AP Calculus BC Quiz: Defining Convergent And Divergent Infinite Series
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Defining Convergent And Divergent Infinite SeriesQuestion 1 of 20

A ball's bounce heights follow hn=3(23)nh_n=3\left(\frac{2}{3}\right)^n meters. Does ∑n=1∞hn\sum_{n=1}^{\infty} h_n converge or diverge?

Diverges by the harmonic series test
Converges because it is a geometric series with ∣r∣<1|r|<1
Diverges because terms do not approach 00
Converges by the integral test
Diverges because it is a geometric series with ∣r∣>1|r|>1
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AP Calculus BC Quiz

AP Calculus BC Quiz: Defining Convergent And Divergent Infinite Series

Practice Defining Convergent And Divergent Infinite Series 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 Defining Convergent And Divergent Infinite Series, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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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.

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Question 1

A ball's bounce heights follow hn=3(23)nh_n=3\left(\frac{2}{3}\right)^n meters. Does ∑n=1∞hn\sum_{n=1}^{\infty} h_n converge or diverge?

  1. Diverges by the harmonic series test
  2. Converges because it is a geometric series with ∣r∣<1|r|<1 (correct answer)
  3. Diverges because terms do not approach 00
  4. Converges by the integral test
  5. Diverges because it is a geometric series with ∣r∣>1|r|>1
Explanation: This question tests your ability to classify infinite series as convergent or divergent. The series ∑n=1∞3(23)n\sum_{n=1}^{\infty} 3\left(\frac{2}{3}\right)^n is a geometric series with first term a=3⋅23=2a = 3 \cdot \frac{2}{3} = 2 and common ratio r=23r = \frac{2}{3}. Since ∣r∣=23<1|r| = \frac{2}{3} < 1, the geometric series converges to a1−r=21−23=6\frac{a}{1-r} = \frac{2}{1-\frac{2}{3}} = 6. Choice E incorrectly claims ∣r∣>1|r| > 1, which would be true if the ratio were 32\frac{3}{2} instead of 23\frac{2}{3}. Remember: geometric series ∑arn\sum ar^n converge when ∣r∣<1|r| < 1 and diverge when ∣r∣≥1|r| \geq 1.

Question 2

A decaying input produces terms an=nn+1a_n=\frac{n}{n+1}. Does ∑n=1∞an\sum_{n=1}^{\infty} a_n converge or diverge?

  1. Converges by the ratio test
  2. Converges because an<1a_n<1 for all nn
  3. Diverges because lim⁡n→∞an≠0\lim_{n\to\infty} a_n\neq 0 (correct answer)
  4. Converges because lim⁡n→∞an=1\lim_{n\to\infty} a_n=1
  5. Diverges by comparison to ∑1n2\sum \frac{1}{n^2}
Explanation: This question tests your ability to classify infinite series as convergent or divergent. For the series ∑n=1∞nn+1\sum_{n=1}^{\infty} \frac{n}{n+1}, we first check if the terms approach zero: lim⁡n→∞nn+1=lim⁡n→∞11+1n=1≠0\lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n}} = 1 \neq 0. By the divergence test (nth term test), if lim⁡n→∞an≠0\lim_{n\to\infty} a_n \neq 0, then ∑an\sum a_n must diverge. The series diverges because we're essentially adding terms that approach 1, giving an infinite sum. Choice D incorrectly suggests that having lim⁡an=1\lim a_n = 1 would make the series converge. Remember: if terms don't approach zero, the series must diverge—this is the first test to try.

Question 3

A physics model uses ∑n=2∞1nln⁡n\sum_{n=2}^{\infty}\frac{1}{n\ln n}. Does this series converge or diverge?

