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Unformatted text preview: Rehman (aar638) HW13 sachse (56620) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the following series ( A ) summationdisplay n =1 2 ln(4 n ) n 2 , ( B ) summationdisplay n = 1 1 + cos(4 n ) n 2 + 1 converge or diverge. 1. A diverges, B converges 2. both series diverge 3. A converges, B diverges 4. both series converge correct Explanation: ( A ) The function f ( x ) = 2 ln 4 x x 2 is continous and positive on [ 1 2 , ); in addi- tion, since f ( x ) = 2 parenleftbigg 1 2 ln4 x x 3 parenrightbigg < on [ 1 2 , ), f is also decreasing on this inter- val. This suggests applying the Integral Test. Now, after Integration by Parts, we see that integraldisplay t 1 f ( x ) dx = 2 bracketleftBig ln(4 x ) x 1 x bracketrightBig t 1 , and so integraldisplay 1 f ( x ) dx = 2(1 + ln 4) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities 1 + cos(4 n ) n 2 + 1 2 n 2 + 1 2 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, series ( B ) converges . keywords: 002 10.0 points Determine the convergence or divergence of the series ( A ) summationdisplay k =2 k 2(ln k ) 2 , and ( B ) summationdisplay k = 1 tan 1 k 1 + k 4 . 1. A diverges, B converges correct 2. both series converge 3. both series diverge 4. A converges, B diverges Explanation: ( A ) By the Divergence Test, a series summationdisplay k = N a k Rehman (aar638) HW13 sachse (56620) 2 will be divergent for each fixed choice of N if lim k a k negationslash = 0 since it is only the behaviour of a k as k thats important. Now, for the given series, N = 2 and a k = k 2(ln k ) 2 . But by LHospitals 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 k = tan 1 k 1 + k 4 , b k = 1 k 4 . For lim k k 4 parenleftBig tan 1 k 1 + k 4 parenrightBig = lim k tan 1 k = 2 . Thus the given series summationdisplay k = 1 tan 1 k 1 + k 4 is convergent if and only if the series summationdisplay k = 1 1 k 4 is convergent. But by the p-series test, this last series converges because p = 4 > 1. Con- sequently, series ( B ) converges . 003 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 divergent but know nothing else about a m and b m ?...
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