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Unformatted text preview: Cheung, Anthony – Homework 11 – Due: Nov 14 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Compare the values of the series A = ∞ X n = 2 3 n 2 . 1 and the improper integral B = Z ∞ 1 3 x 2 . 1 dx. 1. A < B correct 2. A > B 3. A = B Explanation: In the figure 1 2 3 4 5 ... a 2 a 3 a 4 a 5 the bold line is the graph of the function f ( x ) = 3 x 2 . 1 on [1 , ∞ ) and the area of each of the rectan gles is one of the values of a n = 3 n 2 . 1 . Clearly from this figure we see that a 2 = f (2) < Z 2 1 f ( x ) dx, a 3 = f (3) < Z 3 2 f ( x ) dx, while a 4 = f (4) < Z 4 3 f ( x ) dx, a 5 = f (5) < Z 5 4 f ( x ) dx, and so on for all n . Consequently, A = ∞ X n = 2 3 n 2 . 1 < Z ∞ 1 3 x 2 . 1 dx = B . keywords: integral test, pseries 002 (part 1 of 1) 10 points Determine the convergence or divergence of the series ( A ) 1 + 1 8 + 1 27 + 1 64 + 1 125 + ..., and ( B ) ∞ X k = 1 k 3 e k 4 . 1. both series convergent correct 2. A convergent, B divergent 3. A divergent, B convergent 4. both series divergent Explanation: ( A ) The given series has the form 1 + 1 8 + 1 27 + 1 64 + 1 125 + ... = ∞ X n =1 1 n 3 . Cheung, Anthony – Homework 11 – Due: Nov 14 2006, 3:00 am – Inst: David Benzvi 2 This is a pseries with p = 3 > 1, so the series converges. ( B ) The given series has the form ∞ X k = 1 f ( k ) with f defined by f ( x ) = x 3 e x 4 . Note first that f is continuous and positive on [1 , ∞ ); in addition, since f ( x ) = e x 4 (3 x 2 4 x 6 ) < for x > 1, f is decreasing on [1 , ∞ ). Thus we can use the Integral Test. Now, by substitu tion, Z t 1 x 3 e x 4 dx = • 1 4 e x 4 ‚ t 1 , and so Z ∞ 1 x 3 e x 4 dx = 1 4 e . Since the integral converges, the series con verges. This could also be established using the Ratio Test. keywords: 003 (part 1 of 1) 10 points First find a n so that ∞ X n =1 a n = 6 + 3 √ 2 + 2 √ 3 + 3 4 + 6 5 √ 5 + ... and then determine whether the series con verges or diverges. 1. a n = 3 2 n 3 / 2 , series converges 2. a n = 6 n 1 / 2 , series converges 3. a n = 6 n 1 / 2 , series diverges 4. a n = 6 n 3 / 2 , series diverges 5. a n = 3 2 n 3 / 2 , series diverges 6. a n = 6 n 3 / 2 , series converges correct Explanation: By inspection a n = 6 n 3 / 2 . To test for convergence we use the Integral test with f ( x ) = 6 x 3 / 2 . This is a positive, continuous, decreasing function on [1 , ∞ ). Furthermore, Z ∞ 1 f ( x ) dx = lim n →∞ Z n 1 6 x 3 / 2 dx = lim n →∞ • 12 x 1 / 2 ‚ n 1 = 12 , so the improper integral Z ∞ 1 f ( x ) dx is convergent, which by the Integral test means that the series converges ....
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 Spring '08
 RAdin
 Calculus, Mathematical Series, 1 1 k, David Benzvi

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