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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.
 Fall '08
 STAFF
 Math, Calculus

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