Andrew HW 11 - rabinovich(ahr454 HW11 um(55915 This print-out should have 17 questions Multiple-choice questions may continue on the next column or page

# Andrew HW 11 - rabinovich(ahr454 HW11 um(55915 This...

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rabinovich (ahr454) – HW11 – um – (55915) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Let h be a continuous, positive, decreasing function on [3 , ). Compare the values of the series A = 11 n = 4 h ( n ) and the integral B = 11 3 h ( z ) dz . 1. A = B 2. A < B correct 3. A > B Explanation: In the figure 3 4 5 6 7 . . . a 4 a 5 a 6 a 7 the bold line is the graph of h on [3 , ) and the areas of the rectangles the terms in the series n = 4 a n , a n = h ( n ) . Clearly from this figure we see that a 4 = h (4) < 4 3 h ( z ) dz, a 5 = h (5) < 5 4 h ( z ) dz , while a 6 = h (6) < 6 5 h ( z ) dz, a 7 = h (7) < 7 6 h ( z ) dz , and so on. Consequently, A < B . keywords: Szyszko 002 10.0 points Which, if any, of the following series con- verge? A . 1 + 1 4 + 1 9 + 1 16 + . . . B . n = 1 n 1 + n 2 1. neither of them 2. B only 3. A only correct 4. both of them Explanation: A. Series is n =1 1 n 2 . Use f ( x ) = 1 x 2 . Then 1 f ( x ) dx is convergent, so series converges.
rabinovich (ahr454) – HW11 – um – (55915) 2 B. Use f ( x ) = x 1 + x 2 . Then 1 f ( x ) dx is divergent (ln integral), so series diverges. 003 10.0 points Determine whether the series n = 1 6 - 4 n n 3 converges or diverges. 1. series is convergent correct 2. series is divergent Explanation: We check separately the convergence or di- vergence of n = 1 6 n 3 , n = 1 4 n n 3 = n = 1 4 n 5 / 2 . Now both are p-series: the first being of the form n = 1 6 n 3 = n = 1 6 n p , p = 3 , while the second one has the form n = 1 4 n 5 / 2 = n = 1 4 n p , p = 5 2 . Since p > 1 in both cases, each series con- verges. As the di ff erence of convergent series, therefore, the given series is convergent . 004 10.0 points Determine whether the series n = 2 n 4(ln( n )) 2 is convergent or divergent. 1. divergent correct 2. convergent Explanation: By the Divergence Test, a series n = N a n will be divergent for each fixed choice of N if lim n → ∞ a n = 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 4(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 = . Consequently, by the Divergence Test, the given series is divergent . 005 10.0 points If the improper integral 1 1 x p dx converges, which of the following statements is (are) always true? ( A ) n 1 n p converges; ( B ) n 1 n p +1 diverges;
rabinovich (ahr454) – HW11 – um – (55915) 3 ( C ) n 1 n p - 1 converges; ( D ) n 1 n p - 1 diverges; ( E ) n 1 n p +1 converges . 1. A, D and E 2. B and D only 3. A and E only correct 4. A, C and E only 5. A only Explanation: To apply the Integral test we need to start with a function f which is positive, continuous and decreasing on [1 , ). Then the integral test says that the improper integral 1 f ( x ) dx converges if and only if the infinite series n = 1 f ( n ) converges.

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