B To test for convergence of the sequence braceleftbigg e n 6 4 n 5

# B to test for convergence of the sequence

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Unformatted text preview: B . To test for convergence of the sequence braceleftbigg e n + 6 4 n + 5 bracerightbigg we use L’Hospital’s Rule with f ( x ) = e x + 6 , g ( x ) = 4 x + 5 since f ( x ) −→ ∞ , g ( x ) −→ ∞ as x → ∞ . Thus lim n →∞ e n 4 n + 5 = lim x →∞ f ′ ( x ) g ′ ( x ) . But f ′ ( x ) g ′ ( x ) = e x 4 −→ ∞ as n → ∞ . Consequently, only A converges . 012 10.0 points Determine whether the series 3 4 − 3 5 + 3 6 − 3 7 + 3 8 − . . . is conditionally convergent, absolutely con- vergent or divergent. 1. series is divergent 2. series is absolutely convergent 3. series is conditionally convergent cor- rect Explanation: In summation notation, 3 4 − 3 5 + 3 6 − 3 7 + 3 8 − . . . = ∞ summationdisplay n = 1 ( − 1) n − 1 f ( n ) , with f ( x ) = 3 x + 3 . Now the improper integral integraldisplay ∞ 1 f ( x ) dx = integraldisplay ∞ 1 3 x + 3 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 3 n + 3 > 3 n + 1 + 3 = f ( n + 1) , while lim n →∞ 3 n + 3 = 0 . Consequently, by the Alternating Series Test, the given series is conditionally convergent . Version 023 – EXAM 3 – lawn – (55930) 7 keywords: alternating series, Alternating se- ries test, conditionally convergent, absolutely convergent, divergent 013 10.0 points Determine the interval of convergence of the infinite series ∞ summationdisplay k = 0 ( − 1) k ( k + 1)! 7 k (4 x + 9) k . 1. parenleftbigg 1 2 , 4 parenrightbigg 2. ( −∞ , ∞ ) 3. parenleftbigg − 1 2 , 1 2 parenrightbigg 4. series converges only at x = − 9 4 correct 5. parenleftbigg − 4 , − 1 2 parenrightbigg Explanation: There are three possibilities for the interval of convergence of the infinite series ∞ summationdisplay k = 0 a k ( x − a ) k . First set L = lim k →∞ vextendsingle vextendsingle vextendsingle vextendsingle a k +1 a k vextendsingle vextendsingle vextendsingle vextendsingle . Then (i) when L > 0, the interval of convergence is given by parenleftbigg a − 1 L , a + 1 L parenrightbigg ; (ii) when L = 0, the interval of convergence is given by ( −∞ , ∞ ); (iii) when L = ∞ , the interval of conver- gence reduces to the point { a } , i.e. , the series converges only at x = a . For the given series, ∞ summationdisplay k =0 ( − 1) k ( k + 1)! 7 k (4 x + 9) k = ∞ summationdisplay k =0 ( − 1) k ( k + 1)!4 k 7 k parenleftbigg x + 9 4 parenrightbigg k , so the corresponding value of a k is a k = ( − 1) k 4 k ( k + 1)! 7 k . In this case vextendsingle vextendsingle vextendsingle vextendsingle a k +1 a k vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle − ( k + 2)!4 k +1 7 k +1 · 7 k ( k + 1)!4 k vextendsingle vextendsingle vextendsingle vextendsingle = 4( k + 2) 7 . Thus L = lim k →∞ 4( k + 2) 7 = ∞ . Consequently, the series converges only at x = − 9 4 . keywords: 014 10.0 points Determine whether the series ∞ summationdisplay k = 1 4 k ln(3 k ) is convergent or divergent. 1. series diverges correct 2. series converges Explanation: The function f ( x ) = 4 x ln(3 x ) Version 023 – EXAM 3 – lawn – (55930) 8 is continous, positive and decreasing on [1 , ∞ ). By the Integral Test, therefore, the series ∞ summationdisplay k = 1 4 k ln(3 k ) converges if and only if the improper integral integraldisplay ∞ 1 f ( x ) dx = integraldisplay...
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• Fall '11
• Gramlich
• Accounting, Mathematical Series, lim