This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: morris (gmm643) – HW 3 – mann – (54675) 1 This printout should have 20 questions. Multiplechoice 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 !...
View
Full
Document
This note was uploaded on 11/17/2010 for the course PHY 56630 taught by Professor Coker during the Spring '10 term at University of Texas.
 Spring '10
 COKER

Click to edit the document details