  1. Converges because it is smaller than ∑1n\sum \frac{1}{n}
  2. Diverges because it is larger than ∑1n2\sum \frac{1}{n^2}
  3. Converges by comparison to ∑1n3/2\sum \frac{1}{n^{3/2}}
  4. Diverges by the integral test (correct answer)
  5. Converges because 1nln⁡n→0\frac{1}{n\ln n}\to 0
Explanation: This question tests your ability to classify infinite series as convergent or divergent. The series ∑n=2∞1nln⁡n\sum_{n=2}^{\infty}\frac{1}{n\ln n} can be analyzed using the integral test since f(x)=1xln⁡xf(x) = \frac{1}{x\ln x} is positive, continuous, and decreasing for x≥2x \geq 2. The integral ∫2∞1xln⁡xdx\int_2^{\infty} \frac{1}{x\ln x}dx can be evaluated by substitution: let u=ln⁡xu = \ln x, then du=1xdxdu = \frac{1}{x}dx, giving ∫1udu=ln⁡∣u∣=ln⁡(ln⁡x)\int \frac{1}{u}du = \ln|u| = \ln(\ln x). Since lim⁡x→∞ln⁡(ln⁡x)=∞\lim_{x\to\infty} \ln(\ln x) = \infty, the integral diverges, so the series diverges. Choice E incorrectly assumes that terms approaching zero guarantees convergence. Remember: the integral test is powerful for series involving logarithms—if ∫f(x)dx\int f(x)dx diverges, then ∑f(n)\sum f(n) diverges.

Question 4

An error estimate uses ∑n=1∞nn2+4\sum_{n=1}^{\infty} \frac{n}{n^2+4}. Determine whether the series converges or diverges.

  1. Converges by comparison with a pp-series
  2. Diverges by comparison with the harmonic series (correct answer)
  3. Converges by the alternating series test
  4. Converges by the ratio test
  5. Diverges by the root test
Explanation: This problem requires analyzing the series ∑n=1∞nn2+4\sum_{n=1}^{\infty} \frac{n}{n^2+4}. For large nn, the terms behave like nn2=1n\frac{n}{n^2} = \frac{1}{n}, suggesting comparison with the harmonic series. We can verify that lim⁡n→∞n/(n2+4)1/n=lim⁡n→∞n2n2+4=1\lim_{n\to\infty} \frac{n/(n^2+4)}{1/n} = \lim_{n\to\infty} \frac{n^2}{n^2+4} = 1. Since this limit is positive and finite, and the harmonic series ∑1n\sum \frac{1}{n} diverges, our series also diverges by the limit comparison test. The comparison with a p-series (option A) would be incorrect because the series behaves like p=1p=1, not p>1p>1. When the highest powers of nn in numerator and denominator differ by exactly 1, the series typically behaves like the harmonic series.

Question 5

A feedback process accumulates ∑n=1∞(−1)nn2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}. Does the series converge or diverge?

  1. Diverges because it alternates
  2. Diverges by comparison to ∑1n\sum \frac{1}{n}
  3. Converges absolutely by comparison to ∑1n2\sum \frac{1}{n^2} (correct answer)
  4. Converges conditionally by the alternating series test
  5. Diverges by the ratio test
Explanation: This question tests the skill of classifying infinite series as convergent or divergent. The series ∑ (-1)^n / n² has absolute series ∑ 1/n², which converges as a p-series with p=2>1. Since the absolute series converges, the original converges absolutely by the absolute convergence theorem. This implies conditional convergence as well, but absolute is stronger. A tempting distractor is choice D, which claims conditional convergence by the alternating series test, but this fails because absolute convergence is proven, making it absolutely convergent. A transferable strategy for classifying series is to check absolute convergence first for alternating series using tests like comparison or p-series, as it provides more information than conditional alone.

Question 6

A damped oscillation adds displacements ∑n=1∞(−1)n+11n\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}. Does the series converge or diverge?

  1. Diverges because it is a harmonic series
  2. Converges conditionally by the alternating series test (correct answer)
  3. Converges absolutely by comparison to ∑1n2\sum \frac{1}{n^2}
  4. Diverges because the signs alternate
  5. Converges by the geometric series test
Explanation: This question tests the skill of classifying infinite series as convergent or divergent. The series ∑ (-1)^{n+1}/n is the alternating harmonic series, which converges by the alternating series test since the absolute terms 1/n decrease monotonically to 0. However, the absolute series ∑ 1/n is the divergent harmonic series, indicating conditional convergence. This distinction requires checking both conditional and absolute convergence for alternating series. A tempting distractor is choice A, which claims divergence because it is a harmonic series, but this fails as it ignores the alternating signs that enable conditional convergence. A transferable strategy for classifying series is to first check if terms go to 0, then apply tests like alternating series for signs or comparison for positives, and verify absolute convergence if needed.

Question 7

A process alternates adjustments totaling ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}. Does the series converge or diverge?

