# 29_new - EXAM 01 gilbert(55035 This print-out should have...

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EXAM 01 – gilbert – (55035) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine whether the series n X 4 1 converges, and if it does, compute its value. 1. I = 4 9 2. I = 2 9 correct 3. I = 8 9 4. integral doesn’t converge 3 n = 0 5. I = 8 is convergent or divergent, and if convergent, find its sum. 1. convergent, sum = 3 2. convergent, sum = 13 2 3. divergent 4. convergent, sum = 13 2 5. convergent, sum = 6 correct Explanation: The given series is an infinite geometric series X a r n n = 0 with a = 4 and r = 1 3 . But the sum of such a series is a (i) convergent with sum 1 r when |r| < 1, (ii) divergent when |r| ≥ 1. Consequently, the given series is convergent, sum = 6 . 002 10.0 points Determine if the improper integral 8 x I = 3 (9 + x 2 ) 2 dx 27 Explanation: The integral 8 x I = 3 (9 + x 2 ) 2 dx is improper because of the infinite interval of integration. To overcome this, we truncate and consider the limit t lim →∞ I t , I t = 3 t (9 + 8x x 2 ) 2 dx . 2 To evaluate I t , set u = 9 + x . Then du = 2x dx , in which case (9 + 8x x 2 ) 2 dx = 4 u 1 2 du. Thus I t = 3 t (9 + 8x x 2 ) 2 dx = 4 h 9 + 1 x 2 i 3 t = 4 ë 18 1 9 + 1 t 2 o . Consequently, since lim 1 2 = 0, t → ∞ 9 + t we see that I converges and that I = t lim →∞ 3 t (9 + 8x x 2 ) 2 dx = 9 2 .

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EXAM 01 – gilbert – (55035) 2 as n ∞, so series (B) converges. 003 10.0 points Consequently, of the given infinite series, Determine which, if any, of the following series diverge. (6n) n (A) n X =1 n! n (B) n X =1 3 (n + 3) n 1. neither of them 2. A only correct 3. B only 4. both of them Explanation: To check for divergence we shall use either the Ratio test or the Root test which means computing one or other of n lim →∞ a a n+1 n , n lim →∞ |a n | 1/n for each of the given series. (A) The ratio test is the better one to use because a a n+1 = 6 (n + n! 1)! (n + n 1) n n+1 . n Now n! 1 , = (n + 1)! n + 1 while (n + 1) n+1 = (n + 1) n + n 1 n . n n Thus a a n+1 = 6 n + n 1 n 6e > 1 n −→ as n ∞, so series (A) diverges. (B) The root test is the better one to apply because |a n | 1/n = 3 −→ 0 n + 3
only A diverges . 004

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29_new - EXAM 01 gilbert(55035 This print-out should have...

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