Absolute vs. Conditional Convergence - Calculus 2

1. Let $S_p = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p}$. Which of the following statements is true?
   Answer: $S_p$ converges absolutely for $p > 1$ and conditionally for $0 < p \le 1$. When you encounter an alternating series like $S_p = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p}$, you need to analyze both absolute and conditional convergence separately. This requires testing the series without the alternating factor and then applying the Alternating Series Test. First, examine absolute convergence by considering $\sum_{n=1}^{\infty} \left|\frac{(-1)^{n-1}}{n^p}\right| = \sum_{n=1}^{\infty} \frac{1}{n^p}$. This is the famous p-series, which converges absolutely only when $p > 1$. For $0 < p \leq 1$, the p-series diverges. Next, when absolute convergence fails ($0 < p \leq 1$), check conditional convergence using the Alternating Series Test. The terms $\frac{1}{n^p}$ are positive, decreasing, and approach zero as $n \to \infty$ for any $p > 0$. Therefore, the alternating series converges conditionally for $0 < p \leq 1$. For $p > 1$, since we already have absolute convergence, conditional convergence becomes irrelevant—absolute convergence is the stronger condition. Answer choice B incorrectly swaps the conditions for absolute and conditional convergence. Choice C claims only conditional convergence for all $p > 0$, missing that absolute convergence occurs for $p > 1$. Choice D incorrectly claims absolute convergence for all $p > 0$, ignoring that p-series diverge for $p \leq 1$. Study tip: Remember that p-series $\sum \frac{1}{n^p}$ converge only for $p > 1$. For alternating series, always check absolute convergence first, then apply the Alternating Series Test if needed.
2. If $\sum_{n=1}^{\infty} |a_n|$ converges and $\sum_{n=1}^{\infty} a_n$ also converges, which statement about the series $\sum_{n=1}^{\infty} (-1)^n a_n$ is necessarily true?
   Answer: The series $\sum_{n=1}^{\infty} (-1)^n a_n$ converges absolutely since $\sum |(-1)^n a_n| = \sum |a_n|$ converges Since $\sum |a_n|$ converges by hypothesis, and $|(-1)^n a_n| = |(-1)^n| \cdot |a_n| = |a_n|$, we have $\sum |(-1)^n a_n| = \sum |a_n|$ converges. This means $\sum (-1)^n a_n$ converges absolutely. Choice B incorrectly concludes conditional convergence when we have absolute convergence. Choice C is wrong because absolute convergence of a series guarantees convergence of the series itself. Choice D is irrelevant since we've established absolute convergence regardless of sign changes in $a_n$.
3. Suppose $a_n > 0$ for all $n$, and the series $\sum (-1)^n a_n$ converges. What additional condition would guarantee that this convergence is conditional and not absolute?
   Answer: The series $\sum a_n$ diverges. Conditional convergence means the series converges, but not absolutely. We are given that $\sum (-1)^n a_n$ converges. For the convergence to be conditional, the series of absolute values, $\sum |(-1)^n a_n| = \sum a_n$, must diverge. Therefore, the additional condition needed is that $\sum a_n$ diverges. Choices A and B are conditions of the AST, which are necessary for the convergence of $\sum (-1)^n a_n$ but don't say anything about whether the convergence is absolute or conditional. Choice C represents an inconclusive Ratio Test, which is common for such series but doesn't guarantee divergence of $\sum a_n$ (e.g., for $\sum 1/n^2$, the ratio limit is 1 but the series converges).
4. Determine the convergence behavior of the series $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n!}$.
   Answer: The series converges absolutely. When you encounter a series with alternating signs and factorials, you need to test for both absolute and conditional convergence. This series has the form $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n!}$, so start by checking absolute convergence using $\sum_{n=1}^{\infty} \frac{n^2}{n!}$. The Ratio Test works perfectly here since factorials grow extremely rapidly. For $\sum \frac{n^2}{n!}$, compute: $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)^2}{(n+1)!} \cdot \frac{n!}{n^2} = \lim_{n \to \infty} \frac{(n+1)^2}{(n+1) \cdot n^2} = \lim_{n \to \infty} \frac{n+1}{n^2} = 0$ Since this limit is less than 1, the series of absolute values converges, meaning the original alternating series converges absolutely. Choice A is wrong because conditional convergence occurs when a series converges but its absolute values diverge—that's not the case here. Choice B is incorrect because the Ratio Test gives a definitive result of 0, which is conclusive. Choice C is wrong because we've shown the series does converge. The key insight is recognizing that factorials dominate polynomial growth so dramatically that even with the $n^2$ term, absolute convergence occurs. Remember: when you see factorials in denominators, the Ratio Test is usually your best friend, and absolute convergence is often the result due to factorial's explosive growth rate.
