AP Calculus BC Quiz: Determining Absolute Or Conditional Convergence

What this quiz covers

This quiz focuses on Determining Absolute Or Conditional 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.

Question 1

1. Converges absolutely because $\sum \frac{\ln n}{n}$ converges
2. Converges conditionally by the Alternating Series Test but not absolutely (correct answer)
3. Converges absolutely by comparison to $\sum \frac{1}{n^2}$
4. Diverges because $\ln n$ grows without bound
5. Diverges because $\sum (-1)^n$ diverges

Explanation: Assessing convergence types—absolute, conditional, or divergent—is key for physics models using series. Distinguish by checking if ∑|a_n| converges (absolute) or diverges while the alternating converges (conditional). Apply AST: (ln n)/n decreases to zero for n≥3, confirming convergence. The absolute series ∑ (ln n)/n diverges by Integral Test, as ∫ (ln x)/x dx = (ln x)^2/2 → ∞. Claiming absolute convergence by comparison to ∑1/n² fails because (ln n)/n grows slower than 1/n but still diverges. A transferable approach: test absolute with integral or comparison, falling back to AST for conditional if needed.

Question 2

1. Converges conditionally
2. Diverges by comparison to $\sum \frac{1}{n}$
3. Converges absolutely (correct answer)
4. Diverges by the alternating series test
5. Converges absolutely because it is alternating

Explanation: This problem examines $\sum_{n=1}^{\infty} (-1)^n\frac{1}{n^3+n}$. The alternating series test confirms convergence since $\frac{1}{n^3+n}$ decreases to 0. For absolute convergence, we analyze $\sum_{n=1}^{\infty} \frac{1}{n^3+n} = \sum_{n=1}^{\infty} \frac{1}{n(n^2+1)}$. For large $n$, this behaves like $\frac{1}{n^3}$, and by limit comparison with the convergent p-series $\sum \frac{1}{n^3}$, our series converges. Since the absolute value series converges, the original series converges absolutely. Choice A incorrectly suggests only conditional convergence without properly checking the absolute value series. The key: when terms behave like $\frac{1}{n^p}$ with $p>1$ for large $n$, expect absolute convergence.

Question 3

1. Converges absolutely
2. Diverges because terms do not approach $0$
3. Converges conditionally (correct answer)
4. Converges absolutely by comparison to $\sum \frac{1}{n^{3/2}}$
5. Diverges by the ratio test

Explanation: Classifying alternating perturbations like ∑ (-1)^n / (√n +1) involves convergence skills. Absolute: ∑ 1/(√n +1) ≈ ∑ 1/√n diverges (p=1/2<1). Conditional: AST as 1/(√n +1) decreases to 0. Converges conditionally. The distractor of absolute by comparison to ∑ 1/n^{3/2} fails, as it diverges like 1/√n, not converges. Strategy: approximate for large n; if absolute like divergent p-series, test AST for conditional.

Question 4

1. Converges absolutely
2. Diverges by comparison to $\sum \frac{1}{n\ln n}$
3. Converges conditionally (correct answer)
4. Diverges because $\ln n/n$ does not decrease
5. Converges absolutely by the integral test

Explanation: In Fourier-type sums, classifying convergence like for ∑ (-1)^n (ln n)/n is vital. Absolute: ∑ (ln n)/n diverges by integral test, as ∫ (ln x)/x dx = (1/2)(ln x)^2 → ∞. Conditional: AST applies because (ln n)/n decreases to 0 for large n. Thus, conditional. The distractor of divergence because (ln n)/n does not decrease fails, as it does decrease for n > e. Tip: assess absolute with integrals; if divergent, verify decreasing and limit 0 for conditional via AST.

Question 5

1. Diverges
2. Converges conditionally (correct answer)
3. Converges absolutely
4. Converges by the ratio test, so absolutely
5. Converges by the alternating series test, so absolutely

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series ∑ 1/√n converges, which it does not as it is a p-series with p=1/2<1. Since it alternates with terms decreasing to 0, it converges by the alternating series test (AST), hence conditionally. A tempting distractor claims convergence by the ratio test so absolutely, but the ratio test gives limit 1, which is inconclusive. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Question 6

1. Converges conditionally
2. Diverges because it telescopes to infinity
3. Converges absolutely (correct answer)
4. Diverges by the ratio test
5. Converges conditionally by comparison to $\sum \frac{1}{n}$

Explanation: Convergence type classification is key for numerical methods producing series like ∑ (-1)^{n+1} / [n(n+1)]. Check absolute: ∑ 1/[n(n+1)] telescopes to 1, converging. Since absolute converges, the series converges absolutely. No conditional check needed. The distractor of conditional by comparison to ∑ 1/n fails because the absolute series converges, not diverges. Strategy: decompose terms if possible; absolute convergence simplifies to checking the positive series directly.