  1. Diverges by the alternating series test
  2. Converges by the alternating series test (correct answer)
  3. Converges by the harmonic series test
  4. Diverges by the pp-series test
  5. Diverges because terms do not approach 00
Explanation: This problem involves determining convergence of an alternating series. The series ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} is the alternating harmonic series, where terms alternate in sign and have magnitude 1/n1/n. To apply the alternating series test, we check: (1) terms decrease in magnitude: 1n+1<1n\frac{1}{n+1} < \frac{1}{n}, and (2) lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0. Both conditions are satisfied, so the series converges by the alternating series test. Choice A incorrectly claims divergence—the alternating series test never proves divergence, only convergence. For alternating series ∑(−1)nan\sum (-1)^n a_n, check if ana_n decreases to 0 for convergence.

Question 8

An iterative method produces ∑n=1∞1n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)}. Does the series converge or diverge?

  1. Diverges by comparison with the harmonic series
  2. Diverges by the nth-term test
  3. Converges by telescoping (correct answer)
  4. Converges by the ratio test
  5. Diverges by the root test
Explanation: This problem involves the series ∑n=1∞1n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)}, which can be analyzed using partial fractions. We decompose 1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, creating a telescoping series. The partial sums are SN=∑n=1N(1n−1n+1)=(1−12)+(12−13)+...+(1N−1N+1)=1−1N+1S_N = \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + ... + \left(\frac{1}{N} - \frac{1}{N+1}\right) = 1 - \frac{1}{N+1}. As N→∞N \to \infty, we get lim⁡N→∞SN=1\lim_{N\to\infty} S_N = 1, so the series converges to 1. The harmonic series comparison (option A) is incorrect because this series converges while the harmonic series diverges. When you see products of consecutive integers in the denominator, always check for telescoping by partial fractions.

Question 9

A filter output sums ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}. Does the series converge or diverge?

  1. Diverges by the alternating series test
  2. Converges by the pp-series test
  3. Converges by the alternating series test (correct answer)
  4. Diverges by the pp-series test
  5. Diverges because terms do not approach 00
Explanation: This question tests convergence of alternating series. The series ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}} alternates in sign with terms of magnitude 1n\frac{1}{\sqrt{n}}. For the alternating series test: (1) terms decrease: 1n+1<1n\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}, and (2) lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0. Both conditions hold, so the series converges by the alternating series test. Note that the series of absolute values ∑1n\sum \frac{1}{\sqrt{n}} diverges (p-series with p = 1/2), making this conditionally convergent. For alternating series, always check the two conditions of the alternating series test before considering other methods.

Question 10

A series solution includes ∑n=1∞n22n\sum_{n=1}^{\infty}\frac{n^2}{2^n}; determine whether the infinite series converges or diverges.

  1. Diverges by comparison to ∑1n\sum\frac{1}{n}
  2. Converges by the Ratio Test (correct answer)
  3. Diverges by the Root Test
  4. Converges by the Integral Test
  5. Diverges because n2n^2 increases
Explanation: This problem requires classifying a series combining polynomial and exponential terms. The series ∑n=1∞n22n\sum_{n=1}^{\infty}\frac{n^2}{2^n} is ideal for the Ratio Test due to the exponential denominator. Applying the Ratio Test: lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1)2/2n+1n2/2n=lim⁡n→∞(n+1)22n2=12lim⁡n→∞(n+1n)2=12<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)^2/2^{n+1}}{n^2/2^n} = \lim_{n\to\infty}\frac{(n+1)^2}{2n^2} = \frac{1}{2}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^2 = \frac{1}{2} < 1. Since the limit is less than 1, the series converges by the Ratio Test. Choice C suggesting the Root Test could also work, but the Ratio Test is more straightforward here; choice E incorrectly focuses on the numerator growing while ignoring that 2n2^n grows much faster. When exponential terms appear in denominators, they typically dominate polynomial numerators, leading to convergence via the Ratio Test.

Question 11

A population model uses ∑n=1∞1n+1\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+1} as a correction; does the series converge or diverge?

