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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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This note was uploaded on 03/08/2010 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas.
 Fall '09
 GOGOLEV
 Calculus

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