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  • Math Archive: Questions from 2024-03-5

    Estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. Round to six decimal places. ∑ n = 1 ∞ (−1) n + 1 1 2 n A. 0.015625 B. 0.062500 C. 0.031250 D. 0.156250

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    Estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. ∑ n = 1 ∞ (−1) n + 1 t n n, − 1 < t ≤ 1 |Error| < |t 4 4| |Error| < 0.20 |Error| < |t 5| |Error| < |t 5 5| |Error| < |t 3 3| |Error| < |t 4| |Error| < |t 3|

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    Consider the series ∑ n = 2 ∞ x n n(lnn) 10. (a) Find the series' radius and interval of convergence. (b) For what values of x does the series converge absolutely? (c) For what values of x does the series converge conditionally?

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    Use the nth-term test for divergence to show that the series is divergent, or state that the test is inconclusive. ∑ n = 0 ∞ 1 n + 13 Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The test is inconclusive because lim n → ∞ 1 n + 13 = ◻. B. The series diverges because lim n → ∞ 1 n + 13 exists and is equal to ◻. C. The series diverges because lim n → ∞ 1 n + 13 = − ∞ and fails to exist. D. The series diverges because lim n → ∞ 1 n + 13 = ∞ and fails to exist.

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    Use the nth-term test for divergence to show that the series is divergent, or state that the test is inconclusive. ∑ n = 1 ∞ cos ⁡ 11 n Choose the correct answer below. A. The test is inconclusive because lim n → ∞ cos ⁡ 11 n = ◻. B. The series diverges because lim n → ∞ cos ⁡ 11 n = ∞ and fails to exist. C. The series diverges because lim n → ∞ cos ⁡ 11 n exists and is equal to ◻. D. The series diverges because lim n → ∞ cos ⁡ 11 n = − ∞ and fails to exist.

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    Use the nth-term test for divergence to show that the series is divergent, or state that the test is inconclusive. ∑ n = 1 ∞ ln ⁡ 1 n Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The series diverges because lim n → ∞ ln ⁡ 1 n = − ∞ and fails to exist. B. The series diverges because lim n → ∞ ln ⁡ 1 n exists and is equal to ◻. C. The series diverges because lim n → ∞ ln ⁡ 1 n = ∞ and fails to exist. D. The test is inconclusive because lim n → ∞ ln ⁡ 1 n = ◻.

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    Consider the following series. ∑ n = 1 ∞ (−1) n + 1 n 4 (∣error∣ < 0.00005) Show that the series is convergent. Since this series is -- Select--, which condition(s) below show that it converges? (Select all that apply.) lim n → ∞ 1 n 4 = 0 1 (n + 1) 4 < 1 n 4 1 (n + 1) 4 > 1 n 4 lim n → ∞ 1 (n + 1) 4 = 0 How many terms of the series do we need to add in order to find the sum to the indicated accuracy? terms

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    Consider the alternating series. ∑ n = 1 ∞ (−1) n + 1 n Given that the series converges, find the smallest integer k such that the kth partial sum, ∑ n = 1 k (−1) n + 1 n, estimates the series to within |0.01|.

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    The series ∑ n = 1 ∞ (−1) n cos(1/n) is Select one: a. Convergent by Alternating Series Test. b. Divergent by nth-Term Test. c. Divergent by Alternating Series Test. d. Non of them. e. Absolutely Convergent.

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    Consider the Alternating Series S = ∑ n = 1 ∞ (−1) n n 2 Using the Alternating Series Remainder Theorem, find the smallest value of N so that the partial sum SN has error |S − SN| < 0.00025 = 1/4000 (a) N = 323 (b) N = 2 (c) N = 32 (d) N = 63 (e) N = 171

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