1 s n 1 8 7 n 5 2 s k 2 5 8 k ln k 7 k 3 s n 1 7 n n

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1. s n = 1 8 7 n 5 2. s k = 2 5 8 k ln( k ) + 7 k 3. s n = 1 7 n n ( n + 6) n 4. s n = 1 p 7 5 P n 5. s n = 1 p 6 n 8 n + 5 P n correct
lee (jsl2382) – convergence tests and taylor series – tran – (56320) 23 Explanation: We test the convergence of each of the Fve series. (i) ±or the series s k = 2 5 8 k ln( k ) + 7 k the limit comparison test can be used, com- paring it with s k = 2 1 k ln( k ) . Now, after division, k ln( k ) p 5 8 k ln( k ) + 7 k P = 5 8 + 7 ln( k ) . Since 5 8 + 7 ln( k ) −→ 5 8 > 0 as k → ∞ , the limit comparison test applies. But by the Integral test, the series s k = 2 1 k ln( k ) does not converge. Thus s k = 2 5 8 k ln( k ) + 7 k does not converge. (ii) If a series s n a n converges, then a n 0 as n → ∞ . But for the series s n =1 7 n n ( n + 6) n we see that a n = 7 n n ( n + 6) n = 7 p n n + 6 P n = 7 p n + 6 n P n = 7 pp 1 + 6 n P n P 1 . Now p 1 + 6 n P n e 6 as n → ∞ . Thus lim n →∞ a n = 7 e 6 n = 0 . Consequently, s n =1 7 n n ( n + 6) n does not converge. (iii) ±or the series s n = 1 8 7 n 5 the comparison test can be used since 8 7 n 5 8 7 p 1 n P while s n = 1 1 n does not converge because of the Integral test. Thus s n = 1 8 7 n 5 does not converge. (iv) Use of the Comparison test is suggested in dealing with s n = 1 p 6 n 8 n + 5 P n for after division 6 n 8 n + 5 = 6 8 + 5 n .
lee (jsl2382) – convergence tests and taylor series – tran – (56320) 24 Now the inequality 0 < 6 8 + 5 n 3 4 holds for all n 1, so the inequality p 6 n 8 n + 5 P n p 3 4 P n holds for all n 1. But s n = 1 p 3 4 P n is a geometric series whose common ratio is r = 3 4 . Since 0 < r < 1, this geometric series converges. Hence by the Comparison test the series s n = 1 p 6 n 8 n + 5 P n converges. (v) The series s n = 1 p 7 5 P n is a geometric series with r = 7 5 > 1. But then this geometric series does not converge. Consequently, of the Fve given inFnite se- ries, only s n = 1 p 6 n 8 n + 5 P n converges. 039 10.0 points Determine whether the following series 1 1 · 3 3! + 1 · 3 · 5 5! 1 · 3 · 5 · 7 7! + ··· + ( 1) n 1 · 3 · . . . · (2 n 1) (2 n 1)! + is absolutely convergent, conditionally con- vergent, or divergent. 1. absolutely convergent correct 2. conditionally convergent 3. divergent Explanation: The given series is an alternating series s n = 1 ( 1) n 1 a n where a n = 1 · 3 · . . . · (2 n 1) (2 n 1)! . Now (2 n 1)! = 1 · 2 · 3 · 4 · . . . · (2 n 2) · (2 n 1) . Thus after cancellation we see that a n = 1 2 · 4 · 6 · . . . · (2 n 2) 1 2 n for all n > 1. But s n = 2 1 2 n is a geometric series with r = 1 / 2, hence a convergent geometric series. Consequently, by the Comparison Test, the given series is absolutely convergent . 040 10.0 points Determine all values of p for which the series s k = 3 6 k p ln k converges. 1. p < 2
lee (jsl2382) – convergence tests and taylor series – tran – (56320) 25 2. p 1 3. p > 2 4. p > 0 5. p < 0 6. p > 1 correct Explanation: By the Divergence Test, the series s k = 3 6 k p ln k diverges when p < 0 because then lim k →∞ 1 k p ln k = .

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