggests sin⁡2x\sin^2 x must be constant, which isn't a requirement of the integral test. The correct answer is D because the lack of monotonicity invalidates the student's reasoning. Study tip: Always verify all three conditions for the integral test: positive, continuous, AND decreasing. Oscillating functions, even if they involve decreasing factors, typically fail the monotonicity requirement.

Question 18

For the series ∑n=1∞1n1+1/n\sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}, a student attempts to apply the integral test by evaluating ∫1∞1x1+1/xdx\int_1^{\infty} \frac{1}{x^{1+1/x}} dx. What is the primary obstacle in using this approach?

  1. The function f(x)=1x1+1/xf(x) = \frac{1}{x^{1+1/x}} is not positive for x≥1x \geq 1
  2. The function f(x)=1x1+1/xf(x) = \frac{1}{x^{1+1/x}} is not eventually decreasing for x≥1x \geq 1
  3. The function f(x)=1x1+1/xf(x) = \frac{1}{x^{1+1/x}} is not continuous at x=1x = 1
  4. The integral ∫1∞1x1+1/xdx\int_1^{\infty} \frac{1}{x^{1+1/x}} dx is too difficult to evaluate analytically (correct answer)
Explanation: The integral test is a powerful tool for determining series convergence, but it requires that you can actually evaluate or analyze the corresponding integral. Sometimes the biggest obstacle isn't whether the function meets the test's conditions, but whether the integral itself is manageable. Let's examine why option D is correct. The integral ∫1∞1x1+1/xdx\int_1^{\infty} \frac{1}{x^{1+1/x}} dx presents a significant analytical challenge because the exponent 1+1x1+\frac{1}{x} is not constant—it's a function of xx. This creates an integral that doesn't fit standard forms and lacks obvious substitutions or integration techniques. While the integral might converge or diverge, determining which requires advanced methods beyond typical calculus courses, making the integral test impractical here. Now for the incorrect options: Option A is wrong because f(x)=1x1+1/xf(x) = \frac{1}{x^{1+1/x}} is clearly positive for all x≥1x \geq 1 since we're taking the reciprocal of a positive quantity. Option B is incorrect—while checking if the function is eventually decreasing requires some work, this isn't the primary obstacle; the function's behavior can be analyzed using derivatives. Option C is false because the function is continuous at x=1x = 1; as x→1x \to 1, we get 111+1/1=112=1\frac{1}{1^{1+1/1}} = \frac{1}{1^2} = 1, which matches the function's value at x=1x = 1. When applying the integral test, always consider whether the resulting integral is actually solvable with available techniques. If not, explore alternative convergence tests like comparison or ratio tests.

Question 19

Consider two series: ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n} and ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}. Using the integral test, what can be concluded about the convergence of these series?

  1. Both series converge because both corresponding improper integrals converge
  2. Both series diverge because the integrands approach zero too slowly
  3. The first series diverges and the second series converges (correct answer)
  4. The first series converges and the second series diverges
Explanation: When you encounter series with logarithmic terms, the integral test is often your best tool. The key insight is that small changes in the exponent of logarithmic terms can dramatically affect convergence behavior. For the first series ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n}, we test the corresponding integral ∫2∞1xln⁡xdx\int_2^{\infty} \frac{1}{x \ln x} dx. Using the substitution u=ln⁡xu = \ln x (so du=1xdxdu = \frac{1}{x}dx), this becomes ∫ln⁡2∞1udu=ln⁡u∣ln⁡2∞=∞\int_{\ln 2}^{\infty} \frac{1}{u} du = \ln u \Big|_{\ln 2}^{\infty} = \infty. Since the integral diverges, the series diverges. For the second series ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}, we evaluate ∫2∞1x(ln⁡x)2dx\int_2^{\infty} \frac{1}{x (\ln x)^2} dx. Using the same substitution, we get ∫ln⁡2∞1u2du=−1u∣ln⁡2∞=1ln⁡2\int_{\ln 2}^{\infty} \frac{1}{u^2} du = -\frac{1}{u} \Big|_{\ln 2}^{\infty} = \frac{1}{\ln 2}. This integral converges, so the series converges. Answer choice A is wrong because only the second integral converges. Choice B incorrectly claims both diverge—while the rate of approach to zero matters, you must actually compute the integrals to determine convergence. Choice D reverses the correct conclusion. Study tip: Remember that 1nln⁡n\frac{1}{n \ln n} behaves like the harmonic series (diverges), while adding any positive power to the logarithm, like (ln⁡n)2(\ln n)^2, typically ensures convergence. The integral test makes these distinctions precise through substitution techniques.

Question 20

The integral test is applied to ∑n=1∞ln⁡nn2\sum_{n=1}^{\infty} \frac{\ln n}{n^2} (with the convention that ln⁡1=0\ln 1 = 0). To evaluate ∫1∞ln⁡xx2dx\int_1^{\infty} \frac{\ln x}{x^2} dx, integration by parts is used with u=ln⁡xu = \ln x and dv=1x2dxdv = \frac{1}{x^2} dx. What is the value of this improper integral?

  1. The integral diverges because ln⁡x→∞\ln x \to \infty as x→∞x \to \infty
  2. 00
  3. 11 (correct answer)
  4. 12\frac{1}{2}
Explanation: When you encounter an infinite series and need to apply the integral test, you're essentially checking whether the corresponding improper integral converges or diverges to determine the series' behavior. To evaluate ∫1∞ln⁡xx2dx\int_1^{\infty} \frac{\ln x}{x^2} dx using integration by parts with u=ln⁡xu = \ln x and dv=1x2dxdv = \frac{1}{x^2} dx, you get du=1xdxdu = \frac{1}{x} dx and v=−1xv = -\frac{1}{x}. Applying the formula: ∫1∞ln⁡xx2dx=lim⁡t→∞[−ln⁡xx]1t+∫1∞1x2dx\int_1^{\infty} \frac{\ln x}{x^2} dx = \lim_{t \to \infty} \left[-\frac{\ln x}{x}\right]_1^t + \int_1^{\infty} \frac{1}{x^2} dx The first term evaluates to lim⁡t→∞(−ln⁡tt+0)=0\lim_{t \to \infty} \left(-\frac{\ln t}{t} + 0\right) = 0 (since ln⁡t\ln t grows slower than tt). The second integral is ∫1∞1x2dx=1\int_1^{\infty} \frac{1}{x^2} dx = 1. Therefore, the total integral equals 0+1=10 + 1 = 1. Choice A incorrectly focuses on the behavior of ln⁡x\ln x alone without considering that it's divided by x2x^2, which dominates the growth. Choice B likely results from only evaluating the boundary term −ln⁡xx-\frac{\ln x}{x} and ignoring the second integral from integration by parts. Choice D might come from computational errors or confusing this with similar integrals. Remember: integration by parts produces two terms that you must evaluate separately. Don't let the divergent behavior of individual components fool you—the complete calculation determines convergence. Always check that ln⁡tt→0\frac{\ln t}{t} \to 0 as t→∞t \to \infty using L'Hôpital's rule if needed.