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Unformatted text preview: Math 1910: Prelim III Solutions (12/02/2008) Problem 1. (a) ( 7pts) Determine whether the series n =1 ( 1) n 3 n +ln n converges conditionally, converges ab solutely or diverges. Justify your answer. Solution According to the Alternating Series Test, it converges. (Let a n = 1 3 n +ln n . Note that for a n > a n +1 and lim n a n = 0 . ) However, a n = 1 3 n + ln n 1 3 n + n = 1 4 n , which implies that the series is not absolutely convergent; so the convergence is conditional. (b) ( 6pts) Determine whether the series n =1 ( 1) n (1 + 1 n ) n converges conditionally, con verges absolutely or diverges. Justify your answer. Solution (1 + 1 n ) n > 1 for each n so lim n ( 1) n (1 + 1 n ) n negationslash = 0. Therefore the series diverges by the n th term test. (c) ( 7pts) Find all the values of x such that the series n =1 ( x 3) n n converges. Solution If  x 3  < 1, the series  x 3  n is a convergent geometric series. Since  x 3  n n <  x 3  n for all n 1, the original series converges absolutely if 2 < x < 4. If  x 3  > 1, then lim n  x 3  n n = , so it diverges. If x = 4, it is the harmonic series, which diverges. If x = 2, it is the alternating harmonic series and thus converges. Hence, the series converges if 2 x < 4 and diverges if x < 2 and x 4. 1 Problem 2. Decide whether or not each of the following integrals converges. If it converges, determine the value of the integral. (a) ( 10pts) integraldisplay 1 (ln( x )) 3 x 3 dx. Solution This integral converges (this can be shown by the limit comparison test with 1 /x 2 ). To compute its value, we use the substitution: u = 2 ln x , du = 2 x dx, and e u = x 2 ....
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This note was uploaded on 02/25/2009 for the course MATH 1910 taught by Professor Berman during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 BERMAN
 Math

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