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Unformatted text preview: 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
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 153 lminf x n
nÝ q 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 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 v Since
xm yN v
whenever m N we know that the sequence x n cannot b...
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 Fall '08
 STAFF
 Math, Calculus

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