Determining Absolute or Conditional Convergence - AP Calculus BC

A model uses $\sum_{n=2}^\infty \frac{(-1)^n}{n\ln n}$ ; classify its convergence as absolute, conditional, or divergence.
Answer: Converges conditionally Classifying series convergence as absolute, conditional, or divergent is essential for models involving alternating terms like ∑ $(-1)^n$ /(n ln n). Distinguish by testing the absolute series ∑ 1/(n ln n), which diverges via the integral test since ∫ dx/(x ln x) = ln(ln x) goes to infinity. For conditional, the AST applies as 1/(n ln n) decreases to 0 for n ≥ 3. Thus, it converges conditionally. The distractor of absolute convergence by integral test fails because the integral diverges, not converges. A strategy is to evaluate absolute convergence with divergence tests; if divergent, check AST for conditional.
In approximating $\pi$ , consider $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n-1}$ ; determine absolute or conditional convergence.
Answer: Converges conditionally 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 $\sum \frac{1}{2n-1}$ converges, which it does not as it behaves like the harmonic series. Since it alternates with terms 1/(2n-1) decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims convergence by the ratio test so absolutely, but the ratio limit is 1, inconclusive. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.
A numerical method produces $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n(n+1)}$ ; classify its convergence type.
Answer: Converges absolutely 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.
In an error-correction scheme, consider $\sum_{n=2}^\infty \frac{(-1)^n}{n\ln n}$ ; does it converge absolutely, conditionally, or diverge?
Answer: Converges conditionally by the Alternating Series Test but not absolutely Classifying series convergence as absolute, conditional, or divergent is essential for understanding series behavior in calculus. Distinguish by testing if $\sum |a_n|$ converges (absolute) or if the alternating series converges via AST but $\sum |a_n|$ diverges (conditional). Use the Integral Test on $\sum \frac{1}{n \ln n}$ , which diverges since $\int \frac{dx}{x \ln x} = \ln(\ln x) \to \infty$ . Apply AST: terms $\frac{1}{n \ln n}$ decrease to zero for n≥2, confirming convergence. A common distractor might suggest divergence by comparison to $\sum \frac{1}{n^2}$ , but that converges while this absolute series diverges. A reliable strategy is to verify AST conditions for conditional, then test absolute divergence with integral or comparison tests.
Determine whether $\sum_{n=1}^{\infty} (-1)^{n}\frac{1}{\sqrt{n}+\sqrt{n+1}}$ converges absolutely, conditionally, or diverges.
Answer: Converges conditionally 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.
A power-series remainder includes $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n^4}$ ; determine its convergence type.
Answer: Converges absolutely 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 $\sum \frac{1}{n^4}$ converges, which it does ( $p=4>1$ ). Thus, the original series converges absolutely. A tempting distractor suggests divergence by comparison to $\sum \frac{1}{n}$ , but $p=4>1$ ensures convergence. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.
In analyzing a signal, determine whether $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$ converges absolutely, conditionally, or diverges.
Answer: Converges conditionally 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.
Determine the type of convergence of $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n}{n^2+1}$ .
Answer: Converges conditionally This question examines $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n}{n^2+1}$ . For large $n$ , the terms behave like $\frac{n}{n^2} = \frac{1}{n}$ , which decreases to 0, satisfying the alternating series test. To check absolute convergence, we examine $\sum_{n=1}^{\infty} \frac{n}{n^2+1}$ , which behaves like $\sum \frac{1}{n}$ for large $n$ and diverges by limit comparison. Since the alternating series converges but the absolute value series diverges, we have conditional convergence. Choice A incorrectly claims absolute convergence without checking if the absolute value series converges. The strategy: use limit comparison with known series to determine absolute convergence, then classify as conditional if the original converges but the absolute doesn't.
Classify the convergence of $\sum_{n=1}^{\infty} (-1)^n\frac{1}{n^{0.9}}$ as absolute, conditional, or divergent.
Answer: Converges conditionally 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 $\sum \frac{1}{n^{0.9}}$ converges, which it does not ( $p=0.9 < 1$ ). Since it alternates with terms $\frac{1}{n^{0.9}}$ decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims divergence because alternating series always diverge, but AST shows many alternate series converge. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.
To approximate an alternating correction, classify $\sum_{n=1}^\infty \frac{(-1)^{n}}{\sqrt{n}}$ as absolute, conditional, or divergent.
Answer: Converges conditionally Convergence classification aids in approximating alternating corrections, as in ∑ $(-1)^n$ / √n. Test absolute: ∑ 1/√n is a p-series with p=1/2<1, so diverges. Conditional: AST confirms since 1/√n decreases to 0. It converges conditionally. The distractor of divergence by p-series with p=1/2 fails to consider AST for the alternating case. Generally, for absolute vs. conditional, use p-series for absolute; if p≤1 and divergent, check AST conditions.
Classify the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3+1}$ as absolute, conditional, or divergent.
