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Unformatted text preview: baek (sb32676) Review 3 gogolev (57440) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the series 9 13 10 14 + 11 15 12 16 + 13 17 . . . is absolutely convergent, conditionally con vergent or divergent. 1. conditionally convergent 2. divergent correct 3. absolutely convergent Explanation: In summation notation, 9 13 10 14 + 11 15 12 16 + 13 17 . . . = summationdisplay n = 9 a n , with a n given by a n = ( 1) n n n + 4 . However, lim n n n + 4 = 1 , so that as n , a n oscillates between val ues approaching 1 and 1. In particular, therefore, lim n a n negationslash = 0 . Consequently, by the Divergence Test, the series is divergent . 002 10.0 points Find a power series representation for the function f ( t ) = t 4 t + 1 . 1. f ( t ) = summationdisplay n = 0 ( 1) n 2 n t n 2. f ( t ) = summationdisplay n = 0 2 2 n t n 3. f ( t ) = summationdisplay n = 0 2 n t n 4. f ( t ) = summationdisplay n = 0 ( 1) n 2 2 n t n +1 correct 5. f ( t ) = summationdisplay n = 0 2 2 n t n +1 6. f ( t ) = summationdisplay n = 0 ( 1) n 2 n t n +1 Explanation: After simplification, f ( t ) = t 4 t + 1 = t 1 ( 4 t ) . On the other hand, 1 1 x = summationdisplay n = 0 x n . Thus f ( t ) = t braceleftBig summationdisplay n = 0 ( 4 t ) n bracerightBig = t braceleftBig summationdisplay n = 0 ( 1) n 2 2 n t n bracerightBig . Consequently, f ( t ) = summationdisplay n = 0 ( 1) n 2 2 n t n +1 . keywords: 003 10.0 points baek (sb32676) Review 3 gogolev (57440) 2 Determine the interval of convergence of the series summationdisplay n = 1 ( 2) n n + 6 x n . 1. none of the other answers 2. interval convergence = [ 6 , 6) 3. converges only at x = 0 4. interval convergence = ( 2 , 2] 5. interval convergence = ( , ) 6. interval convergence = bracketleftBig 1 6 , 1 6 parenrightBig 7. interval convergence = parenleftBig 1 2 , 1 2 bracketrightBig cor rect Explanation: The given series has the form summationdisplay n = 1 a n x n with a n = ( 2) n n + 6 . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = if it converges for all x , while if R > 0, (iii) it converges when  x  < R , and (iv) diverges when  x  > R . But lim n vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n 2( n + 6) n + 7 = 2 . By the Ratio Test, therefore, the given series converges when  x  < 1 / 2 and diverges when  x  > 1 / 2. On the other hand, at the points x = 1 / 2 and x = 1 / 2 the series reduces to summationdisplay n = 1 1 n + 6 , summationdisplay n = 1 ( 1) n n + 6 respectively. Now, by the pseries Test with p = 1, the first of these series diverges, while by the Alternating Series Test, the second...
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 Fall '09
 GOGOLEV
 Calculus

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