homework 12 - fh2825 Homework 12 Radin (58415) 1 This...

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Unformatted text preview: fh2825 Homework 12 Radin (58415) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points If a n , b n , and c n satisfy the inequalities < a n c n b n , for all n , what can we say about the series ( A ) : summationdisplay n =1 a n , ( B ) : summationdisplay n = 1 b n if we know that the series ( C ) : summationdisplay n = 1 c n is divergent but know nothing else about a n and b n ? 1. ( A ) converges , ( B ) need not converge 2. ( A ) diverges , ( B ) need not diverge 3. ( A ) converges , ( B ) diverges 4. ( A ) diverges , ( B ) converges 5. ( A ) diverges , ( B ) diverges 6. ( A ) need not diverge , ( B ) diverges correct Explanation: Lets try applying the Comparison Test: (i) if < a n c n , summationdisplay n c n diverges , then the Comparison Test is inconclusive be- cause summationdisplay a n could diverge, but it could con- verge - we cant say precisely without further restrictions on a n ; (ii) while if < c n b n , summationdisplay n c n diverges , then the Comparison Test applies and says that summationdisplay b n diverges. Consequently, what we can say is ( A ) need not diverge , ( B ) diverges . 002 10.0 points Determine whether the series summationdisplay n = 1 sin 2 n n 2 + 6 converges or diverges. 1. series is divergent 2. series is convergent correct Explanation: Note first that the inequalities < sin 2 n n 2 + 6 1 n 2 + 6 1 n 2 hold for all n 1. On the other hand, by the p-series test the series summationdisplay n = 1 1 n 2 is convergent since p = 2 > 1. Thus, by the comparison test, the given series is convergent . 003 10.0 points Which of the following infinite series con- verges? 1. summationdisplay n = 1 6 n n ( n + 5) n fh2825 Homework 12 Radin (58415) 2 2. summationdisplay n = 1 7 6 n 5 3. summationdisplay n = 1 parenleftbigg 5 n 7 n + 5 parenrightbigg n correct 4. summationdisplay n = 1 parenleftbigg 6 5 parenrightbigg n 5. summationdisplay k =2 5 7 k ln k + 6 k Explanation: We test the convergence of each of the five series. (i) For the series summationdisplay k =2 5 7 k ln k + 6 k the limit comparison test can be used, com- paring it with summationdisplay k =2 1 k ln k . Now, after division, k ln k parenleftbigg 5 7 k ln k + 6 k parenrightbigg = 5 7 + 6 ln k . Since 5 7 + 6 ln k 5 7 > as k , the limit comparison test applies. But by the Integral test, the series summationdisplay k =2 1 k ln k does not converge. Thus summationdisplay k =2 5 7 k ln k + 6 k does not converge. (ii) If a series n a n converges, then a n as n . But for the series summationdisplay n =1 6 n n ( n + 5) n we see that a n = 6 n n ( n + 5) n = 6 parenleftbigg n n + 5 parenrightbigg n = 6 parenleftbigg n + 5 n parenrightbigg n = 6 braceleftbiggparenleftbigg 1 + 5 n parenrightbigg n bracerightbigg...
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homework 12 - fh2825 Homework 12 Radin (58415) 1 This...

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