Question 7

1. Converges absolutely
2. Diverges because $\sum \frac{1}{\sqrt{n}}$ diverges
3. Converges conditionally (correct answer)
4. Diverges because terms do not approach $0$
5. Converges absolutely by the $p$-test with $p=\tfrac12$

Explanation: This question asks about $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$. The alternating series test applies since $\frac{1}{\sqrt{n}}$ decreases to 0. For absolute convergence, we check $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$, which is a p-series with $p=\frac{1}{2}<1$, so it diverges. Since the alternating series converges but the absolute value series diverges, we have conditional convergence. Choice E incorrectly suggests that a p-series with $p=\frac{1}{2}$ proves absolute convergence, but p-series diverge when $p\leq 1$. Remember: for alternating p-series $\sum \frac{(-1)^n}{n^p}$, you get absolute convergence when $p>1$ and conditional convergence when $0<p\leq 1$.

Question 8

1. Diverges
2. Converges conditionally
3. Converges absolutely (correct answer)
4. Converges by comparison to $\sum \frac{1}{n}$, so diverges
5. Converges by the alternating series test, so conditionally

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series $\sum \frac{1}{n^2 + \cos n}$ converges, which it does by comparison to $\sum \frac{1}{n^2}$ since $\cos n$ is bounded. Thus, the original series converges absolutely. A tempting distractor suggests convergence by comparison to $\sum \frac{1}{n}$ so diverges, but it converges unlike $\sum \frac{1}{n}$. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Question 9

1. Converges conditionally
2. Diverges
3. Converges absolutely (correct answer)
4. Converges by comparison to $\sum \frac{1}{n}$, so diverges
5. Converges by the ratio test, so conditionally

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series $\sum \sqrt{n}/(n^2 + 3)$ converges, which it does as it behaves like $\sum 1/n^{3/2}$ ($p=3/2>1$). Thus, the original series converges absolutely. A tempting distractor suggests convergence by comparison to $\sum 1/n$ so diverges, but it converges unlike $\sum 1/n$. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Question 10

1. Converges absolutely (correct answer)
2. Converges conditionally
3. Diverges because $\ln(n+1)$ grows slowly
4. Diverges by comparison to $\sum \frac{1}{n\ln(n+1)}$
5. Diverges because terms do not approach $0$

Explanation: This question requires analyzing $\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n(\ln(n+1))^2}$. For absolute convergence, we check if $\sum_{n=1}^{\infty} \frac{1}{n(\ln(n+1))^2}$ converges. Using the integral test with substitution $u = \ln(x+1)$, we get $\int_1^{\infty} \frac{1}{x(\ln(x+1))^2}dx$, which converges because it behaves like $\int \frac{1}{u^2}du$ after substitution. Since the absolute value series converges, the original series converges absolutely. Choice B incorrectly suggests only conditional convergence, not recognizing that the squared logarithm in the denominator provides enough decay for absolute convergence. The key insight: powers of logarithms greater than 1 in the denominator often lead to absolute convergence.

Question 11

1. Converges absolutely
2. Converges conditionally (correct answer)
3. Diverges because terms do not approach $0$
4. Diverges by the ratio test
5. Converges by the geometric series test

Explanation: This problem tests your ability to classify series convergence as absolute, conditional, or divergent. A series converges absolutely if the series of absolute values converges, and conditionally if the original series converges but the absolute value series diverges. For the alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$, we check absolute convergence by examining $\sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series that diverges. However, the original alternating series converges by the alternating series test since $\frac{1}{n}$ decreases to 0. Choice C incorrectly claims the terms don't approach 0, when clearly $\lim_{n\to\infty} \frac{(-1)^{n+1}}{n} = 0$. The key strategy: if an alternating series converges but its absolute value series diverges, it converges conditionally.

Question 12

1. Converges conditionally
2. Diverges by the integral test
3. Converges absolutely (correct answer)
4. Diverges because terms do not approach $0$
5. Converges only for even $n$

Explanation: This problem asks about $\sum_{n=1}^{\infty} (-1)^n\frac{n!}{n^n}$. To determine convergence type, we first check if $\frac{n!}{n^n} \to 0$. By Stirling's approximation or the ratio test, $\lim_{n\to\infty} \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \lim_{n\to\infty} \frac{n^n}{(n+1)^n} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n = \frac{1}{e} < 1$. This means the absolute value series $\sum \frac{n!}{n^n}$ converges by the ratio test, so the original alternating series converges absolutely. Choice A incorrectly suggests only conditional convergence, missing that the factorial grows much slower than $n^n$. The principle: when ratio test gives a limit less than 1, the series converges absolutely regardless of alternating signs.

