1873_solutions

# That the sequence has a limit x that satisfies the

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Unformatted text preview: interval u, Ý is a neighborhood of the partial limit x of x n . c. Prove that the sequence x n cannot be frequently in the interval v, Ý . Hint: Choose an upper bound  of x n . Now use this theorem. If x n were frequently in the interval v, Ý then it would have to have a partial limit in this interval, contradicting the fact that x is the largest partial limit of x n . 5. Suppose that x n is a sequence of real numbers, that x is a real number and that, whenever u  x  v, the sequence x n is frequently in the interval u, Ý but is not frequently in the interval v, Ý . Prove that x  lmsup x n . nÝ The fact that x n must be frequently in the interval u, v whenever u  x  v tells us that x n is frequently in every neighborhood of x and so x must be a partial limit of x n . To show that x is the largest partial limit of x n we shall show that no number larger than x can be a partial limit of x n . Suppose that x  p and choose a number v between x and p. Since the interval v, Ý is a neighborhood of p and x n is not frequently in v, Ý we conclude that p can’t be a partial limit of xn . 6. Suppose that x n is a bounded sequence of real numbers and that, for each integer n in the domain of this sequence we have defined y n  sup x m m n . Prove that the sequence y n is decreasing and that its limit is the lower limit of the sequence x n . For each positive integer n, it follows from the fact that xm m n  1 xm m n that sup x m m n sup x m m n  1 and so the sequence y n must be decreasing. Since every lower bound of x n is also a lower bound of y n , the sequence y n must be bounded below. Therefore y n is convergent. We define y to be the limit of y n . To show that y is lmsup n Ý x n we shall use Exercise 5. Suppose that u  y  v. Using the fact that y n y as n Ý we choose N such that the inequality y n  v holds whenever n N. u yn y 175 v Since xm yN  v whenever m N we know that the sequence x n cannot be frequently in the interval v, Ý . On the other hand, given any integer n N it follows from the fact that u  y n  sup x m m n that there exists an integer m n such that u  x m . u yn y xm v Thus the set of integers m for which u  x m is unbounded above and we have shown that x n is frequently in the interval u, Ý . 7. State and prove an analogue of the preceding exercise for lower limits. Suppose that x n is a bounded sequence that that, for each n we define y n  inf x m m n . Then the sequence y n is decreasing and the limit of the sequence y n is lminf n Ý x n . This assertion can be proved by a mirror image of the proof that was used in Exercise 6 and it can also be obtained from the statement of Exercise 6, in view of Exercise 3b. 8. Given that zn  xn  yn for every positive integer n, prove that lmsup z n nÝ lmsup x n  lmsup y n . nÝ nÝ Solution: To obtain a contradiction we assume that lmsup z n  lmsup x n  lmsup y n . nÝ Now we choose a number nÝ nÝ  0 such that lmsup z n  lmsup x n  lmsup y n  . nÝ nÝ n...
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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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