1873_solutions

# A n and b n in a metric space x are eventually close

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Unformatted text preview: define x n  n and y n  1/n 2 for each n. b. x n y n 6 We define x n  n and y n  6/n for each n. c. x n y n Ý We define x n  n 2 and y n  1/n for each n. d. The sequence x n y n is bounded but has no limit. We define x n  n and y n  1 n /n for each n. 7. Given two sequences x n and y n of real numbers such that both of the sequences x n and x n  y n are convergent, is it true that the sequence y n must be convergent? Yes. Since yn  xn  yn xn for each n it follows at once that lim x lim lim y  n Ý x n  y n nÝ n nÝ n Perhaps this exercise should have been given in R k . 8. Given that x n is a sequence of real numbers and that x n 0, prove that x1  x2  x3    xn 0. n Solution: Suppose that  0. Using the fact that x n 0 as n Ý, choose an integer N 1 such that the inequality |x n |  /2 holds whenever n N 1 . Whenever n N 1 we see that x 1  x 2  x 3    x n  x 1  x 2  x 3    x N 1  x N 1  x N 1 1    x n n n n x N 1  x N 1 1    x n x 1  x 2  x 3    x N1  . n n Now we choose an integer N 2 such that 165 N2 2|x 1  x 2  x 3    x N 1 | and we define N to be the larger of the two numbers N 1 and N 2 . Then whenever n N we have x N 1  x N 1 1    x n x 1  x 2  x 3    x N1 x1  x2  x3    xn 0  n n n x 1  x 2  x 3    x N1 |x N 1 1 |  |x N 1 2 |    |x n |  n N2 n N1   . n 2 2 9. Given that x n is a sequence of real numbers, that x is a real number and that x n x1  x2  x3    xn x. n Solution: Since x n as n as n x 0 as n Ý we have x1 x  x2 x  x3 n Ý. Therefore x1  x2  x3    xn n Ý. x x1 x    xn x  x2 x  x3 n x x, prove that 0 x    xn x 0 10. Given that x n and y n are sequences of real numbers and that x n y n 0, prove that x n and y n have the same set of partial limits. Since x n  x n y n  y n and y n  x n x n y n for each n the present result follows at once from Exercise 1. 11. Suppose that x n and y n are sequences of real numbers, that x n y n a partial limit of at least one of the sequences x n and y n . Prove that xn 1 yn as n 0 and that the number 0 fails to be Ý. Solution: From Exercise 10 we know that x n and y n have the same sets of partial limits and we know, therefore, that 0 is not a partial limit of either of these two sequences. Using the fact that 0 is not a partial limit of y n , choose an integer N 1 and a number   0 such that the inequality |y n |  holds whenever n N 1 . For every n N 1 we see that |x n y n | xn 1  xn yn yn yn  and so the fact that xn 1 yn as n Ý follows from the sandwich theorem. 12. Give an example to show that the requirement in Exercise 11 that 0 not be a partial limit of at least one of the two sequences is really needed. We define x n  2/n and y n  1/n for each n and observe that, even though x n y n 0 as n Ý, x n /y n 2 as n Ý. 13. Suppose that x n and y n are sequences of real numbers, that x n /y n 1 and that at least one of the sequences x n and y n is bounded. Prove that x n y n 0. Give an example to...
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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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