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Unformatted text preview: Spring 2006 Math 152 Exam 3A: 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 π n 2 n + 1 = lim n →∞ 1 + 1 n 2 1 sin π 2 + 1 n = sin π 2 = 1 . 2. (d) 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 pseries ( 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 pseries ( p = 1 4 ≤ 1). Accordingly, ∑ ( 1 ) n 4 √ n converges, but is not absolutely convergent. (We say it is conditionally convergent .) 3. (b) 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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 Spring '08
 Teitler
 Math, Mathematical Series, lim

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