1873_solutions

2 the assertion p 1 is true now suppose that n is 2

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Unformatted text preview: teger n 1, and then show that the sequence x 2n 1 is increasing and that the sequence x 2n is decreasing and that these two sequences have the same limit 2 1. Solution: For every positive number x we define Whenever 0  t  x we see that g x  2x . 5  2x 173 g t  2x 5  2x 2t  5  2t xt 0 5  2x 5  2t and so the function g is strictly increasing. Since x 1  x 3 we have g x 1  g x 3 from which we deduce that x 3  x 5 . Continuing in this way we see that the sequence x 2n 1 is increasing. Since x 2  x 4 we have g x 2  g x 4 from which deduce that x 4  x 5 . Continuing in this way we see that the sequence x 2n is decreasing. Therefore the sequences x 2n 1 and x 2n are convergent. If we write x  n Ý x 2n 1 lim gx then it follows from the identity x 2 n 1  2  x 2 n 1 5  2x 2n 1 that x  2x 5  2x and we can see that the only positive solution of this equation is 2 lim x n Ý 2n 1  2 1. Thus 1 and we can see similarly that lim x n Ý 2n c. Deduce that x n 2 1 as n  2 1. Ý. Exercises on Upper and Lower Limits 1. Prove that if x n is a sequence of real numbers then x n has a limit if and only if lmsup x n  lminf x n . nÝ nÝ Solution: This exercises follows at once from an earlier theorem. 2. Prove that a sequence x n of real numbers is bounded above if and only if lmsup x n  Ý. nÝ We know that a sequence is bounded above if and only if Ý is not a partial limit of the sequence. There a sequence is bounded above if and only if its largest partial limit is not Ý. 3. Suppose that x n is a sequence of real numbers. a. Prove that if x is a partial limit of the sequence x n then the number x is a partial limit of the sequence xn . if Ý is a partial limit of a sequence x n then x n is unbounded above and so x n is unbounded below, making Ý a partial limit of x n . Now suppose that a real number x is a partial limit of a given sequence x n . To show that the number x is a partial limit of the sequence x n , suppose that  0. We know that there are infinitely many integers n for which x  xn  x  and for each of these integers n we have x  xn  x  . Therefore the sequence x n is frequently in the interval x , x  and we conclude that x is a partial limit of x n . b. Prove that 174 lmsup x n  nÝ lminf x n . nÝ Since lminf n Ý x n is a partial limit of x n we know from part a that lminf n Ý x n is a partial limit of the sequence x n . Now given any partial limit q of x n , we know that q, being a partial limit of x n , cannot be less than lminf n Ý x n . In other words, whenever q is a partial limit of x n we have q lminf x n nÝ which gives us q Therefore lminf n Ý xn lminf x n . nÝ is the largest partial limit of the sequence xn . 4. Suppose that x n is a sequence of real numbers, that x  lmsup x n nÝ and that u  x  v. a. Prove that the sequence x n must be bounded above. The given inequality u  x  v tells us that x Ý and so x n is bounded above. b. Prove that the sequence x n must be frequently in the interval u, Ý . The...
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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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