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HW 4-solutions

# HW 4-solutions - morris(gmm643 – HW 3 – mann –(54675...

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Unformatted text preview: morris (gmm643) – HW 3 – mann – (54675) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Sections 12.7, 12.8 001 10.0 points Which of the following properties does the series ∞ summationdisplay n = 1 ( − 4) n (2 n )! have? 1. conditionally convergent 2. absolutely convergent correct 3. divergent Explanation: The given series can be written as ∞ summationdisplay n = 1 a n with a n defined by a n = ( − 4) n (2 n )! = ( − 4 n ) 1 · 2 · 3 . . . (2 n − 1) · 2 n . In this case, vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 4 (2 n + 1)(2 n + 2) −→ as n → ∞ . Consequently, by the Ratio Test, the given series is absolutely convergent . 002 10.0 points Determine whether the series ∞ summationdisplay k = 1 2 e − k k ! is absolutely convergent, conditionally con- vergent, or divergent. 1. absolutely convergent 2. divergent correct 3. conditionally convergent Explanation: The given series has the form ∞ summationdisplay k =1 a k where a k = 2( k !) e k . But then vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = ( k + 1)! k ! e k e k +1 = k + 1 e . Thus lim k →∞ vextendsingle vextendsingle vextendsingle vextendsingle a k +1 a k vextendsingle vextendsingle vextendsingle vextendsingle = ∞ . Consequently, by the Ratio Test, the given series is divergent . 003 10.0 points Determine whether the series ∞ summationdisplay k = 3 5 k (ln k ) 6 converges or diverges. 1. diverges 2. converges correct Explanation: The given series can be written in the form ∞ summationdisplay k =1 f ( n ) where f is the function defined on (1 , ∞ ) by f ( x ) = 5 x (ln x ) 6 . morris (gmm643) – HW 3 – mann – (54675) 2 Now f is positive and decreasing on (1 , ∞ ); in addition, integraldisplay t 3 f ( x ) dx = bracketleftbigg − 1 (ln x ) 5 bracketrightbigg t 3 = 1 (ln 3) 5 − 1 (ln t ) 5 , in which case, integraldisplay ∞ 3 f ( x ) dx = lim t →∞ integraldisplay t 3 f ( x ) dx = lim t →∞ braceleftBig 1 (ln3) 5 − 1 (ln t ) 5 bracerightBig = 1 (ln 3) 5 . Consequently, by the Integral Test, the given series converges . 004 10.0 points Determine whether the series ∞ summationdisplay m = 1 cos( m ) converges or diverges. 1. diverges correct 2. converges Explanation: Since lim n →∞ cos( m ) does not exist, in particular, lim n →∞ cos( m ) negationslash = 0 , the Divergence Test says that the given series diverges . keywords: 005 10.0 points Determine whether the series ∞ summationdisplay k = 1 3 k k 2 k ! converges or diverges. 1. converges correct 2. diverges Explanation: The given series can be written as ∞ summationdisplay k =1 a k , a k = 3 k k 2 k !...
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