1873_solutions

Test and is there to lead into parts b and c since

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Unformatted text preview: erge absolutely. a. b. 2n n The given series must be divergent if |3x 2 |  1 because, in this case, the expression 3x 2 n |3x 2 | n fails to approach 0 as n Ý. In the event that |3x 2 |  1, the series n n converges by a simple application of d’Alembert’s test. We are left with the case |3x 2 |  1 1n which is which occurs when x  1 or x  1. When x  1 the given series becomes n 3 3 1 conditionally convergent and when x  1 it becomes which is divergent. n Thus the series converges absolutely when 1  x  1, converges conditionally when x  1 3 3 and diverges otherwise. 3x log x n n This series converges absolutely when |log x |  1, converges conditionally when log x  1 and diverges otherwise. In other words, the series converges absolutely when 1  x  e, e converges conditionally when x  1 and diverges otherwise. e 341 c. 1 nxn log n x For each n 2 we define an  1 nxn log n x and observe that x log n  |x |. log n  1 a n converges absolutely when |x |  1. If |x |  1 then, since a n fails to approach 0 Therefore as n Ý, the series a n diverges. When x  1 the series is log n which diverges. When 1n x  1 the series is which is conditionally convergent. log n n lim aa 1 nÝ n d.  n Ý|x | lim 3n ! x n 2n ! n! For each n we define 3n ! x n 2n ! n! an  and observe that a n 1 an 3n  3 3n  2 3n  1 2n  2 2n  1 n  1  |x | and so n lim aa 1 n nÝ  27|x | . 4 3n ! x n converges absolutely when |x |  2n ! n! a n fails to approach 0 as n Ý, the series a n diverges. 4 Now suppose x  27 . The series a n becomes 3n ! 4 n 2n ! n! 27 n and in this case 3n  3 3n  2 3n  1 4 n lim lim n 1 aa 1  n Ý n 1 nÝ n 2n  2 2n  1 n  1 27 Therefore the series 4 27 . If |x |  4 27 then, since  1 1 2 Thus it follows from the criterion for the nth term to approach 0 that a n decreases to 0 as a n is divergent. n Ý. In spite of this fact, the series 4 Finally, suppose that x  27 . The series a n becomes 3n ! 4 n 1n 2n ! n! 27 n 3n ! x n which converges by Dirichlet’s test. Therefore the series converges absolutely 2n ! n! 4 4 when |x |  27 , converges conditionally when x  27 and diverges for other values of x. e. nnxn n! For each n we define nn an  n x n! and observe that a n 1 an and so 342  |x | 1 1 n e n n lim aa 1  e|x |. n n n x n converges absolutely when |x |  Therefore the series n! to approach 0 as n Ý, the series a n diverges. nÝ Now suppose that x  1 e . The series a n becomes 1 e . If |x |  1 e then, since a n fails n n . Since e n n! n1 n1 lim n 1 nÝ a n 1 an  n Ýn lim 1 e n1 n1 ! nn e n n! 1 2 we know from Raabe’s test that a n is divergent but we also know from the criterion for the nth term to approach 0 that a n decreases to 0 as n Ý. n Finally suppose that x  1 . The series 1 n n which converges by a n becomes e n n! e Dirichlet’s test. nxn n Therefore the series converges if |x |  1 , converges conditionally if x  1 and e e n! diverges for other values of x. 5. Find the values of x...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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