MATH 301 Homework 4 - lim x n 6 = 0 ? Give a...

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Math 301, Homework 4 For problems below do not use any convergence theorem, use only de&- nitions; i.e. give direct proofs by an N argument. All sequences are in R : 1. Suppose ( x n ) is a convergent sequence and ( y n ) is another sequence that y n = x n for all n ± 1000 : Prove that ( y n ) convergent. 2. Suppose ( x n ) is convergent and ; and ( y n ) is some bounded sequence. Hint: Since ( y n ) is bounded, there is some M so that j y n j ² M for all n: (a) If lim x n = 0 ; prove that ( x n y n ) is also convergent. (b) Is ( x n y n ) still convergent if
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Unformatted text preview: lim x n 6 = 0 ? Give a counterexample. 3. The sequence ( x n ) satis&es the following condition: for every positive integer n;k with n ± k 2 ; we have j x n j < 1 2 k : Find one (speci&c) integer m so that j x m j < 1 999 . Prove your answer. 4. Suppose ( x n ) is a convergent : (a) If x n > for all n; is it always true that lim x n > 0? Can you &nd a counterexample? (b) If lim x n 2 [0 ; 1] ; Show that the set A = & k : ± ± x k & 1 2 ± ± > 3 5 ² is &nite. 1...
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This note was uploaded on 06/25/2010 for the course MATH MATH 301 taught by Professor Alberterkip during the Fall '08 term at Sabancı University.

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