HW13 - Math 191 11.3: 2 converges; a geometric series with...

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Math 191 HW 13 Solutions 11.3: 2 converges; a geometric series with r = 1 e < 1 11.3: 6 converges; X n =1 - 2 n n = - 2 X n =1 1 n 3 / 2 , which is a convergent p-series ( p = 3 2 ). 11.3: 18 diverges; lim n →∞ a n = lim n →∞ ± 1 + 1 n ² n = e 6 = 0 11.3: 28 diverges by the Integral Test; Z 1 x x 2 + 1 dx ; ³ u = x 2 + 1 du = 2 xdx ´ 1 2 Z 2 du 4 = lim b →∞ ³ 1 2 ln u ´ b 2 = lim b →∞ 1 2 (ln b - ln2) = 11.3: 39 a. Z 2 dx x (ln x ) p ; ³ u = ln x du = dx x ´ Z ln 2 u - p du = lim b →∞ ³ u - p +1 - p + 1 ´ b ln 2 = lim b →∞ ( 1 1 - p )[ b - p +1 - (ln2) - p +1 ] = µ 1 p - 1 (ln2) - p +1 , p > 1 , p < 1 the improper integral converges if p > 1 and diverges if p < 1. For p = 1; Z 2 dx x ln x = lim b →∞ [ln(ln x )] b 2 = lim b →∞ [ln(ln b ) - ln(ln2)] = , so the improper integral diverges if p = 1. b. Since the series and the integral converges or diverges together, X n =2 1 n (ln n ) p converges if and only if
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This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell University (Engineering School).

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HW13 - Math 191 11.3: 2 converges; a geometric series with...

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