Exam03vr92 - Cheung Anthony Exam 3 Due Dec 5 2006 11:00 pm...

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Cheung, Anthony – Exam 3 – Due: Dec 5 2006, 11:00 pm – Inst: David Benzvi 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine whether the series X n = 2 ( - 1) n n 7 ln n is conditionally convergent, absolutely con- vergent, or divergent. 1. series is absolutely convergent 2. series is conditionally convergent 3. series is divergent correct Explanation: By the Divergence Test, a series X n = N ( - 1) n a n will be divergent for each fixed choice of N if lim n → ∞ a n 6 = 0 since it is only the behaviour of a n as n → ∞ that’s important. Now, for the given series, N = 2 and a n = n 7 ln n . But by L’Hospital’s Rule, lim x → ∞ x ln x = lim x → ∞ 1 1 /x = . Consequently, by the Divergence Test, the given series is divergent . keywords: 002 (part 1 of 1) 10 points Which one of the following properties does the series X n = 3 ( - 1) n 7 n (ln n ) n have? 1. absolutely convergent correct 2. conditionally convergent 3. divergent Explanation: The given series can be written in the form X n = 3 ( - 1) n 7 n (ln n ) n = X n = 3 ( - 1) n a n with a n = 7 n (ln n ) n > 0 . Now 0 < a n +1 a n = 7(ln n ) n (ln( n + 1)) n +1 = 7 ln n ln( n + 1) · n n 1 ln( n + 1) o < 7 ln( n + 1) . Consequently, lim n → ∞ a n +1 a n = 0 . In view of the Ratio Test, therefore, the series X n = 3 fl fl fl ( - 1) n 7 n (ln n ) n fl fl fl converges, so the given series is absolutely convergent .
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Cheung, Anthony – Exam 3 – Due: Dec 5 2006, 11:00 pm – Inst: David Benzvi 2 keywords: 003 (part 1 of 1) 10 points Which one of the following properties does the series X n = 1 ( - 7) n n ! have? 1. conditionally convergent 2. absolutely convergent correct 3. divergent Explanation: The given series has the form X n = 1 a n , a n = ( - 7) n n ! . But then fl fl fl a n +1 a n fl fl fl = 7( n !) ( n + 1)! = 7 n + 1 , in which case lim n → ∞ fl fl fl a n +1 a n fl fl fl = 0 < 1 . Consequently, by the Ratio test, the given series is absolutely convergent . keywords: alternating series, absolutely con- vergent, divergent, conditionally convergent, Ratio Test 004 (part 1 of 1) 10 points Which one of the following properties does the series X n = 3 ( - 1) n - 1 n - 2 n 2 + n - 4 have? 1. absolutely convergent 2. divergent 3. conditionally convergent correct Explanation: The given series has the form X n = 3 ( - 1) n - 1 n - 1 n 2 + n - 4 = X n = 3 ( - 1) n - 1 f ( n ) where f is defined by f ( x ) = x - 2 x 2 + x - 4 . Notice that x 2 + x - 4 > 0 on [3 , ), so the terms in the given series are defined for all n 3. On the other hand, x - 2 > 0 on (2 , ), so x > 2 = f ( x ) > 0 . Now, by the Quotient Rule, f 0 ( x ) = ( x 2 + x - 4) - ( x - 2)(2 x + 1) ( x 2 + x - 4) 2 = - x 2 - 4 x + 2 ( x 2 + x - 4) 2 ; in particular, f is decreasing on [6 , ). Thus by the Limit Comparison Test and the p -series Test with p = 1, we see that the series X n = 6 f ( n ) diverges, so the given series fails to be abso- lutely convergent. But n 6 = f ( n ) > f ( n + 1) ,
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Cheung, Anthony – Exam 3 – Due: Dec 5 2006, 11:00 pm – Inst: David Benzvi 3 while lim x → ∞ f ( x ) = 0 .
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