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Unformatted text preview: doesnt use Theorem 2.4.6.) Recall that we already know from our discussion of Section 2.4 that the series is divergent for p = 1 and convergent for p = 2. (a) Suppose p < p . Use the Comparison Test to show that if 1 /n p is convergent then 1 /n p is also convergent. (b) Show that the pseries diverges for any p 1. (c) Consider p > 1. (1) Dene the sequence ( b n ) by b 1 = 1, and b n = Z n n1 1 x p dx for n 1 . Explain why 1 n p b n . (You are free to use properties of integrals that you know from calculus.) (2) The partial sums for n =1 b n are s n = b 1 + + b n = 1 + Z n 1 1 x p dx. Show that s n 1 + 1 p1 . (Just calculate the integral!) (3) Explain why ( s n ) and n =1 b n both converge. (4) Conclude that n =1 1 /n p converges....
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This note was uploaded on 11/07/2009 for the course MATH 3224 at Virginia Tech.
 '08
 KBHANNSGEN
 Calculus

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