1873_solutions

# 1873_solutions

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Unformatted text preview: 5 must be a partial limit of x n , suppose that x 4, 5 and suppose that  0. Since the interval x , x   must contain infinitely many members of the set Q 4, 5 we know that the condition x n x , x   must hold for infinitely many positive integers n and so x must be a partial limit of x n . Finally we observe that if x R 1 Þ 4, 5 then the open set R 1 Þ 4, 5 is a neighborhood of x and that x n fails to be frequently (or ever) in R 1 Þ 4, 5 and so x can’t be a partial limit of x n . 6. Given that x n  3  2n 5n for every positive integer n, prove that x n 2 as n Ý. We begin by observing that if n is a positive integer then 3  2n 2  7 5n 5n Now suppose that  0. The inequality 3  2n 2  5n says that 7 5n which holds when 5n  1 7 in other words n 7 5 With these inequalities in mind we choose an integer N such that N 7 5 and we observe that whenever n is an integer and n N we have 3  2n 2  . 5n 135 7. Given that 1 2n 1 n 2 1 xn  if n is even if n is odd prove that x n 0 as n Ý. We observe that if n is a positive integer then 1. n  0. Choose an integer N such that N  1/ . We observe that whenever n 1. |x n 0 |  x n  1 n N 0 Now suppose that xn 8. Suppose that x n is a sequence of real numbers and that x equivalent: a. x n x as n N, R. Prove that the following conditions are Ý. b. For every number 5 ,x  5 .  0 the sequence x n is eventually in the interval x To prove that condition a implies condition b we assume that x n x as n Ý. Suppose that  0 and, using the fact that x n x as n Ý, choose N such that the condition xn x ,x  will hold whenever n N. Then, whenever n N we have x ,x  x 5 ,x  5 . xn Now to prove that condition b implies condition a we assume that condition b holds. To prove that condition a holds, suppose that  0. Using the fact that /5 is a positive number, we choose an integer N such that the condition xn holds whenever n ,x  5 5 5 N we have x N. Thus for every n xn x 5 5 5 ,x  5 5 x ,x  . 9. Prove that n 2  3n  1 2n 2  n  4 1 2 as n Ý. We begin by observing that if n is any positive integer then 5n  5 . n 2  3n  1 5n 2 1 2 4n 4n 2 2n 2  n  4 4n 2  2n  8 Now suppose that  0. The inequality n 2  3n  1 1 2 2n 2  n  4 will hold when 5 4n which requires that n 5 4 Choose an integer N such that N  5/ 4 and observe that, whenever n N we have 5 5. n 2  3n  1 1 2 4n 4N 2n 2  n  4 10. For each positive integer n, if n can be written in the form 136 n  2m3k where m and k are postive integers and 0 1 then we define xn  m . k Otherwise we define x n  0. Prove that the set of partial limits of the sequence x n is 0, 1 . Since the range of x n is the set of all rational numbers in the interval 0, 1 , every neighborhood of a number x in the interval 0, 1 must contain infinitely many members of the range of x n and must, therefore, contain the number x n for infinitely many integers n. Thus every member of the interval 0, 1 is a partial limit of x n . If x is any number in...
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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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