Alternating Series Test - Calculus 2

1. If the Alternating Series Test is used to establish the convergence of $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$, what is the minimum number of terms required to approximate the sum with an error less than 0.01?
   Answer: 99 The Alternating Series Estimation Theorem states that the error of the Nth partial sum, $|S - S_N|$, is less than or equal to the magnitude of the first neglected term, $b_{N+1}$. We want $|S - S_N| < 0.01$. We need to find $N$ such that $b_{N+1} < 0.01$. Here, $b_n = 1/n$, so we need $b_{N+1} = \frac{1}{N+1} < 0.01 = \frac{1}{100}$. This inequality becomes $N+1 > 100$, which means $N > 99$. The smallest integer value for $N$ is 100, but the question asks for the minimum number of terms in the sum, which is $N$. To ensure the error is less than $b_{N+1}$, we need to sum up to the $N$th term. If we sum $N=99$ terms, the error is less than $b_{100} = 1/100 = 0.01$. So, 99 terms are required.
2. Let $b_n$ be a positive, decreasing sequence with limit 0. The sum of the series $S = \sum_{n=1}^{\infty} (-1)^{n+1}b_n$ is approximated by the second partial sum, $S_2 = b_1 - b_2$. Which of the following must be true about the true sum $S$?
   Answer: $b_1 - b_2 < S < b_1$ The series converges by the AST. The sequence of partial sums oscillates around the true sum $S$. The first term is $b_1$, so the first partial sum is $S_1 = b_1$. The second term is $-b_2$, so $S_2 = b_1 - b_2$. Since the terms are decreasing, $b_1 - b_2 > 0$. The odd partial sums are overestimates and the even partial sums are underestimates. Therefore, $S_2 < S < S_1$. Substituting the values for the partial sums gives $b_1 - b_2 < S < b_1$.
3. Which statement correctly describes the series $\sum_{n=2}^{\infty} (-1)^n \frac{\ln(n^2)}{n}$?
   Answer: The series converges conditionally. The series can be written as $\sum_{n=2}^{\infty} (-1)^n \frac{2\ln n}{n}$. Let $b_n = \frac{2\ln n}{n}$. Using L'Hôpital's Rule, $\lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac{2/n}{1} = 0$. To check if $b_n$ is decreasing, consider the function $f(x) = \frac{2\ln x}{x}$. Its derivative is $f'(x) = \frac{2(1-\ln x)}{x^2}$, which is negative for $x > e$. Thus, the terms are eventually decreasing. By the AST, the series converges. For absolute convergence, we examine $\sum \frac{2\ln n}{n}$. Since $\ln n > 1$ for $n > e$, we have $\frac{2\ln n}{n} > \frac{2}{n}$. Since $\sum \frac{2}{n}$ diverges, $\sum \frac{2\ln n}{n}$ diverges by comparison. Therefore, the series converges conditionally.
4. The series $\sum_{n=2}^{\infty} (-1)^n \frac{1}{\ln(n)}$ is being analyzed for convergence. What can be concluded?
   Answer: The series converges conditionally by the Alternating Series Test since $\frac{1}{\ln(n)}$ decreases to 0 The Alternating Series Test applies here: (1) $\lim_{n \to \infty} \frac{1}{\ln(n)} = 0$, and (2) $\frac{1}{\ln(n)}$ is decreasing since $\ln(n)$ is increasing. However, $\sum_{n=2}^{\infty} \frac{1}{\ln(n)}$ diverges by the integral test ($\int_2^{\infty} \frac{1}{\ln(x)} dx$ diverges), so the convergence is conditional. Choice B incorrectly concludes the alternating series diverges. Choice C wrongly claims absolute convergence. Choice D misapplies the divergence test - having limit 0 doesn't cause divergence.
