HW3 - hussain (tsh476) HW03 rusin (55565) 1 This print-out...

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Unformatted text preview: hussain (tsh476) HW03 rusin (55565) 1 This print-out should have 20 questions. Multiple-choice 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 2 ln(3 k ) k 2 , ( B ) summationdisplay k = 1 1 + cos(3 k ) k 2 + 4 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 ) = 2 ln 3 x x 2 is continous and positive on [ 2 3 , ); in addi- tion, since f ( x ) = 2 parenleftbigg 1 2 ln3 x x 3 parenrightbigg < on [ 2 3 , ), 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 = 2 bracketleftBig ln(3 x ) x 1 x bracketrightBig t 1 , and so integraldisplay 1 f ( x ) dx = 2(1 + ln 3) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities 1 + cos(3 k ) k 2 + 4 2 k 2 + 4 2 k 2 hold for all n 1. On the other hand, by the p-series 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 2 n 3 / 2 B . 1 + 1 4 + 1 9 + 1 16 + . . . C . summationdisplay n =1 3 n 2 + 1 1. none of them 2. all of them correct 3. A and C only 4. C only hussain (tsh476) HW03 rusin (55565) 2 5. A only 6. A and B only 7. B and C only 8. B 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 ) = 2 x 3 / 2 . Then integraldisplay 1 f ( x ) dx is convergent. B. Use f ( x ) = 1 x 2 . Then integraldisplay 1 f ( x ) dx is convergent. C. Use f ( x ) = 3 x 2 + 1 . Then integraldisplay 1 f ( x ) dx is convergent (tan 1 integral). keywords: convergent, Integral test, 003 10.0 points Determine whether the series summationdisplay n = 2 n 6(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 thats important. Now, for the given series, N = 2 and a n = n 6(ln n ) 2 ....
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HW3 - hussain (tsh476) HW03 rusin (55565) 1 This print-out...

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