solution_pd3f

# solution_pd3f - suleimenov(bs26835 – HW03 – rusin...

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Unformatted text preview: suleimenov (bs26835) – 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 4 ln(2 k ) k 2 , ( B ) ∞ summationdisplay k =1 1 + sin(2 k ) k 2 + 1 converge or diverge. 1. A diverges, B converges 2. both series converge correct 3. both series diverge 4. A converges, B diverges Explanation: ( A ) The function f ( x ) = 4 ln 2 x x 2 is continous and positive on [1 , ∞ ); in addi- tion, since f ′ ( x ) = 4 parenleftbigg 1 − 2 ln2 x x 3 parenrightbigg < on [1 , ∞ ), 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 = 4 bracketleftBig − ln(2 x ) x − 1 x bracketrightBig t 1 , and so integraldisplay ∞ 1 f ( x ) dx = 4(1 + ln 2) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities ≤ 1 + sin(2 k ) k 2 + 1 ≤ 2 k 2 + 1 ≤ 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 2 / 3 B . 1 + 1 √ 2 + 1 √ 3 + 1 2 + 1 √ 5 + . . . C . ∞ summationdisplay n =1 3 n 2 + 1 1. A and C only 2. A only 3. A and B only 4. B and C only suleimenov (bs26835) – HW03 – rusin – (55565) 2 5. none of them 6. all of them 7. C only correct 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 2 / 3 . Then integraldisplay ∞ 1 f ( x ) dx is divergent. B. Use f ( x ) = 1 x 1 / 2 . Then integraldisplay ∞ 1 f ( x ) dx is divergent. 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 5(ln n ) 2 is convergent or divergent. 1. divergent correct 2. convergent 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 03/20/2011 for the course M 408 C taught by Professor Treisman during the Spring '07 term at University of Texas.

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solution_pd3f - suleimenov(bs26835 – HW03 – rusin...

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