HW13-solutions - nguyen(avn365 HW13 staron(52970 This...

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nguyen (avn365) – HW13 – staron – (52970) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Determine whether the series summationdisplay k =1 ( 1) k 1 cos 2 ( k ) 4 k is absolutely convergent, conditionally con- vergent or divergent Determine whether the series 2 3 2 4 + 2 5 2 6 + 2 7 . . . is conditionally convergent, absolutely con- vergent or divergent. 1. absolutely convergent correct 2. divergent 3. conditionally convergent Explanation: To check for absolute convergence we have to decide if the series summationdisplay k =1 cos 2 ( k ) 4 k is convergent. For this we can use the Com- parison Test with a k = cos 2 ( k ) 4 k , b k = 1 4 k . For then 0 a k b k , since 0 cos 2 ( k ) 1. Thus the series summationdisplay k =1 cos 2 ( k ) 4 k converges if the series summationdisplay k =1 1. series is divergent 2. series is absolutely convergent 3. series is conditionally convergent cor- rect Explanation: In summation notation, 2 3 2 4 + 2 5 2 6 + 2 7 . . . = summationdisplay n =1 ( 1) n 1 f ( n ) , with f ( x ) = 2 x + 2 . Now the improper integral integraldisplay 1 f ( x ) dx = integraldisplay 1 2 x + 2 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 2 n + 2 > 2 n + 1 + 2 = f ( n + 1) , while lim n →∞ 2 n + 2 = 0 . Consequently, by the Alternating Series Test, the given series is conditionally convergent . 1 4 k converges. But this last series is a geometric series with r = 1 4 < 1 , hence convergent. Consequently, the given series is absolutely convergent . 002 10.0points
nguyen (avn365) – HW13 – staron – (52970) 2 keywords: alternating series, Alternating se- ries test, conditionally convergent, absolutely convergent, divergent 003 10.0points To apply the root test to an infinite series summationdisplay n a n the value of ρ = lim n → ∞ | a n | 1 /n has to be determined. Compute the value of ρ for the series summationdisplay n =1 5 n + 2 n parenleftbigg 7 4 parenrightbigg n . 1. ρ = 4 7 2. ρ = 8 7 3. ρ = 7 2 4. ρ = 35 4 5. ρ = 7 4 correct Explanation: After division, 5 n + 2 n = 5 parenleftbigg 1 + 2 5 n parenrightbigg , so ( a n ) 1 /n = parenleftbigg 5 parenleftbigg 1 + 2 5 n parenrightbiggparenrightbigg 1 /n 7 4 . But lim n → ∞ 5 1 /n parenleftbigg 1 + 2 5 n parenrightbigg 1 /n = 1 as n → ∞ . Consequently, ρ = 7 4 . 004 10.0points Determine whether the series summationdisplay n =0 ( 3) n (2 n )! is absolutely convergent, conditionally con- vergent, or divergent. 1. absolutely convergent correct 2. divergent 3. conditionally convergent Explanation: We use the Ratio Test with a n = ( 3) n (2 n )! . For then vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle ( 3) n +1 (2 n + 2)! (2 n )! ( 3) n vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle ( 3) n +1 (2 n + 2)! (2 n )! ( 3) n vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle 3 (2 n + 1)(2 n + 2) vextendsingle vextendsingle vextendsingle . Thus lim n → ∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n → ∞ 3 (2 n + 1)(2 n + 2) = 0 < 1 . Consequently, the series is absolutely convergent . 005 10.0points Which one of the following properties does the series summationdisplay n =1 ( 1) n n + 2 have?
nguyen (avn365) – HW13 – staron – (52970) 3 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: This series is not absolutely convergent as shown: summationdisplay n =1 ( 1) n n + 2 = | summationdisplay n =1 ( 1) n n + 2 | = summationdisplay n =1 (1) n n + 2 summationdisplay n =1 1 n + 2 where p 1, this series is divergent.

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