5. Consider the series $\sum_{n=1}^{\infty} (-1)^n (\sqrt{n+1} - \sqrt{n})$. What is the behavior of this series?
   Answer: Converges conditionally. Let $b_n = \sqrt{n+1} - \sqrt{n}$. We can rationalize this expression: $b_n = (\sqrt{n+1} - \sqrt{n}) \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{n+1-n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$. The sequence $\{b_n\}$ is positive, decreasing, and $\lim_{n\to\infty} b_n = 0$. By the AST, the series converges. To test for absolute convergence, we examine $\sum b_n = \sum \frac{1}{\sqrt{n+1} + \sqrt{n}}$. We can use the Limit Comparison Test with the divergent p-series $\sum \frac{1}{\sqrt{n}}$. The limit of the ratio of terms is $1/2$. Since the p-series diverges, $\sum b_n$ also diverges. Thus, the series converges conditionally.
6. Consider the series $\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n+1} - \sqrt{n}}{n}$. To determine convergence using the Alternating Series Test, what must be verified?
   Answer: That $\lim_{n \to \infty} \frac{\sqrt{n+1} - \sqrt{n}}{n} = 0$ and the sequence is eventually decreasing, both of which hold First, simplify: $\sqrt{n+1} - \sqrt{n} = \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$. So the general term is $\frac{1}{n(\sqrt{n+1} + \sqrt{n})} \sim \frac{1}{2n^{3/2}}$. For the Alternating Series Test: (1) $\lim_{n \to \infty} \frac{1}{n(\sqrt{n+1} + \sqrt{n})} = 0$, and (2) this sequence is decreasing. Both conditions hold, so the series converges. Choice B doesn't verify the decreasing condition properly. Choice C discusses absolute convergence but doesn't address the Alternating Series Test. Choice D gives the right simplification but doesn't complete the AST verification.
7. The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4}$ converges to a sum $S$. If $S$ is approximated by the 4th partial sum, $S_4$, the Alternating Series Estimation Theorem states that the absolute error $|S - S_4|$ is bounded by what value?
   Answer: The magnitude of the 5th term, $1/625$. The Alternating Series Estimation Theorem states that for a convergent alternating series satisfying its conditions, the absolute error in approximating the total sum $S$ by the Nth partial sum $S_N$ is less than or equal to the magnitude of the first neglected term, $b_{N+1}$. In this case, $N=4$ and $b_n = 1/n^4$. The error bound is $|S - S_4| \le b_{4+1} = b_5 = 1/5^4 = 1/625$.
8. Given that $b_n > 0$, $\lim_{n\to\infty} b_n = 0$, and the series $\sum_{n=1}^\infty (-1)^n b_n$ diverges. What must be true about the sequence $\{b_n\}$?
   Answer: The sequence $\{b_n\}$ is not eventually monotonically decreasing. The Alternating Series Test has two conditions for convergence: (1) $\lim_{n\to\infty} b_n = 0$ and (2) $\{b_n\}$ is eventually decreasing. The problem states that condition (1) is met, but the series still diverges. This implies that the AST is not satisfied, so condition (2) must fail. Therefore, the sequence $\{b_n\}$ cannot be eventually monotonically decreasing.
9. Which of the following alternating series diverges?
   Answer: $\sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n+1}$ When you encounter questions about alternating series convergence, your primary tool is the Alternating Series Test (Leibniz criterion). This test states that an alternating series $\sum (-1)^n a_n$ converges if: (1) the terms $a_n$ are decreasing, and (2) $\lim_{n \to \infty} a_n = 0$. However, if the limit condition fails, the series diverges by the Divergence Test. Let's examine each series by checking whether the terms approach zero: For choice A: $\sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n+1}$, we have $\lim_{n \to \infty} \frac{n+1}{2n+1} = \lim_{n \to \infty} \frac{1 + 1/n}{2 + 1/n} = \frac{1}{2} \neq 0$. Since the terms don't approach zero, this series diverges by the Divergence Test. Choice B: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\ln(n+1)}$ has terms $\frac{1}{\ln(n+1)} \to 0$ as $n \to \infty$, and these terms are decreasing for large $n$. This converges by the Alternating Series Test. Choice C: $\sum_{n=1}^{\infty} (-1)^n e^{-n}$ has terms $e^{-n} \to 0$ rapidly, and they're clearly decreasing. This converges. Choice D: $\sum_{n=1}^{\infty} \frac{\sin((n+1/2)\pi)}{n}$ can be rewritten as $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, which is the alternating harmonic series. Since $\frac{1}{n} \to 0$ and the terms decrease, this converges. Therefore, A is the only divergent series. Study tip: Always check the limit of terms first—if they don't approach zero, the series diverges immediately, regardless of the alternating pattern.