  1. Converges absolutely
  2. Converges conditionally
  3. Converges by the ratio test
  4. Diverges by comparison to a pp-series with p=12p=\tfrac12 (correct answer)
  5. Diverges by the alternating series test
Explanation: This question examines ∑1/(√n+1), testing comparison with p-series for convergence analysis. For large n, √n+1 behaves like √n, so terms behave like 1/√n = 1/n^(1/2). Using limit comparison with the divergent p-series ∑1/n^(1/2) (p = 1/2 < 1): lim[1/(√n+1)]/(1/√n) = lim[√n/(√n+1)] = lim[1/(1+1/√n)] = 1. Since this limit is positive and finite, and ∑1/n^(1/2) diverges, our series also diverges. Students might think the +1 in the denominator helps convergence, but it doesn't change the growth rate enough. When square roots appear with added constants, use limit comparison with the corresponding p-series to determine convergence.

Question 12

In a fractal perimeter model, consider ∑n=1∞ln⁡nn3/2\sum_{n=1}^{\infty}\frac{\ln n}{n^{3/2}}; does this series converge or diverge?

  1. Converges by comparison to ∑1n4/3\sum\frac{1}{n^{4/3}} (correct answer)
  2. Diverges by comparison to ∑1n\sum\frac{1}{n}
  3. Diverges by the Ratio Test
  4. Converges by the Alternating Series Test
  5. Diverges because ln⁡n\ln n grows without bound
Explanation: This problem tests convergence classification for series combining logarithmic and power functions. For the series ∑n=1∞ln⁡nn3/2\sum_{n=1}^{\infty}\frac{\ln n}{n^{3/2}}, we can use the Limit Comparison Test with ∑1n4/3\sum\frac{1}{n^{4/3}}. We need to verify that lim⁡n→∞ln⁡n/n3/21/n4/3=lim⁡n→∞ln⁡n⋅n4/3n3/2=lim⁡n→∞ln⁡nn1/6\lim_{n\to\infty}\frac{\ln n/n^{3/2}}{1/n^{4/3}} = \lim_{n\to\infty}\frac{\ln n \cdot n^{4/3}}{n^{3/2}} = \lim_{n\to\infty}\frac{\ln n}{n^{1/6}}. By L'Hôpital's rule: lim⁡n→∞ln⁡nn1/6=lim⁡n→∞1/n(1/6)n−5/6=lim⁡n→∞6n1/6=0\lim_{n\to\infty}\frac{\ln n}{n^{1/6}} = \lim_{n\to\infty}\frac{1/n}{(1/6)n^{-5/6}} = \lim_{n\to\infty}\frac{6}{n^{1/6}} = 0. Since this limit is 0 and ∑1n4/3\sum\frac{1}{n^{4/3}} converges (p-series with p=4/3>1p = 4/3 > 1), our series also converges. Choice B incorrectly suggests divergence; while ln⁡n\ln n grows, it grows slowly enough that n3/2n^{3/2} in the denominator ensures convergence. For series with ln⁡n\ln n in the numerator, remember that logarithms grow more slowly than any positive power of nn.

Question 13

A cost model includes ∑n=1∞3nn2+1\sum_{n=1}^{\infty}\frac{3n}{n^2+1}. Does this infinite series converge or diverge?

  1. Converges by comparison to ∑1n2\sum \frac{1}{n^2}
  2. Diverges by comparison to ∑1n\sum \frac{1}{n} (correct answer)
  3. Converges because 3nn2+1→0\frac{3n}{n^2+1}\to 0
  4. Converges because it is geometric
  5. Diverges because it is a pp-series with p=2p=2
Explanation: This question tests your ability to classify infinite series as convergent or divergent. For the series ∑n=1∞3nn2+1\sum_{n=1}^{\infty}\frac{3n}{n^2+1}, we analyze the behavior for large nn: 3nn2+1∼3nn2=3n\frac{3n}{n^2+1} \sim \frac{3n}{n^2} = \frac{3}{n}. By limit comparison with ∑1n\sum\frac{1}{n} (harmonic series), we compute lim⁡n→∞3nn2+11n=lim⁡n→∞3n2n2+1=3\lim_{n\to\infty} \frac{\frac{3n}{n^2+1}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{3n^2}{n^2+1} = 3. Since this limit is positive and finite, and ∑1n\sum\frac{1}{n} diverges, our series also diverges. Choice C incorrectly assumes that terms approaching zero guarantees convergence. Remember: use limit comparison when terms behave like a known series for large nn.

