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hw13_solution

# hw13_solution - padilla(tp5647 HW13 Gilbert(56650 This...

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padilla (tp5647) – HW13 – Gilbert – (56650) 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 m =1 2 ln(3 m ) m 2 , ( B ) summationdisplay m =1 1 + sin(3 m ) m 2 + 2 converge or diverge. 1. both series converge correct 2. A converges, B diverges 3. A diverges, B converges 4. both series diverge 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 ln 3 x x 3 parenrightbigg < 0 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 0 1 + sin(3 m ) m 2 + 2 2 m 2 + 2 2 m 2 hold for all n 1. On the other hand, by the p -series 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 n =2 n 3(ln n ) 2 , and ( B ) summationdisplay n =1 tan 1 n 2 + n 2 . 1. both series converge 2. both series diverge 3. A converges, B diverges 4. A diverges, B converges correct Explanation: ( A ) By the Divergence Test, a series summationdisplay n = N a n

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padilla (tp5647) – HW13 – Gilbert – (56650) 2 will be divergent for each fixed choice of N if lim n → ∞ a n negationslash = 0 since it is only the behaviour of a n as n → ∞ that’s important. Now, for the given series, N = 2 and a n = n 3(ln n ) 2 . But by L’Hospital’s 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 n = tan 1 n 2 + n 2 , b n = 1 n 2 . For lim n → ∞ n 2 parenleftBig tan 1 n 2 + n 2 parenrightBig = lim n → ∞ tan 1 n = π 2 . Thus the given series summationdisplay n =1 tan 1 n 2 + n 2 is convergent if and only if the series summationdisplay n =1 1 n 2 is convergent. But by the p -series 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 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 ) diverges 2. ( A ) need not converge , ( B ) converges 3. ( A ) diverges , ( B ) converges 4. ( A ) converges , ( B ) converges 5. ( A ) diverges , ( B ) diverges 6. ( A ) converges , ( B ) need not converge correct 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
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