5. Let $p_n$ denote the $n$-th prime number. The series $\sum_{n=1}^\infty \frac{(-1)^n}{p_n}$ converges. What is the nature of its convergence?
   Answer: It converges conditionally, as $\sum \frac{1}{p_n}$ is a known divergent series. When you encounter an alternating series with terms involving primes, you need to determine whether it converges absolutely or conditionally. This requires checking both the original alternating series and the series of absolute values. The series $\sum_{n=1}^\infty \frac{(-1)^n}{p_n}$ is alternating with terms $\frac{1}{p_n}$. Since prime numbers increase without bound, $\frac{1}{p_n} \to 0$, and the terms decrease monotonically. By the Alternating Series Test, this series converges. However, for absolute convergence, we must examine $\sum_{n=1}^\infty \left|\frac{(-1)^n}{p_n}\right| = \sum_{n=1}^\infty \frac{1}{p_n}$. This is the famous series of reciprocals of primes, which Euler proved diverges. Since the alternating series converges but the series of absolute values diverges, the convergence is conditional. Choice A correctly identifies this reasoning. Choice B is wrong because even though primes grow quickly, they don't grow fast enough to make $\sum \frac{1}{p_n}$ converge—the growth rate of primes is similar to $n \ln n$, which still leads to divergence. Choice C incorrectly claims the original series diverges, but the Alternating Series Test guarantees convergence. Choice D falsely suggests absolute convergence by comparison with $\sum \frac{1}{n^2}$, but since $\sum \frac{1}{p_n}$ diverges, no such comparison works. Remember: for alternating series, always check both conditional and absolute convergence separately. Many series converge conditionally but not absolutely, especially those involving primes or harmonic-like terms.
6. The series $\sum_{n=1}^{\infty} \frac{(-1)^n (n^2 + 1)}{n^3 + 2n^2 + n}$ requires careful analysis. What is the most precise classification of its convergence?
   Answer: Converges conditionally since the alternating series test succeeds but the comparison with $\sum \frac{1}{n}$ shows divergence For large $n$, $\frac{n^2 + 1}{n^3 + 2n^2 + n} \sim \frac{n^2}{n^3} = \frac{1}{n}$. Since $\sum \frac{1}{n}$ diverges, by limit comparison test, $\sum \frac{n^2 + 1}{n^3 + 2n^2 + n}$ diverges, so absolute convergence fails. However, $\frac{n^2 + 1}{n^3 + 2n^2 + n}$ is positive, decreasing for sufficiently large $n$, and approaches 0, so the alternating series test confirms convergence. Therefore, the series converges conditionally. Choice A misapplies the harmonic series comparison. Choice C ignores that the decreasing condition eventually holds. Choice D incorrectly identifies the asymptotic behavior.
7. Consider the series $\sum_{n=2}^{\infty} \frac{(-1)^n}{n(\ln n)(\ln(\ln n))^2}$. What is the correct convergence classification?
   Answer: Converges absolutely by integral test since $\int_3^{\infty} \frac{dx}{x(\ln x)(\ln(\ln x))^2}$ converges To test absolute convergence, consider $\sum \frac{1}{n(\ln n)(\ln(\ln n))^2}$. Using the integral test with substitution $u = \ln(\ln x)$, so $du = \frac{dx}{x \ln x}$: $\int_3^{\infty} \frac{dx}{x(\ln x)(\ln(\ln x))^2} = \int_{\ln(\ln 3)}^{\infty} \frac{du}{u^2} = \left[-\frac{1}{u}\right]_{\ln(\ln 3)}^{\infty} = \frac{1}{\ln(\ln 3)} < \infty$. Since the integral converges, the series converges absolutely. Choice B incorrectly states the integral diverges. Choice C misunderstands the convergence criteria. Choice D uses vague reasoning without proper analysis.
8. The series $\sum_{n=1}^\infty a_n$ is conditionally convergent. What can be concluded about the convergence of the series $\sum_{n=1}^\infty a_n^2$?