Answer: Converges absolutely This problem tests classification of $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3+1}$ . To check absolute convergence, we examine $\sum_{n=1}^{\infty} \frac{1}{n^3+1}$ . Since $n^3+1 \sim n^3$ for large $n$ , we can compare to $\sum \frac{1}{n^3}$ , which is a convergent p-series with $p=3>1$ . By the limit comparison test, $\lim_{n\to\infty} \frac{\frac{1}{n^3+1}}{\frac{1}{n^3}} = \lim_{n\to\infty} \frac{n^3}{n^3+1} = 1$ , so both series have the same convergence behavior. Choice A incorrectly compares to the divergent harmonic series, missing that the denominator grows as $n^3$ , not $n$ . Remember: when a series converges absolutely, the alternating factor becomes irrelevant for convergence classification.
For an error-correction term, decide whether $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}$ converges absolutely, conditionally, or diverges.
Answer: Converges absolutely 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/ $n^2$ converges, which it does as a p-series with p=2>1. Thus, the original series converges absolutely. A tempting distractor claims convergence by the harmonic series so diverges, but the harmonic series is ∑ 1/n, not ∑ 1/ $n^2$ which converges. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.
Determine whether the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ converges absolutely, conditionally, or diverges.
Answer: Converges conditionally 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.
A series in a report is $\sum_{n=1}^{\infty} (-1)^{n}\frac{1}{n(\ln(n+1))^2}$ ; classify its convergence.
Answer: Converges absolutely 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.
Classify $\sum_{n=1}^{\infty} (-1)^n\frac{n!}{n^n}$ as absolutely convergent, conditionally convergent, or divergent.
Answer: Converges absolutely 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.
A series for calibrating bias is $\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{n\sqrt{n}}$ ; which choice is correct?
Answer: Converges absolutely since $\sum \frac{1}{n^{3/2}}$ converges Classifying convergence types is key for bias calibration using series. Absolute if ∑| $a_n$ | converges; conditional if only the alternating does. With terms 1/ $n^{3/2}$ , p=3/2>1 ensures ∑| $a_n$ | converges. Hence, absolute convergence. The distractor saying it diverges because p=3/2 makes it harmonic is wrong, as harmonic has p=1. Practically, assess the exponent of | $a_n$ | for p>1 absolute convergence; if ≤1, explore AST for conditional.
Classify the convergence of $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\ln n}$ as absolute, conditional, or divergent.
Answer: Converges conditionally This problem asks you to classify the convergence of $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\ln n}$ . For absolute convergence, we check if $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$ converges. By the integral test, $\int_2^{\infty} \frac{1}{x\ln x}dx = \ln(\ln x)|_2^{\infty} = \infty$ , so the absolute value series diverges. However, the original alternating series converges by the alternating series test because $\frac{1}{n\ln n}$ decreases to 0. Choice E incorrectly compares to $\sum \frac{1}{n^2}$ , but $\frac{1}{n\ln n} > \frac{1}{n^2}$ for large $n$ , making this comparison invalid. The pattern: when an alternating series passes the alternating series test but fails absolute convergence, it converges conditionally.
To check stability, classify $\sum_{n=1}^{\infty} (-1)^n\frac{\sqrt{n}}{n+1}$ as absolutely convergent, conditionally convergent, or divergent.
Answer: Converges conditionally 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 $∑ √n/(n+1)$ converges, which it does not as it behaves like $∑ 1/√n$ $(p=1/2<1)$ . Since it alternates with terms √n/(n+1) decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims convergence by comparison to $∑ 1/n^2$ so absolutely, but it behaves like $∑ 1/√n$ , not $∑ 1/n^2$ . Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.
Determine the type of convergence of $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ : absolute, conditional, or divergent.
Answer: Converges absolutely This question requires determining whether the alternating series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ converges absolutely, conditionally, or diverges. To test for absolute convergence, we examine $\sum_{n=1}^{\infty} \frac{1}{n^2}$ , which is a p-series with $p=2>1$ , so it converges. Since the series of absolute values converges, the original series converges absolutely—it doesn't matter that it's alternating. The original series also converges by the alternating series test, but absolute convergence is the stronger condition. Choice A incorrectly suggests only conditional convergence, missing that the absolute value series converges. Remember: when the absolute value series converges, always classify as absolutely convergent, not conditionally convergent.
Determine whether $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}}$ converges absolutely, conditionally, or diverges.
Answer: Converges conditionally but not absolutely. This question requires determining absolute versus conditional convergence for an alternating p-series. The series of absolute values is $\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}$ , which diverges as a p-series with $p = \frac{1}{2} < 1$ . However, the alternating series satisfies both conditions of the Alternating Series Test: $\frac{1}{\sqrt{n}}$ decreases monotonically and approaches 0 as $n\to\infty$ . Choice C mistakenly claims the p-series converges, but p-series only converge when $p > 1$ . The series converges by the Alternating Series Test but diverges absolutely, making it conditionally convergent—a key pattern is that alternating p-series with $0 < p \leq 1$ always converge conditionally.

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