1873_solutions

A n suppose that 0 choose an integer n such that the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hat xy is a partial limit of the sequence x n y n , suppose that  0. Using the fact that x n x as n Ý we choose an integer N such that the inequality |x n x |  2|y |  1 holds whenever n N. Since there are only finitely many positive integers n  N, and y is a partial limit of the sequence y n , there must be infinitely many integers n N for which the inequality |y n y |  2p holds. For each of these infinitely many integers n we have 140 |x n y n xy |  |x n y n |x n y n p|y n p xny  xny xy | x n y |  |x n y y |  |y ||x n 2p  |y | xy | x| 2|y |  1 . 4. Give an example of two sequences x n and y n and a partial limit x of x n and a partial limit y of y n such that x  y fails to be a partial limit of the sequence x n  y n . Define x n  y n  1 n for each n. We observe that the numbers 1 and 1 are partial limits of x n and y n respectively but that 1  1 fails to be a partial limit of x n  y n . 5. Give an example of two divergent sequences x n and y n such that the sequence x n  y n is convergent. We define x n  1 n and y n  1 n 1 for each n. Note that the sequence x n  y n is the sequence with constant value 0. 6. Give an example of two sequences x n and y n such that x n 0 and y n Ý and a. x n y n 0 We 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 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 2|x 1  x 2  x 3    x N 1 | N2 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 nN 1 . 2 2 141 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...
View Full Document

Ask a homework question - tutors are online