# hw6sol - Math 131A - Section 2 Spring 2010 Homework 6 The...

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Math 131A - Section 2 Spring 2010 Homework 6 The following problems were graded for this assignment: 14.7 (2 points), 14.12 (2 points), 15.6 (2 points), 17.9(b,c) (2 points), 17.10 (b,c) (2 points), 17.13 (3 points), 17.14 (2 points). 14.2.(a). For n 2, we have that n - 1 n 2 = 1 n - 1 n 2 1 n - 1 2 n = 1 2 n . Since 1 2 n diverges, the comparison test shows n - 1 n 2 diverges. (e). Note that | ( n +1) 2 ( n +1)! n ! n 2 | = n +1 n 2 , so lim sup | a n +1 a n | = 0 < 1. By the ratio test, n 2 n ! converges. (g). Note that | n +1 2 n +1 2 n n | = n +1 2 n , so lim sup | a n +1 a n | = 1 2 < 1. By the ratio test, n 2 n converges. 14.3.(a). By the ratio test this converges since lim sup | a n +1 a n | = lim sup | q n ! ( n +1)! | = lim sup q 1 n +1 = 0 < 1. (b). Note that | 2+cos n 3 n | ≤ 2+ | cos n | 3 n 3 1 3 n . Since 3 1 3 n converges, by the comparison test, we have the 2+cos n 3 n converges. (e). Let s n = sin( 9 ), and consider the subsequence ( s 9 n +1 ). Note that this a non-zero constant sequence, so lim sup | s n | 6 = 0. By 14.5, we have that the series s n does not converge. 14.12. Let ( a n ) be a sequence with lim inf | a n | = 0. Then there is a subequence ( a n k ) such that lim k →∞ | a n k | = 0. Hence, there exists n k 1 such that | a n k 1 | < 1 2 . Having chosen n k 1 < ··· < n kj , select n kn +1 > n kj such that | a n kj +1 | < 1 2 j +1 , which can be done since lim

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## This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.

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hw6sol - Math 131A - Section 2 Spring 2010 Homework 6 The...

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