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Unformatted text preview: Pham, Quoc Homework 12 Due: Apr 17 2007, 3:00 am Inst: Eric Katerman 1 This printout should have 22 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 If a n , b n , and c n satisfy the inequalities < b n c n a n , for all n , what can we say about the series ( A ) : X n = 1 a n , ( B ) : X n = 1 b n if we know that the series ( C ) : X n = 1 c n is divergent but know nothing else about a n and b n ? 1. ( A ) need not diverge , ( B ) diverges 2. ( A ) diverges , ( B ) diverges 3. ( A ) diverges , ( B ) converges 4. ( A ) converges , ( B ) need not converge 5. ( A ) converges , ( B ) diverges 6. ( A ) diverges , ( B ) need not diverge correct Explanation: Lets try applying the Comparison Test: (i) if < b n c n , X n c n diverges , then the Comparison Test is inconclusive be cause X b n could diverge, but it could con verge  we cant say precisely without further restrictions on b n ; (ii) while if < c n a n , X n c n diverges , then the Comparison Test applies and says that X a n diverges. Consequently, what we can say is ( A ) diverges , ( B ) need not diverge . keywords: Comparison Test, conceptual 002 (part 1 of 1) 10 points Determine whether the series X n = 1 tan 1 n 1 + n 4 converges or diverges. 1. series is convergent correct 2. series is divergent Explanation: We apply the Limit Comparison Test with a n = tan 1 n 1 + n 4 , b n = 1 n 4 . For lim n n 4 tan 1 n 1 + n 4 = lim n tan 1 n = 2 . Thus the given series X n = 1 tan 1 n 1 + n 4 is convergent if and only if the series X n = 1 1 n 4 Pham, Quoc Homework 12 Due: Apr 17 2007, 3:00 am Inst: Eric Katerman 2 is convergent. But by the pseries test, this last series converges because p = 4 > 1. Con sequently, the given series is convergent . keywords: 003 (part 1 of 1) 10 points Determine whether the series X n = 1 2 + n + n 3 5 + n 2 + n 8 converges or diverges. 1. series is convergent 2. series is divergent correct Explanation: We apply the Limit Comparison Test with a n = 2 + n + n 3 5 + n 2 + n 8 , b n = 1 n . For then lim n a n b n = lim n 2 n + n 2 + n 4 5 + n 2 + n 8 = lim n 2 n 3 + 1 n 2 + 1 r 5 n 8 + 1 n 6 + 1 = 1 > . Thus the given series X n = 1 2 + n + n 3 5 + n 2 + n 8 converges if and only if the series X n = 1 1 n converges. But, by the pseries test (or be cause the harmonic series diverges), this last series diverges because p = 1. Consequently, the series is divergent . keywords: 004 (part 1 of 1) 10 points Which of the following series converge(s)?...
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 Spring '08
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