Question 13

1. Converges absolutely
2. Converges conditionally (correct answer)
3. Diverges because terms do not approach $0$
4. Diverges by the ratio test
5. Converges absolutely by the alternating series test

Explanation: This problem tests your ability to classify series convergence as absolute, conditional, or divergent. The series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ is the alternating harmonic series, which converges by the alternating series test since $\frac{1}{n}$ decreases to 0. To check absolute convergence, we examine $\sum_{n=1}^{\infty} \left|\frac{(-1)^{n-1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series and diverges. Since the original series converges but the absolute value series diverges, we have conditional convergence. Choice E incorrectly claims the alternating series test proves absolute convergence, but this test only establishes convergence of the alternating series itself. Remember: a series converges conditionally when it converges but its absolute value series diverges.

Question 14

1. Diverges by the alternating series test
2. Converges conditionally
3. Converges absolutely (correct answer)
4. Diverges because it is a $p$-series with $p=2$
5. Converges absolutely by the ratio test only

Explanation: This problem asks about the convergence of $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}$. The alternating series test confirms convergence since $\frac{1}{n^2}$ decreases to 0. For absolute convergence, we check $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a p-series with $p=2>1$, so it converges. Since the absolute value series converges, the original series converges absolutely. Choice D incorrectly claims divergence for a p-series with $p=2$, but p-series converge when $p>1$. The fundamental principle: a series converges absolutely when its absolute value series converges, making the alternating nature irrelevant to the convergence classification.

Question 15

1. Converges absolutely
2. Diverges because $\ln n$ grows too slowly
3. Converges conditionally (correct answer)
4. Diverges by the alternating series test
5. Converges absolutely by the integral test

Explanation: This question requires determining the convergence type of $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\ln n}$. For the alternating series test, we need $\frac{1}{n\ln n}$ to be decreasing and approach 0, which it does since both $n$ and $\ln n$ increase. To check absolute convergence, we examine $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$, which diverges by the integral test since $\int_2^{\infty} \frac{1}{x\ln x}dx = \ln(\ln x)|_2^{\infty} = \infty$. Therefore, the series converges conditionally. Choice B incorrectly suggests that slow growth of $\ln n$ causes divergence, when actually the alternating nature ensures convergence. The key insight: when an alternating series converges but its absolute value series diverges, you have conditional convergence.

Question 16

1. Converges absolutely
2. Converges conditionally (correct answer)
3. Diverges
4. Converges by comparison to $\sum \frac{1}{n^2}$, so absolutely
5. Diverges because terms do not go to $0$

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series ∑ 1/(√n + √(n+1)) converges, which it does not as it behaves like ∑ 1/(2√n) (p=1/2<1). Since it alternates with terms 1/(√n + √(n+1)) decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims divergence because terms do not go to 0, but they do approach 0. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Question 17

1. Converges absolutely
2. Converges conditionally (correct answer)
3. Diverges
4. Converges by comparison to $\sum \frac{1}{n^2}$, so absolutely
5. Converges by the ratio test, so absolutely

Explanation: Determining absolute or conditional convergence is essential for analyzing infinite series in calculus. Absolute convergence requires the series of absolute terms to converge, while conditional means only the original converges. Tests involve checking ∑ |a_n| with comparison or integral, and if divergent, applying the alternating series test to the original. Here, the absolute series ∑ ln(n+1)/√n diverges by comparison to ∑ 1/√n (p=1/2<1), but the original converges by AST since ln(n+1)/√n decreases to 0. A common distractor is thinking it converges absolutely by ratio test, but the ratio limit is 1, inconclusive. Always verify absolute convergence first; if it fails, assess conditional via AST for alternating series with decreasing terms to 0.

Question 18

1. Converges conditionally
2. Diverges
3. Converges absolutely (correct answer)
4. Converges by the alternating series test, so conditionally
5. Converges by comparison to $\sum \frac{1}{n\ln n}$, so diverges

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series ∑ 1/(n (ln(n+1))^2) converges, which it does by integral test as ∫ dx/(x (ln x)^2) converges. Thus, the original series converges absolutely. A tempting distractor suggests convergence by comparison to ∑ 1/(n ln n) so diverges, but the extra (ln n)^2 in denominator makes it converge. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Question 19

1. Converges absolutely
2. Diverges
3. Converges conditionally (correct answer)
4. Converges by comparison to $\sum \frac{1}{n^2}$, so absolutely
5. Converges by the ratio test, so absolutely

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series $\sum \frac{1}{n + \sin n}$ converges, which it does not as it behaves like $\sum \frac{1}{n}$. Since it alternates with terms $\frac{1}{n + \sin n}$ decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims convergence by the ratio test so absolutely, but the ratio is inconclusive due to oscillation. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Question 20

1. Diverges
2. Converges absolutely
3. Converges conditionally (correct answer)
4. Converges by comparison to $\sum \frac{1}{n^2}$, so absolutely
5. Diverges because terms do not alternate

Explanation: Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series ∑ 1/(n+1) converges, which it does not (harmonic series). Since it alternates with terms 1/(n+1) decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims divergence because terms do not alternate, but they do alternate. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.