10. Consider the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\arctan(n)}{n^{3/2}}$. Which conclusion about this series is justified?
    Answer: The series converges absolutely since $\arctan(n) < \frac{\pi}{2}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges Since $\arctan(n) < \frac{\pi}{2}$ for all n, we have $\frac{\arctan(n)}{n^{3/2}} < \frac{\pi/2}{n^{3/2}}$. Since $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges (p-series with p = 3/2 > 1), the series $\sum_{n=1}^{\infty} \frac{\arctan(n)}{n^{3/2}}$ converges by direct comparison. Therefore, the alternating series converges absolutely. Choice A is wrong because limit comparison with $\sum \frac{1}{n^{1/2}}$ (which diverges) would be incorrect - we'd need $\sum \frac{1}{n^{3/2}}$. Choice B discusses conditional convergence but the convergence is actually absolute. Choice D incorrectly claims non-decreasing behavior when absolute convergence makes this irrelevant.
11. The Alternating Series Test can be used to show convergence for which of the following series?
    Answer: $\sum_{n=1}^{\infty} (-1)^n \frac{n}{e^n}$ A) This series is $0 + 2/2 + 0 + 2/4 + ...$ which is the harmonic series, not an alternating series. AST does not apply. B) The limit of the terms $b_n = n/(n+1)$ is 1, not 0. Series diverges by nth-Term Test. AST does not show convergence. C) The terms $b_n = (2+\sin n)/n^2$ are not guaranteed to be decreasing due to the $\sin n$ term. AST cannot be directly applied. D) Let $b_n = n/e^n$. $\lim_{n\to\infty} n/e^n = 0$ by L'Hôpital's Rule. The function $f(x) = x/e^x$ has derivative $f'(x) = (1-x)/e^x$, which is negative for $x>1$. So, the terms are decreasing for $n\ge 2$. Both conditions of the AST are met.
12. Consider $\sum_{n=1}^{\infty} (-1)^n a_n$ where $a_n = \frac{n^2 + 3n + 1}{n^3 + n^2 + 5}$. Which statement about applying the Alternating Series Test is most accurate?
    Answer: The test confirms convergence since $a_n \to 0$ and $a_n$ is eventually decreasing as verified by the derivative First, $\lim_{n \to \infty} \frac{n^2 + 3n + 1}{n^3 + n^2 + 5} = \lim_{n \to \infty} \frac{n^2(1 + 3/n + 1/n^2)}{n^3(1 + 1/n + 5/n^3)} = \lim_{n \to \infty} \frac{1}{n} \cdot \frac{1 + 3/n + 1/n^2}{1 + 1/n + 5/n^3} = 0$. For large n, $a_n \approx \frac{1}{n}$ which is decreasing. More rigorously, $f(x) = \frac{x^2 + 3x + 1}{x^3 + x^2 + 5}$ has $f'(x) = \frac{(2x+3)(x^3+x^2+5) - (x^2+3x+1)(3x^2+2x)}{(x^3+x^2+5)^2}$. For large x, the numerator behaves like $2x^4 - 3x^4 = -x^4 < 0$, so f is eventually decreasing. Choice B incorrectly evaluates the limit. Choice C wrongly claims we can't verify decreasing. Choice D misunderstands the analysis.
13. If $S_N = \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^2}$ is used to approximate $S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}$, what is the relationship between the true sum $S$ and the partial sum $S_N$?
