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Unformatted text preview: hat xy is a partial limit of the sequence x n y n , suppose that 0.
Using the
fact that x n x as n Ý we choose an integer N such that the inequality
x n x 
2y  1
holds whenever n N. Since there are only finitely many positive integers n N, and y is a partial
limit of the sequence y n , there must be infinitely many integers n N for which the inequality
y n y 
2p
holds. For each of these infinitely many integers n we have 140 x n y n xy  x n y n
x n y n
py n
p xny xny xy  x n y  x n y
y  y x n 2p y  xy 
x 2y  1 . 4. Give an example of two sequences x n and y n and a partial limit x of x n and a partial limit y of y n such
that x y fails to be a partial limit of the sequence x n y n .
Define x n y n 1 n for each n. We observe that the numbers 1 and 1 are partial limits of x n
and y n respectively but that 1 1 fails to be a partial limit of x n y n .
5. Give an example of two divergent sequences x n and y n such that the sequence x n y n is convergent.
We define x n 1 n and y n 1 n 1 for each n. Note that the sequence x n y n is the sequence
with constant value 0.
6. Give an example of two sequences x n and y n such that x n 0 and y n Ý and a. x n y n 0
We define x n n and y n 1/n 2 for each n.
b. x n y n 6
We define x n n and y n 6/n for each n.
c. x n y n Ý
We define x n n 2 and y n 1/n for each n.
d. The sequence x n y n is bounded but has no limit.
We define x n n and y n 1 n /n for each n.
7. Given two sequences x n and y n of real numbers such that both of the sequences x n and x n y n are
convergent, is it true that the sequence y n must be convergent?
Yes. Since
yn xn yn
xn
for each n it follows at once that
lim x
lim
lim y n Ý x n y n
nÝ n
nÝ n
8. Given that x n is a sequence of real numbers and that x n 0, prove that
x1 x2 x3 xn
0.
n Solution: Suppose that 0. Using the fact that x n 0 as n Ý, choose an integer N 1
such that the inequality x n  /2 holds whenever n N 1 . Whenever n N 1 we see that
x 1 x 2 x 3 x n x 1 x 2 x 3 x N 1 x N 1 x N 1 1 x n
n
n
n
x N 1 x N 1 1 x n
x 1 x 2 x 3 x N1
.
n
n
Now we choose an integer N 2 such that
2x 1 x 2 x 3 x N 1 
N2
and we define N to be the larger of the two numbers N 1 and N 2 . Then whenever n N we have
x N 1 x N 1 1 x n
x 1 x 2 x 3 x N1
x1 x2 x3 xn 0
n
n
n
x 1 x 2 x 3 x N1
x N 1 1  x N 1 2  x n 
n
N2
n nN 1
.
2
2 141 9. Given that x n is a sequence of real numbers, that x is a real number and that x n
x1 x2 x3 xn
x.
n Solution: Since x n
as n as n x 0 as n Ý we have
x1 x x2 x x3
n Ý. Therefore
x1 x2 x3 xn
n
Ý. x x1 x xn x x2 x x3
n x x, prove that 0 x xn x 0 10. Given that x n and y n are sequences of real numbers and that x n y n 0, prove that x n and y n have
the same set of partial limits.
Since x n x n y n y n...
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 Fall '08
 STAFF
 Math, Calculus

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