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2005a_x3a_sols - Spring 2005 Math 152 Exam 3A Solutions c...

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Spring 2005 Math 152 Exam 3A: Solutions Mon, 02/May c 2005, Art Belmonte 1. (e) Examine the corresponding series . X n = 0 a n = X n = 0 ( - 10 ) n n ! = e - 10 Since this series converges, we must have lim n →∞ a n = 0. Alternatively, look at ln | a n | . As n → ∞ , we have ln 10 n n ! = n ln 10 - n X k = 1 ln k = n X k = 1 ( ln 10 - ln k ) → -∞ . Thus lim | a n | = lim e ln | a n | = 0. Hence lim a n = 0. 2. (c) The series ( - 1 ) n e 1 / n diverges by the Test for Divergence since lim n →∞ a n = lim n →∞ ( - 1 ) n e 1 / n 6= 0. Indeed, lim inf a n = - 1 and lim sup a n = + 1. 3. (d) Compute a few partial sums of this telescoping series until it’s clear what’s happening. Now s 1 = cos 1 2 - cos 1 3 , s 2 = cos 1 2 - cos 1 4 , s 3 = cos 1 2 - cos 1 5 , and in general, s n = cos 1 2 - cos 1 n + 2 . Hence as n → ∞ , we have s n cos 1 2 - cos 0 = cos 1 2 - 1. 4. (d) This series converges via the Geometric Series Theorem. X n = 1 3 ( 2 ) 2 n 5 n + 1 = X n = 1 3 ( 4 ) 5 2 4 5 n - 1 = 12 / 25 1 - 4 5 = 12 / 25 1 / 5 = 12 5 5. (b) For all real x we have x cos x 3 = x X n = 0 ( - 1 ) n ( x 3 ) 2 n ( 2 n ) ! = X n = 0 ( - 1 ) n x 6 n + 1 ( 2 n ) ! . 6. (c) Since the power series c n x n , centered at a = 0, converges at x = 3 and diverges at x = 5, we know that the radius of convergence R is at least 3 and at most 5.
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