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CalcHW12Answers - nav277 – Homework 12 – Odell...

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Unformatted text preview: nav277 – Homework 12 – Odell – (58340) 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 m , b m , and c m satisfy the inequalities < a m ≤ c m ≤ b m , for all m , what can we say about the series ( A ) : ∞ summationdisplay m =1 a m , ( B ) : ∞ summationdisplay m =1 b m if we know that the series ( C ) : ∞ summationdisplay m = 1 c m is convergent but know nothing else about a m and b m ? 1. ( A ) need not converge , ( B ) converges 2. ( A ) converges , ( B ) converges 3. ( A ) diverges , ( B ) converges 4. ( A ) diverges , ( B ) diverges 5. ( A ) converges , ( B ) diverges 6. ( A ) converges , ( B ) need not converge correct Explanation: Let’s try applying the Comparison Test: (i) if < a m ≤ c m , summationdisplay m c m converges , then the Comparison Test applies and says that summationdisplay a m converges; (ii) while if < c m ≤ b m , summationdisplay m c m converges , then the Comparison Test is inconclusive be- cause summationdisplay b m could converge, but it could di- verge - we can’t say precisely without further restrictions on b m . Consequently, what we can say is ( A ) converges , ( B ) need not converge . 002 10.0 points Determine whether the series ∞ summationdisplay n = 1 cos 2 n n 2 + 6 converges or diverges. 1. series is convergent correct 2. series is divergent Explanation: Note first that the inequalities < cos 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 parenleftbigg 5 4 parenrightbigg n nav277 – Homework 12 – Odell – (58340) 2 2. ∞ summationdisplay k =2 4 7 k ln k + 5 k 3. ∞ summationdisplay n = 1 5 n n ( n + 4) n 4. ∞ summationdisplay n = 1 7 5 n − 4 5. ∞ summationdisplay n = 1 parenleftbigg 4 n 7 n + 4 parenrightbigg n correct Explanation: We test the convergence of each of the five series. (i) For the series ∞ summationdisplay k =2 4 7 k ln k + 5 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 4 7 k ln k + 5 k parenrightbigg = 4 7 + 5 ln k . Since 4 7 + 5 ln k −→ 4 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 4 7 k ln k + 5 k does not converge. (ii) If a series ∑ n a n converges, then a n −→ as n → ∞ . But for the series ∞ summationdisplay n =1 5 n n ( n + 4) n we see that a n = 5 n n ( n + 4) n = 5 parenleftbigg n n + 4 parenrightbigg n = 5 parenleftbigg n + 4 n parenrightbigg − n = 5 braceleftbiggparenleftbigg 1 + 4 n parenrightbigg n bracerightbigg...
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CalcHW12Answers - nav277 – Homework 12 – Odell...

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