sample_midtermI_21c_solutions3

sample_midtermI_21c_solutions3 - Calculus Math 21C, Fall...

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Calculus Math 21C, Fall 2010 Solutions to Sample Questions: Midterm I 1. Do the following sequences { a n } converge or diverge as n → ∞ ? Give reasons for your answer. If a sequence converges, find its limit. (a) a n = cos n n ; (b) a n = n ln n ; (c) a n = n 2 + 1 - n. Solution. (a) We have - 1 n cos n n 1 n . Since - 1 n 0 , 1 n 0 as n → ∞ , the ‘sandwich’ theorem implies that cos n/n 0 as n → ∞ . So this sequence converges to 0. (b) Since x → ∞ and ln x → ∞ as x → ∞ , l’Hˆospital’s rule implies that lim n →∞ n ln n = lim x →∞ x ln x = lim x →∞ 1 / (2 x ) 1 /x = lim x →∞ x 2 = . So this sequence diverges to . 1
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(c) We have lim n →∞ ( n 2 + 1 - n ) = lim n →∞ ± n 2 + 1 - n ²± n 2 + 1 + n ² n 2 + 1 + n = lim n →∞ ( n 2 + 1) - n 2 n 2 + 1 + n = lim n →∞ 1 n 2 + 1 + n = lim n →∞ 1 n ³ 1 1 + 1 /n 2 + 1 ´ = 0 So this sequence converges to 0. 2
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Do the following series converge absolutely, converge conditionally, or diverge? Give reasons for your answer. (a) n =1 ( - 1) n +1 n ; (b) n =1 sin n n 2 ; (c) n =1 ( - 1) n sin n Solution. (a) The series of absolute values, n =1 1 n is a p -series with p = 1 / 2, which diverges since p < 1. Thus, the series is not absolutely convergent. Since u n = 1 / n is a decreasing positive sequence with u n 0 as n → ∞ , the alternating series test implies that the series converges. So the series is conditionally convergent. (b) The series of absolute values n =1 | sin n | n 2 converges by the comparison test, since 0 | sin n | n 2 1 n 2 and n =1 1 n 2 is a convergent p -series (with p = 2 > 1). So the series is absolutely convergent. (c) The limit lim n →∞ ( - 1) n sin n does not exist (since, for example, we can find arbitrarily large even integers m , n such that sin m > 1 / 2 and sin n < - 1 / 2). So the series diverges by the n th term test. 3
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sample_midtermI_21c_solutions3 - Calculus Math 21C, Fall...

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