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Homework 12

# Homework 12 - tgo72 Homework 12 Gompf(58370 This print-out...

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tgo72 – Homework 12 – Gompf – (58370) 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 0 < 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 ) converges , ( B ) need not converge correct 2. ( A ) diverges , ( B ) diverges 3. ( A ) converges , ( B ) diverges 4. ( A ) need not converge , ( B ) converges 5. ( A ) converges , ( B ) converges 6. ( A ) diverges , ( B ) converges Explanation: Let’s try applying the Comparison Test: (i) if 0 < a m c m , summationdisplay m c m converges , then the Comparison Test applies and says that summationdisplay a m converges; (ii) while if 0 < 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 k =1 cos 2 k k 2 + 7 converges or diverges. 1. series is convergent correct 2. series is divergent Explanation: Note first that the inequalities 0 < cos 2 k k 2 + 7 1 k 2 + 7 1 k 2 hold for all n 1. On the other hand, by the p -series test the series summationdisplay n =1 1 k 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 3 2 parenrightbigg n

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tgo72 – Homework 12 – Gompf – (58370) 2 2. summationdisplay n =1 3 n n ( n + 4) n 3. summationdisplay n =1 parenleftbigg 4 n 5 n + 2 parenrightbigg n correct 4. summationdisplay k =2 2 5 k ln k + 3 k 5. summationdisplay n =1 5 3 n 2 Explanation: We test the convergence of each of the five series. (i) For the series summationdisplay k =2 2 5 k ln k + 3 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 2 5 k ln k + 3 k parenrightbigg = 2 5 + 3 ln k . Since 2 5 + 3 ln k −→ 2 5 > 0 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 2 5 k ln k + 3 k does not converge. (ii) If a series n a n converges, then a n −→ 0 as n → ∞ . But for the series summationdisplay n =1 3 n n ( n + 4) n we see that a n = 3 n n ( n + 4) n = 3 parenleftbigg n n + 4 parenrightbigg n = 3 parenleftbigg n + 4 n parenrightbigg n = 3 braceleftbiggparenleftbigg 1 + 4 n parenrightbigg n bracerightbigg 1 .
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