2006a_x3b_sols

2006a_x3b_sols - Spring 2006 Math 152 Exam 3B: Solutions...

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Unformatted text preview: Spring 2006 Math 152 Exam 3B: Solutions Mon, 01/May c 2006, Art Belmonte Notes The notation b n 0 means 0 < b n + 1 b n and lim b n = 0; i.e., the sequence { b n } decreases to zero in the limit. GFF [Generalized Fun Fact]: Let p ( n ) = m k = c k n k be a polynomial in n with c m > 0. Then lim n n p ( n ) = 1. This boosts the power of the Root Test, which was optionally covered in some classes. (In most classes, only the Ratio Test was covered.) 1. (b) Algebraic manipulation gives lim n n 2 + 1 n 2 sin n 2 n + 1 = lim n 1 + 1 n 2 1 sin 2 + 1 n = sin 2 = 1 . 2. (a) Only the series (- 1 ) n 4 n converges, but not absolutely. I. The series (- 1 ) n n 8 converges absolutely since | a n | = 1 n 8 , a convergent p-series ( p = 8 > 1). II. The series (- 1 ) n + 1 n 5 + 2 n 3 diverges by the Test for Divergence since lim a n = 0. (Indeed, the limit does not exist since lim sup a n = and lim inf a n = - .) III. The alternating series (- 1 ) n 4 n converges by the Alternating Series Test since b n = | a n | = 1 4 n 0. Note, however, that | a n | = 1 n 1 / 4 is a divergent p-series ( p = 1 4 1). Accordingly, (- 1 ) n 4 n converges, but is not absolutely convergent. (We say it is conditionally convergent .) 3. (c) We have 1 1- (- x 4 ) = n =- x 4 n = n = (- 1 ) n x 4 n , provided- x 4 < 1 or | x | < 1....
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2006a_x3b_sols - Spring 2006 Math 152 Exam 3B: Solutions...

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