Question 14

To model repeated fees, a plan uses ∑n=1∞1n\sum_{n=1}^{\infty}\frac{1}{n}. Does this infinite series converge or diverge?

  1. Converges by the ratio test
  2. Converges because 1n→0\frac{1}{n}\to 0
  3. Diverges because it is a pp-series with p=1p=1 (correct answer)
  4. Converges because it is a pp-series with p=1p=1
  5. Diverges by comparison to ∑1n2\sum \frac{1}{n^2}
Explanation: This question tests your ability to classify infinite series as convergent or divergent. The series ∑n=1∞1n\sum_{n=1}^{\infty}\frac{1}{n} is the harmonic series, which is a p-series with p=1p = 1. For p-series ∑1np\sum\frac{1}{n^p}, the series converges when p>1p > 1 and diverges when p≤1p \leq 1. Since p=1p = 1, the harmonic series diverges. Choice B incorrectly assumes that because the terms 1n→0\frac{1}{n} \to 0, the series must converge, but this is only a necessary condition, not sufficient. Remember: the harmonic series is the classic example of a divergent series whose terms approach zero.

Question 15

A sequence of deposits is modeled by ∑n=1∞(nn+1)n\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^n. Does the series converge or diverge?

  1. Converges because it is geometric
  2. Diverges because (nn+1)n→1e≠0\left(\frac{n}{n+1}\right)^n\to \frac{1}{e}\neq 0 (correct answer)
  3. Converges because (nn+1)n→0\left(\frac{n}{n+1}\right)^n\to 0
  4. Converges by comparison to ∑1n2\sum \frac{1}{n^2}
  5. Diverges by the ratio test since the limit equals 1e<1\frac{1}{e}<1
Explanation: This question tests your ability to classify infinite series as convergent or divergent. For the series ∑n=1∞(nn+1)n\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^n, we need to find lim⁡n→∞(nn+1)n\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n. Rewriting as (11+1n)n=1(1+1n)n\left(\frac{1}{1+\frac{1}{n}}\right)^n = \frac{1}{\left(1+\frac{1}{n}\right)^n}, and since lim⁡n→∞(1+1n)n=e\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e, we get lim⁡n→∞(nn+1)n=1e≈0.368≠0\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n = \frac{1}{e} \approx 0.368 \neq 0. By the divergence test, since terms don't approach zero, the series diverges. Choice E incorrectly interprets the ratio test—having a limit less than 1 would indicate convergence, not divergence. Remember: always check if lim⁡an=0\lim a_n = 0 first; if not, the series must diverge.

Question 16

An algorithm's error terms are ∑n=1∞52n\sum_{n=1}^{\infty}\frac{5}{2^n}. Does this infinite series converge or diverge?

  1. Diverges because it is geometric with ∣r∣>1|r|>1
  2. Converges because it is a geometric series with ∣r∣<1|r|<1 (correct answer)
  3. Diverges because 52n→5\frac{5}{2^n}\to 5
  4. Converges by the harmonic series test
  5. Diverges by the integral test
Explanation: This question tests your ability to classify infinite series as convergent or divergent. The series ∑n=1∞52n=5∑n=1∞(12)n\sum_{n=1}^{\infty}\frac{5}{2^n} = 5\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n is a geometric series with first term a=52a = \frac{5}{2} and common ratio r=12r = \frac{1}{2}. Since ∣r∣=12<1|r| = \frac{1}{2} < 1, the geometric series converges to 5⋅121−12=55 \cdot \frac{\frac{1}{2}}{1-\frac{1}{2}} = 5. Choice A incorrectly claims ∣r∣>1|r| > 1, perhaps confusing the base 2 in the denominator with the ratio. Remember: rewrite series in the form ∑arn\sum ar^n to identify geometric series and check if ∣r∣<1|r| < 1.

Question 17

A computer sums ∑n=1∞(56)n\sum_{n=1}^{\infty} \left(\frac{5}{6}\right)^n to estimate a limit. Does the series converge or diverge?