    Answer: $S > S_N$ for all even $N$, and $S < S_N$ for all odd $N$. For an alternating series where $b_n$ is decreasing, the partial sums oscillate around the true sum $S$. The first term is $a_1 = 1/1^2 = 1 > 0$. The odd partial sums $S_1, S_3, ...$ form a decreasing sequence that converges to $S$ from above, so $S_N > S$ for odd $N$. The even partial sums $S_2, S_4, ...$ form an increasing sequence that converges to $S$ from below, so $S_N < S$ for even $N$. Rearranging these inequalities gives $S < S_N$ for odd $N$ and $S > S_N$ for even $N$.
14. Which of the following statements is true for the series $\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n^{1/2} + 1}$?
    Answer: The series converges conditionally. The term $\cos(n\pi)$ is equivalent to $(-1)^n$. So the series is $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n} + 1}$. This is an alternating series with $b_n = \frac{1}{\sqrt{n} + 1}$. Since $\lim_{n\to\infty} b_n = 0$ and $b_n$ is a decreasing sequence, the series converges by the Alternating Series Test. To check for absolute convergence, we examine $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + 1}$. By the Limit Comparison Test with the divergent p-series $\sum \frac{1}{\sqrt{n}}$ (where $p=1/2$), this series diverges. Since the original series converges but not absolutely, it converges conditionally.
15. Let $f(x)$ be a positive, continuous, and decreasing function for $x \ge 1$ such that $\int_1^\infty f(x) dx$ diverges. What can be concluded about the series $\sum_{n=1}^\infty (-1)^n f(n)$?
    Answer: It must converge conditionally, provided $\lim_{x\to\infty} f(x) = 0$. Let $b_n = f(n)$. The conditions given are that $b_n$ is positive and decreasing. For the Alternating Series Test to apply, we also need $\lim_{n\to\infty} b_n = 0$. The problem implicitly suggests this by framing it as a standard convergence problem. If we assume $\lim_{n\to\infty} f(n) = 0$, the series $\sum (-1)^n f(n)$ converges by the AST. The Integral Test states that since $\int_1^\infty f(x) dx$ diverges, the series of absolute values, $\sum f(n)$, must also diverge. A series that converges but does not converge absolutely is conditionally convergent. The condition $\lim_{x\to\infty} f(x) = 0$ is necessary for this conclusion.
16. Consider the series $\sum_{n=1}^{\infty} (-1)^n \frac{n^2}{n^3+10}$. What is its behavior?
    Answer: It converges conditionally. When you encounter an alternating series like $\sum_{n=1}^{\infty} (-1)^n \frac{n^2}{n^3+10}$, you need to test for both conditional and absolute convergence. This requires checking the Alternating Series Test and examining whether the series converges when you remove the alternating signs. For the Alternating Series Test, let $b_n = \frac{n^2}{n^3+10}$. First, check if $\lim_{n \to \infty} b_n = 0$: dividing numerator and denominator by $n^3$ gives $\lim_{n \to \infty} \frac{1}{1 + 10/n^3} = 0$. Next, verify that $b_n$ is eventually decreasing. For large $n$, $b_n \approx \frac{1}{n}$, which decreases. Since both conditions are satisfied, the alternating series converges. To test for absolute convergence, examine $\sum_{n=1}^{\infty} \frac{n^2}{n^3+10}$. Using the Limit Comparison Test with $\sum \frac{1}{n}$: $\lim_{n \to \infty} \frac{n^2/(n^3+10)}{1/n} = \lim_{n \to \infty} \frac{n^3}{n^3+10} = 1$. Since this limit is positive and finite, and $\sum \frac{1}{n}$ diverges, the series of absolute values diverges. Therefore, the series converges conditionally. Option B is wrong because the Alternating Series Test only requires $b_n$ to be eventually decreasing, not for all $n \geq 1$. Option C incorrectly concludes absolute convergence when the comparison actually shows divergence. Option D is wrong because $\lim_{n \to \infty} (-1)^n \frac{n^2}{n^3+10} = 0$, so the nth-Term Test is inconclusive. Study tip: Always check both conditional and absolute convergence for alternating series—convergence of the alternating series doesn't guarantee absolute convergence.
