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Unformatted text preview: byrne (cmb3744) – HW03 – rusin – (55565) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Sorry to be late getting these questions posted – I took my own advice to go play in the snow, and I hope you did too! Due date is still Wednesday evening. 001 10.0 points Determine whether the following series ( A ) ∞ summationdisplay k =1 3 ln(4 k ) k 2 , ( B ) ∞ summationdisplay k =1 1 + sin(4 k ) k 2 + 1 converge or diverge. 1. A converges, B diverges 2. both series converge correct 3. A diverges, B converges 4. both series diverge Explanation: ( A ) The function f ( x ) = 3 ln 4 x x 2 is continous and positive on [ 1 2 , ∞ ); in addi tion, since f ′ ( x ) = 3 parenleftbigg 1 − 2 ln4 x x 3 parenrightbigg < on [ 1 2 , ∞ ), f is also decreasing on this inter val. This suggests applying the Integral Test. Now, after Integration by Parts, we see that integraldisplay t 1 f ( x ) dx = 3 bracketleftBig − ln(4 x ) x − 1 x bracketrightBig t 1 , and so integraldisplay ∞ 1 f ( x ) dx = 3(1 + ln 4) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities ≤ 1 + sin(4 k ) k 2 + 1 ≤ 2 k 2 + 1 ≤ 2 k 2 hold for all n ≥ 1. On the other hand, by the pseries test the series ∞ summationdisplay k = 1 1 k 2 is convergent since p = 2 > 1. Thus, by the comparison test, series ( B ) converges . keywords: 002 10.0 points Which of the following series are convergent: A . ∞ summationdisplay n = 1 3 n + 1 B . ∞ summationdisplay n = 1 2 n 2 / 3 C . 1 + 1 4 + 1 9 + 1 16 + . . . 1. A and B only 2. all of them 3. B and C only 4. C only correct byrne (cmb3744) – HW03 – rusin – (55565) 2 5. none of them 6. B only 7. A only 8. A and C only Explanation: By the Integral test, if f ( x ) is a positive, decreasing function, then the infinite series ∞ summationdisplay n =1 f ( n ) converges if and only if the improper integral integraldisplay ∞ 1 f ( x ) dx converges. Thus for the three given series we have to use an appropriate choice of f . A. Use f ( x ) = 3 x + 1 . Then integraldisplay ∞ 1 f ( x ) dx is divergent (log integral). B. Use f ( x ) = 2 x 2 / 3 . Then integraldisplay ∞ 1 f ( x ) dx is divergent. C. Use f ( x ) = 1 x 2 . Then integraldisplay ∞ 1 f ( x ) dx is convergent. keywords: convergent, Integral test, 003 10.0 points Determine whether the series ∞ summationdisplay n = 2 n 5(ln n ) 2 is convergent or divergent. 1. convergent 2. divergent correct Explanation: By the Divergence Test, a series ∞ summationdisplay n = N a n will be divergent for each fixed choice of N if lim n →∞ a n negationslash = 0 since it is only the behaviour of a n as n → ∞ that’s important. Now, for the given series, N = 2 and a n = n 5(ln n ) 2 ....
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This note was uploaded on 05/08/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.
 Spring '09
 Chu

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