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Unformatted text preview: lawrence (cdl678) – Homework 3 – Odell – (56280) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the convergence or divergence of the series ( A ) ∞ summationdisplay n = 2 n 5(ln n ) 2 , and ( B ) ∞ summationdisplay n = 1 tan − 1 n 2 + n 4 . 1. A diverges, B converges correct 2. both series diverge 3. A converges, B diverges 4. both series converge Explanation: ( A ) 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 . But by L’Hospital’s Rule applied twice, lim x →∞ x (ln x ) 2 = lim x →∞ 1 (2 ln x ) /x = lim x →∞ x 2 ln x = lim x →∞ 1 2 /x = ∞ . By the Divergence Test, therefore, series ( A ) diverges . ( B ) We apply the Limit Comparison Test with a n = tan − 1 n 2 + n 4 , b n = 1 n 4 . For lim n →∞ n 4 parenleftBig tan − 1 n 2 + n 4 parenrightBig = lim n →∞ tan − 1 n = π 2 . Thus the given series ∞ summationdisplay n = 1 tan − 1 n 2 + n 4 is convergent if and only if the series ∞ summationdisplay n = 1 1 n 4 is convergent. But by the p-series test, this last series converges because p = 4 > 1. Con- sequently, series ( B ) converges . 002 10.0 points If a k , b k , and c k satisfy the inequalities < b k ≤ c k ≤ a k , for all k , what can we say about the series ( A ) : ∞ summationdisplay k = 1 a k , ( B ) : ∞ summationdisplay k = 1 b k if we know that the series ( C ) : ∞ summationdisplay k = 1 c k lawrence (cdl678) – Homework 3 – Odell – (56280) 2 is divergent but know nothing else about a k and b k ? 1. ( A ) converges , ( B ) need not converge 2. ( A ) diverges , ( B ) converges 3. ( A ) need not diverge , ( B ) diverges 4. ( A ) converges , ( B ) diverges 5. ( A ) diverges , ( B ) need not diverge correct 6. ( A ) diverges , ( B ) diverges Explanation: Let’s try applying the Comparison Test: (i) if < b k ≤ c k , summationdisplay k c k diverges , then the Comparison Test is inconclusive be- cause summationdisplay b k could diverge, but it could con- verge - we can’t say precisely without further restrictions on b k ; (ii) while if < c k ≤ a k , summationdisplay k c k diverges , then the Comparison Test applies and says that summationdisplay a k diverges. Consequently, what we can say is ( A ) diverges , ( B ) need not diverge . 003 10.0 points Which, if any, of the following series converge? ( A ) ∞ summationdisplay k = 2 1 k (ln k ) 2 + 3 ( B ) ∞ summationdisplay n = 5 parenleftBig 2 3 parenrightBig n 1. A and B correct 2. neither A nor B 3. A but not B 4. B but not A Explanation: ( A ) Since lim k →∞ k (ln k ) 2 k (ln k ) 2 + 3 = 1 , the Limit Comparison Test ensures that the series ∞ summationdisplay k = 2 1...
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This note was uploaded on 04/09/2010 for the course M 56280 taught by Professor Odell during the Spring '10 term at University of Texas at Austin.

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Calc hw 3 - lawrence(cdl678 – Homework 3 – Odell –(56280 1 This print-out should have 30 questions Multiple-choice questions may continue on

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