2005a_x3b_sols

# 2005a_x3b_sols - Spring 2005 Math 152 Exam 3B: Solutions c...

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Spring 2005 Math 152 Exam 3B: Solutions Mon, 02/May c ± 2005, Art Belmonte 1. (b) 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 →∞ ,wehave 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) 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. Accordingly, it must be true that series converges at x = 2, but diverges at x = 6. (For x = 4, the series may converge or it may diverge.) 3. (a) 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 ² n - 1 = 12 / 25 1 - 4 5 = 12 / 25 1 / 5 = 12 5 4. (d) We have X n = 1 a n = lim n →∞ s n = lim n →∞ 4 n - 5 2 + n = lim n →∞ 4 - 5 n 2 n + 1 = 4 . 5. (e) At x = π 3 we have f ( x ) = cos x = 1 / 2 f 0 ( x ) =- sin x 3 / 2 f 00 ( x ) cos x 1 / 2 f 000 ( x ) = sin x = 3 / 2 Therefore, T 3 ( x ) = 3 X n = 0 f ( n ) ( π 3 ) n ! ³ x - π 3 ´ n = 1 2 - 3 2 ( x - π 3 ) - 1 4 ( x - π 3 ) 2 + 3 12 ( x - π 3 ) 3 6. (a) 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.

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## This note was uploaded on 03/25/2008 for the course MATH 152 taught by Professor Teitler during the Spring '08 term at Texas A&M.

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2005a_x3b_sols - Spring 2005 Math 152 Exam 3B: Solutions c...

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