0 l n k 1 ak may either converge or diverge 1k 2

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Unformatted text preview: converges and ∞ Thus, for k ≥ N , a2 < ak and so k (b) 1/k 4 converges and 51. 0 < L − n k =1 ak may either converge or diverge. 1/k 2 converges; ∞ k =n+1 1/k diverges. f ( k ) = L − Sn = f (k ) < n f (x) dx 53. N = 3. 55. (a) Set f (x) = x1/4 − ln x. Then f (x) = 1 −3/4 1 1 1/4 x −= (x − 4). Since f (e12 ) = e3 − 12 > 0 and f (x) > 0 for x > e12 , we know that k 1/4 > 4 x 4x 1 ln k 1 ln k ln k and therefore 5/4 > 3/2 for sufficiently large k . Since is a convergent p-series, converges by the basic comparison test. k k k 5/4 k 3/2 ln x 1 (b) By L opital’s rule lim ’H ˆ =0 x→∞ x3/2 x5/4 SECTION 11.3 1. converges; ratio test 3. converges; root test 5. diverges; ak → 0 / 7. diverges; limit comparison √ 11. diverges; limit comparison 1/ k 13. diverges; ratio test 15. converges; comparison 1/k 3/2 1/k 2 19. diverges; integral test 1/k 3/2 21. diverges; ak → e...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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