Pre-Calc Homework Solutions 385

Pre-Calc Homework Solutions 385 - Section 9.4 40. This is a...

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Unformatted text preview: Section 9.4 40. This is a geometric series with first term a sin x sin x common ratio . Since 2 2 385 1 and 46. 1 for all x, the interval . 2 sin x n 1 (2n s1 s2 s3 sn S of convergence is Sum a 1 r 1 1 sin x 2 x 3 (3 (3 3 6 1)(2n 3 3 1) n 1 3 2n 1 2n 3 1 1) 1) 3 2n 1 1 1 2 3 5 3 5 3 3 5 3 5 3 7 3 3 7 41. Almost, but the Ratio Test won't determine whether there is convergence or divergence at the endpoints of the interval. 42. (a) For k a1 ... N, it's obvious that ak N, ak ... ... aN aN a1 ... cN 1 a1 ... aN n N 1 n lim sn 3 cn. ... For all k a1 a1 a1 ... 47. ak n 1 (2n s1 s2 s3 sn S aN ... aN ck 1 n N 1 cn (b) Since all of the an are nonnegative, the partial sums of the series form a nondecreasing sequence of real numbers. Part (a) shows that the sequence is bounded above, so it must converge to a limit. 43. (a) For k d1 ... N, it's obvious that dk N, dk ... ... dN dN d1 ... aN 1 40n 1)2(2n 1)2 5 5 9 5 5 5 9 9 5 5 5 9 9 5 5 (2n 1)2 n 1 5 (2n 1)2 (2n 5 1)2 5 25 5 25 5 5 25 5 25 5 49 5 5 49 n lim sn 2n (n 1 1)2 1 4 1 4 1 4 5 d1 ... dN n N 1 48. an. ... n 1n s1 2 n 1 1 4 1 4 1 n2 1 (n 1)2 For all k d1 d1 d1 ... 1 1 1 1 dN ... dN ak 1 dk s2 s3 sn S n N 1 1 9 1 9 1 1 9 1 9 1 16 1 1 16 an 1 (n 1)2 (b) If an converged, that would imply that dn was also convergent. 44. Answers will vary. 4 45. 3)(4n n 1 (4n 1 s1 1 5 1 s2 1 5 1 s3 1 5 1 sn 1 4n 1 n lim sn 1 1 1 1 lim sn 1 2 1 1 49. s1 1) n 1 1 9 1 9 1 4n 3 4n 1 1 s2 s3 1 2 1 2 1 1 3 1 3 1 1 3 1 3 1 4 2 1 2 1 n 1 1 4 1 5 1 5 1 1 9 1 9 1 13 sn 1 1 13 S 50. s1 s2 s3 1 n S n lim sn 1 sn S 1 1 ln 3 ln 2 1 1 1 ln 3 ln 2 ln 4 1 1 1 ln 3 ln 2 ln 4 1 1 ln 5 ln 2 1 1 ln(n 2) ln 2 1 lim sn ln 2 n 1 ln 3 1 ln 3 1 ln 4 1 ln 5 1 ln 2 1 ln 4 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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