   Answer: It may converge or it may diverge. If $\sum a_n$ is conditionally convergent, we know $\lim_{n\to\infty} a_n = 0$. This implies that for large $n$, $|a_n| < 1$, and thus $a_n^2 < |a_n|$. However, this comparison is not sufficient to guarantee convergence of $\sum a_n^2$. Let's consider two examples. 1. Let $a_n = \frac{(-1)^n}{\sqrt{n}}$. $\sum a_n$ converges conditionally. Then $\sum a_n^2 = \sum \frac{1}{n}$, which is the divergent harmonic series. 2. Let $a_n = \frac{(-1)^n}{n}$. $\sum a_n$ converges conditionally. Then $\sum a_n^2 = \sum \frac{1}{n^2}$, which is a convergent p-series. Since we have found examples where $\sum a_n^2$ converges and examples where it diverges, we must conclude that it may converge or it may diverge.
9. Let $\sum a_n$ be a series that converges to a sum $L$. If the terms of the series are rearranged to form a new series $\sum b_n$, under which condition must $\sum b_n$ also converge to $L$?
   Answer: The series $\sum a_n$ converges absolutely. This question refers to the Riemann Rearrangement Theorem. The theorem states that if a series is absolutely convergent, then any rearrangement of its terms results in a series that converges to the same sum. If a series is conditionally convergent, its terms can be rearranged to converge to any real number, or to diverge. Therefore, absolute convergence is the condition required to guarantee that the sum remains unchanged after rearrangement.
10. Let $a_n = \frac{(-1)^n}{n^p}$. The series $\sum a_n$ converges conditionally if and only if:
    Answer: $0 < p \le 1$ This is an alternating p-series. The series of absolute values is $\sum |a_n| = \sum \frac{1}{n^p}$, which is a standard p-series. It converges if $p > 1$ and diverges if $p \le 1$. For the original series $\sum \frac{(-1)^n}{n^p}$ to converge, we can use the Alternating Series Test. The conditions are that $b_n = \frac{1}{n^p}$ must be positive, decreasing, and approach 0. All these conditions are met if $p > 0$. Conditional convergence occurs when the series converges ($p > 0$) but does not converge absolutely ($p \le 1$). The intersection of these two conditions is $0 < p \le 1$.
11. How does the series $\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{\sqrt{n^2+1}}$ behave?
    Answer: It converges conditionally. First, recognize that $\cos(n\pi) = (-1)^n$. The series is $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n^2+1}}$. To check for absolute convergence, consider $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+1}}$. Using the Limit Comparison Test with the divergent harmonic series $\sum \frac{1}{n}$, we find $\lim_{n\to\infty} \frac{1/\sqrt{n^2+1}}{1/n} = \lim_{n\to\infty} \frac{n}{\sqrt{n^2+1}} = 1$. Since the limit is a finite positive number, and $\sum \frac{1}{n}$ diverges, the series of absolute values also diverges. Thus, the original series does not converge absolutely. To check for conditional convergence, we apply the Alternating Series Test. Let $b_n = \frac{1}{\sqrt{n^2+1}}$. The terms are positive, $\lim_{n\to\infty} b_n = 0$, and the terms are decreasing. Therefore, the series converges. Since it converges but not absolutely, it converges conditionally.
12. For which values of the real number $p$ does the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^p + 3}$ converge conditionally?
    Answer: $1 < p \le 2$ First, test for absolute convergence using the Limit Comparison Test on $\sum \frac{n}{n^p + 3}$ with the p-series $\sum \frac{n}{n^p} = \sum \frac{1}{n^{p-1}}$. The limit of the ratio of terms is 1. The p-series $\sum \frac{1}{n^{p-1}}$ converges if $p-1 > 1$, which means $p > 2$. So, the original series converges absolutely for $p > 2$. Next, test for convergence using the Alternating Series Test. The terms $b_n = \frac{n}{n^p + 3}$ must approach 0, which requires $p > 1$. The terms must also be decreasing for large $n$, which is also true for $p > 1$. Thus, the series converges for all $p > 1$. Conditional convergence occurs when the series converges but not absolutely. This corresponds to the interval $p > 1$ (for convergence) but not $p > 2$ (for absolute convergence), which is the interval $1 < p \le 2$.