  1. Diverges by the nth-term test
  2. Converges as a geometric series (correct answer)
  3. Diverges by the ratio test
  4. Converges by the integral test
  5. Diverges by comparison with a pp-series
Explanation: This problem involves classifying the geometric series ∑n=1∞(56)n\sum_{n=1}^{\infty} \left(\frac{5}{6}\right)^n. A geometric series ∑arn\sum ar^n converges when ∣r∣<1|r| < 1 and diverges when ∣r∣≥1|r| \geq 1. Here, the common ratio is r=56r = \frac{5}{6}, and since ∣56∣=56<1|\frac{5}{6}| = \frac{5}{6} < 1, the series converges. The sum of this convergent geometric series is a1−r=5/61−5/6=5/61/6=5\frac{a}{1-r} = \frac{5/6}{1-5/6} = \frac{5/6}{1/6} = 5. The nth-term test (option A) only tells us when a series must diverge, not when it converges. For any series with constant ratio between consecutive terms, immediately recognize it as geometric and check if ∣r∣<1|r| < 1.

Question 18

A model adds adjustments ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n\ln n}. Determine whether the series converges or diverges.

  1. Converges by the ratio test
  2. Converges by comparison with a pp-series
  3. Diverges by the integral test (correct answer)
  4. Converges by the alternating series test
  5. Diverges by the root test
Explanation: This problem requires determining convergence of the series ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n\ln n}. To analyze this series, we can apply the integral test by evaluating ∫2∞1xln⁡xdx\int_2^{\infty} \frac{1}{x\ln x} dx. Using the substitution u=ln⁡xu = \ln x, we get du=1xdxdu = \frac{1}{x}dx, transforming the integral to ∫ln⁡2∞1udu=ln⁡∣u∣∣ln⁡2∞\int_{\ln 2}^{\infty} \frac{1}{u} du = \ln|u|\big|_{\ln 2}^{\infty}. This integral diverges to infinity, so by the integral test, the series diverges. The ratio test (option A) would be inconclusive here as the limit equals 1. Remember that 1nln⁡n\frac{1}{n\ln n} is a classic example of a divergent series that decreases more slowly than the harmonic series.

Question 19

A sequence of updates is ∑n=1∞n!nn\sum_{n=1}^{\infty} \frac{n!}{n^n}. Does the series converge or diverge?

  1. Diverges by the nth-term test
  2. Converges by the ratio test (correct answer)
  3. Diverges by comparison with a geometric series
  4. Converges by the integral test
  5. Diverges by comparison with a pp-series
Explanation: This problem involves the series ∑n=1∞n!nn\sum_{n=1}^{\infty} \frac{n!}{n^n}, which requires the ratio test. Computing lim⁡n→∞an+1an=lim⁡n→∞(n+1)!/(n+1)n+1n!/nn=lim⁡n→∞(n+1)⋅nn(n+1)n+1=lim⁡n→∞nn(n+1)n\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \lim_{n\to\infty} \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{n^n}{(n+1)^n}. This simplifies to lim⁡n→∞(nn+1)n=lim⁡n→∞(1−1n+1)n=1e<1\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n = \lim_{n\to\infty} \left(1 - \frac{1}{n+1}\right)^n = \frac{1}{e} < 1. Since the ratio test limit is less than 1, the series converges. The nth-term test (option A) would only check if terms go to zero, not prove convergence. For series involving factorials and exponentials, the ratio test is typically the most effective approach.

Question 20

A sensor correction uses the infinite series ∑n=1∞3n2+1\sum_{n=1}^{\infty} \frac{3}{n^2+1}. Does the series converge or diverge?

  1. Converges by comparison with a pp-series (correct answer)
  2. Diverges by comparison with a pp-series
  3. Converges by the alternating series test
  4. Diverges by the ratio test
  5. Converges by the integral test
Explanation: This problem tests your ability to classify infinite series as convergent or divergent. The series ∑n=1∞3n2+1\sum_{n=1}^{\infty} \frac{3}{n^2+1} has terms that behave like 3n2\frac{3}{n^2} for large nn, since the '+1' becomes negligible compared to n2n^2. We can use the comparison test with the convergent p-series ∑3n2\sum \frac{3}{n^2} (where p=2>1p=2>1). Since 3n2+1<3n2\frac{3}{n^2+1} < \frac{3}{n^2} for all n≥1n \geq 1, and the comparison series converges, our original series also converges. The integral test (option E) would also work but requires more computation than the direct comparison. When terms resemble 1np\frac{1}{n^p}, always check if you can compare with a p-series first.