17. Let $\sum_{n=1}^{\infty} (-1)^n b_n$ be a series with $b_n > 0$. If $\lim_{n\to\infty} b_n = L$ where $L$ is a finite positive number, what can be definitively concluded about the series?
    Answer: The series diverges by the nth-Term Test for Divergence. If $\lim_{n\to\infty} b_n = L > 0$, then the limit of the terms of the series, $\lim_{n\to\infty} (-1)^n b_n$, does not exist because the terms oscillate between values close to $L$ and $-L$. Since the limit of the terms is not 0, the series diverges by the nth-Term Test for Divergence. The Alternating Series Test is not merely inconclusive; a definitive conclusion of divergence can be made.
18. What is the behavior of the series $\sum_{n=1}^{\infty} (-1)^n \left( \frac{\pi}{2} - \arctan(n) \right)$?
    Answer: The series converges conditionally. Let $b_n = \frac{\pi}{2} - \arctan(n)$. As $n \to \infty$, $\arctan(n) \to \pi/2$, so $\lim_{n\to\infty} b_n = 0$. Since $\arctan(x)$ is an increasing function, $b_n$ is a decreasing sequence. By the AST, the series converges. To test for absolute convergence, we examine $\sum b_n$. We can use the identity $\frac{\pi}{2} - \arctan(n) = \text{arccot}(n)$. For large $n$, $\text{arccot}(n) \approx 1/n$. Using the Limit Comparison Test with the divergent harmonic series $\sum 1/n$, we find that $\sum b_n$ diverges. Therefore, the series converges conditionally.
19. What is the behavior of the series $\sum_{n=1}^{\infty} \sin\left(\frac{(2n-1)\pi}{2}\right) \frac{1}{\sqrt{n}}$?
    Answer: The series converges conditionally. When you encounter a series with trigonometric terms, you need to analyze both the trigonometric behavior and apply appropriate convergence tests. This series combines a trigonometric function with a power term, making it a candidate for alternating series analysis. First, let's examine $\sin\left(\frac{(2n-1)\pi}{2}\right)$. For odd values of $2n-1$, this gives us $\sin(\pi/2) = 1$, $\sin(3\pi/2) = -1$, $\sin(5\pi/2) = 1$, and so on. The pattern alternates: $1, -1, 1, -1, ...$, so our series becomes $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{\sqrt{n}}$ This is indeed an alternating series where the terms $\frac{1}{\sqrt{n}}$ decrease monotonically to zero. By the Alternating Series Test, the series converges. However, to check absolute convergence, we examine $\sum_{n=1}^{\infty} \left|\sin\left(\frac{(2n-1)\pi}{2}\right)\right| \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ This is a p-series with $p = 1/2 < 1$, which diverges. Since the original series converges but its absolute value diverges, the series converges conditionally. Option B is wrong because this clearly is an alternating series. Option C incorrectly suggests that oscillation alone causes divergence—oscillation can actually help convergence in alternating series. Option D is wrong because absolute convergence requires the p-series to converge, which it doesn't. Strategy tip: For series with trigonometric terms, always evaluate the trig function at specific values to identify patterns, then check both conditional and absolute convergence separately.
20. The series $\sum_{n=1}^{\infty} a_n$ converges conditionally. Which of the following statements about the series $\sum_{n=1}^{\infty} (a_n + |a_n|)$ must be true?
    Answer: It diverges. If $\sum a_n$ converges conditionally, then $\sum a_n$ converges and $\sum |a_n|$ diverges. Consider the new series $\sum (a_n + |a_n|)$. If this series were to converge, then since $\sum a_n$ converges, their difference must also converge. The difference is $\sum (a_n + |a_n|) - \sum a_n = \sum |a_n|$. This would imply that $\sum |a_n|$ converges, which contradicts the definition of conditional convergence. Therefore, the series $\sum (a_n + |a_n|)$ must diverge.

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