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Unformatted text preview: adams (bda255) HW13 Radin (56635) 1 This printout should have 18 questions. Multiplechoice 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 m = 1 2 ln(3 m ) m 2 , ( B ) summationdisplay m = 1 1 + sin(3 m ) m 2 + 1 converge or diverge. 1. A converges, B diverges 2. both series diverge 3. A diverges, B converges 4. both series converge correct Explanation: ( A ) The function f ( x ) = 2 ln 3 x x 2 is continous and positive on [ 2 3 , ); in addi tion, since f ( x ) = 2 parenleftbigg 1 2 ln3 x x 3 parenrightbigg < on [ 2 3 , ), 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(3 x ) x 1 x bracketrightBig t 1 , and so integraldisplay 1 f ( x ) dx = 2(1 + ln 3) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities 1 + sin(3 m ) m 2 + 1 2 m 2 + 1 2 m 2 hold for all n 1. On the other hand, by the pseries test the series summationdisplay m =1 1 m 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 m = 2 m 3(ln m ) 2 , and ( B ) summationdisplay m =1 tan 1 m 5 + m 2 . 1. both series diverge 2. A converges, B diverges 3. both series converge 4. A diverges, B converges correct Explanation: ( A ) By the Divergence Test, a series summationdisplay m = N a m adams (bda255) HW13 Radin (56635) 2 will be divergent for each fixed choice of N if lim m a m negationslash = 0 since it is only the behaviour of a m as m thats important. Now, for the given series, N = 2 and a m = m 3(ln m ) 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 m = tan 1 m 5 + m 2 , b m = 1 m 2 . For lim m m 2 parenleftBig tan 1 m 5 + m 2 parenrightBig = lim m tan 1 m = 2 . Thus the given series summationdisplay m =1 tan 1 m 5 + m 2 is convergent if and only if the series summationdisplay m =1 1 m 2 is convergent. But by the pseries test, this last series converges because p = 2 > 1. Con sequently, series ( B ) converges . 003 10.0 points If a m , b m , and c m satisfy the inequalities < b m c m a 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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This note was uploaded on 04/30/2009 for the course UNKNOWN 408 L taught by Professor Radin during the Spring '09 term at SUNY Adirondack.
 Spring '09
 RAdin
 Calculus, Inequalities

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