13. The series $\sum_{n=1}^\infty \frac{(-1)^n (n+1)}{n^2+1}$ is best described as:
    Answer: Conditionally convergent by the Alternating Series Test and Limit Comparison Test. When you encounter an alternating series like this one, you need to test for both conditional and absolute convergence. An alternating series converges conditionally if the series itself converges but the series of absolute values diverges. For the series $\sum_{n=1}^\infty \frac{(-1)^n (n+1)}{n^2+1}$, first apply the Alternating Series Test. You need: (1) the terms $\frac{n+1}{n^2+1}$ must decrease, and (2) $\lim_{n \to \infty} \frac{n+1}{n^2+1} = 0$. The limit equals zero since the denominator grows faster than the numerator. For large n, the terms do decrease, so the series converges by the Alternating Series Test. Next, test for absolute convergence by examining $\sum_{n=1}^\infty \frac{n+1}{n^2+1}$. Using the Limit Comparison Test with $\sum \frac{1}{n}$: $\lim_{n \to \infty} \frac{\frac{n+1}{n^2+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n(n+1)}{n^2+1} = 1$. Since this limit is positive and finite, and $\sum \frac{1}{n}$ diverges, the series of absolute values diverges. Therefore, the series is conditionally convergent. Option B is wrong because the Ratio Test would show convergence to 1, indicating inconclusive results, not absolute convergence. Option C fails because $\lim_{n \to \infty} \frac{(-1)^n (n+1)}{n^2+1} = 0$. Option D is incorrect because comparison with $\sum \frac{1}{n^2}$ doesn't work—the terms behave like $\frac{1}{n}$, not $\frac{1}{n^2}$. Study tip: Always check both conditional and absolute convergence for alternating series—use limit comparison to find the right comparison series based on the dominant terms.
14. Given that $\sum_{n=1}^\infty a_n$ converges, but $\sum_{n=1}^\infty (-1)^n a_n$ diverges. What can be concluded about the series $\sum a_n$?
    Answer: It converges conditionally. Let $\sum a_n$ be the series. We are given that it converges. So it must be either absolutely or conditionally convergent. If $\sum a_n$ were absolutely convergent, then $\sum |a_n|$ would converge. This would imply that $\sum |(-1)^n a_n| = \sum |a_n|$ also converges, meaning $\sum (-1)^n a_n$ must be absolutely convergent, and therefore convergent. However, we are told that $\sum (-1)^n a_n$ diverges. This creates a contradiction. Therefore, the initial assumption that $\sum a_n$ converges absolutely must be false. Since $\sum a_n$ converges, but not absolutely, it must converge conditionally.
15. Consider the series $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n}$$ Which of the following statements accurately describes its convergence?
    Answer: The series converges conditionally. To test for absolute convergence, we examine the series $\sum_{n=2}^{\infty} |\frac{(-1)^n}{n \ln n}| = \sum_{n=2}^{\infty} \frac{1}{n \ln n}$. We can use the Integral Test. The integral $\int_2^\infty \frac{1}{x \ln x} dx$ with substitution $u = \ln x$ becomes $\int_{\ln 2}^\infty \frac{1}{u} du = [\ln|u|]_{\ln 2}^\infty$, which diverges. Thus, the series does not converge absolutely. To test for conditional convergence, we use the Alternating Series Test (AST). Let $b_n = \frac{1}{n \ln n}$. The terms are positive, $\lim_{n \to \infty} b_n = 0$, and the terms are decreasing since $f(x) = x \ln x$ is an increasing function for $x \ge 2$. Since the conditions for AST are met, the series converges. Because the series converges but does not converge absolutely, it converges conditionally.
16. If the series $\sum_{n=1}^{\infty} a_n$ converges conditionally, what can be concluded about the series $\sum_{n=1}^{\infty} a_n^2$?
    Answer: The convergence of $\sum a_n^2$ cannot be determined without additional information about the $a_n$ Conditional convergence of $\sum a_n$ means $\sum a_n$ converges but $\sum |a_n|$ diverges. However, this gives no direct information about $\sum a_n^2$. For example: (1) If $a_n = \frac{(-1)^n}{\sqrt{n}}$, then $\sum a_n$ converges conditionally, but $\sum a_n^2 = \sum \frac{1}{n}$ diverges. (2) If $a_n = \frac{(-1)^n}{n}$, then $\sum a_n$ converges conditionally, and $\sum a_n^2 = \sum \frac{1}{n^2}$ converges. Choice A assumes convergence without justification. Choice B incorrectly assumes divergence. Choice D misunderstands convergence classification.
17. If $\sum_{n=1}^\infty a_n$ converges conditionally, which statement about $\sum_{n=1}^\infty |a_n|$ must be true?
    Answer: $\sum_{n=1}^\infty |a_n|$ diverges to infinity. The definition of conditional convergence is that the series $\sum a_n$ converges, but the series of absolute values $\sum |a_n|$ diverges. Since the terms $|a_n|$ are all non-negative, if the series $\sum |a_n|$ diverges, its partial sums must increase without bound. Therefore, $\sum |a_n|$ must diverge to infinity.
18. For the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \sqrt{n}}{n + \sqrt{n} + 1}$, determine its convergence classification after careful analysis.
    Answer: Converges conditionally since the alternating series converges but $\sum \frac{\sqrt{n}}{n + \sqrt{n} + 1}$ diverges As $n \to \infty$, $\frac{\sqrt{n}}{n + \sqrt{n} + 1} \sim \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$. Since $\sum \frac{1}{\sqrt{n}}$ diverges (p-series with $p = 1/2 < 1$), the series of absolute values diverges. However, $\frac{\sqrt{n}}{n + \sqrt{n} + 1}$ is decreasing for large $n$ and approaches 0, so by the alternating series test, the original series converges. This gives conditional convergence. Choice A incorrectly states that $\sum \frac{1}{\sqrt{n}}$ converges. Choice B ignores the alternating series test. Choice D has wrong asymptotic behavior.
19. Consider the series $S = \sum_{n=1}^\infty (-1)^n a_n$, where $a_n = \int_n^{n+1} \frac{1}{x} dx$. Which statement is correct?
    Answer: The series S converges conditionally. First, evaluate the term $a_n$: $a_n = \int_n^{n+1} \frac{1}{x} dx = [\ln|x|]_n^{n+1} = \ln(n+1) - \ln(n) = \ln\left(\frac{n+1}{n}\right)$. The series is $\sum_{n=1}^\infty (-1)^n \ln(1 + \frac{1}{n})$. Let's check for absolute convergence: $\sum \ln(1 + \frac{1}{n})$. Using the Limit Comparison Test with $\sum \frac{1}{n}$, we have $\lim_{n\to\infty} \frac{\ln(1+1/n)}{1/n} = 1$ (a standard limit, provable with L'Hopital's rule on $\frac{\ln(1+x)}{x}$ as $x\to 0$). Since $\sum \frac{1}{n}$ diverges, the series of absolute values diverges. Now, apply the AST to the original series. Let $b_n = \ln(1+1/n)$. The terms are positive. $\lim_{n\to\infty} \ln(1+1/n) = \ln(1) = 0$. The terms are decreasing since $1+1/n$ is a decreasing function and $\ln(x)$ is an increasing function. Thus, the series converges by AST. Since it converges but not absolutely, it is conditionally convergent.
20. The series $\sum_{n=1}^{\infty} \frac{(-1)^n \ln(n^2 + 1)}{n \sqrt{n^2 + 1}}$ requires sophisticated analysis. What is its convergence classification?
    Answer: Converges absolutely since $\frac{\ln(n^2 + 1)}{n \sqrt{n^2 + 1}} \sim \frac{2\ln n}{n^2}$ and $\sum \frac{\ln n}{n^2}$ converges When analyzing the convergence of an alternating series with complex terms, you need to examine both absolute and conditional convergence by studying the behavior of the terms as n approaches infinity. For this series, the key insight is finding the asymptotic behavior of $\frac{\ln(n^2 + 1)}{n \sqrt{n^2 + 1}}$. As n becomes large, $n^2 + 1 \approx n^2$, so $\ln(n^2 + 1) \approx \ln(n^2) = 2\ln n$ and $\sqrt{n^2 + 1} \approx n$. Therefore, $\frac{\ln(n^2 + 1)}{n \sqrt{n^2 + 1}} \sim \frac{2\ln n}{n \cdot n} = \frac{2\ln n}{n^2}$. Since $\sum \frac{\ln n}{n^2}$ converges (this is a known result from integral test or comparison with $\sum \frac{1}{n^{1.5}}$), the series $\sum \frac{2\ln n}{n^2}$ also converges. By the limit comparison test, our original series converges absolutely, making answer D correct. Answer A incorrectly assumes logarithmic growth prevents absolute convergence, but the $n^2$ denominator dominates. Answer B contains a logical contradiction—if the asymptotic gives absolute convergence, that doesn't contradict the alternating nature; it simply means the series converges absolutely. Answer C wrongly applies the alternating series test; while $\ln(n^2 + 1)$ grows without bound, the full term $\frac{\ln(n^2 + 1)}{n \sqrt{n^2 + 1}}$ approaches zero. Study tip: For series with logarithmic terms, always check if higher-order polynomial terms in the denominator can overcome the logarithmic growth—logarithms grow slower than any positive power of n.

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