Solutions, Homework #4
Section 17.
6. (a)
Counterexample:
(
s
n
) = (1
,

1
,
1
,

1
,
· · ·
);
(
t
n
) = (

1
,
1
,

1
,
1
,
· · ·
).
(b) Counterexample:
the example in part (a) will work here, too.
(c) Suppose (
s
n
)
→
A
and (
s
n
+
t
n
)
→
B
. Then
t
n
= (
s
n
+
t
n
)

s
n
→
B

A
.
(d) Counterexample:
(
s
n
) =
parenleftbigg
1
,
1
2
,
1
3
,
· · ·
,
1
n
· · ·
parenrightbigg
, s
n
→
0, (
t
n
) = ((

1)
n
) =
(

1
,
1
,

1
,
1
, . . .
) does not converge. But (
s
n
t
n
) =
parenleftbigg
(

1)
n
n
parenrightbigg
→
0.
9. (
s
n
)
and (
t
n
) are sequences such that
s
n
≤
t
n
for all
n
.
(a) Choose any number
M
∈
R
. Since
lim
n
→∞
s
n
= +
∞
,
there is a positive
integer
N
such that
s
n
> M
for all
n > N
. Since
t
n
≥
s
n
for all
n
, it
follows that
t
n
> M
for all
n > N
. Therefore
lim
n
→∞
t
n
= +
∞
:
(b) Choose any number
M
∈
R
. Since
lim
n
→∞
t
n
=
∞
,
there is a positive
integer
N
such that
t
n
< M
for all
n > N
. Since
s
n
≤
t
n
for all
n
, it
follows that
s
n
< M
for all
n > N
. Therefore
lim
n
→∞
s
n
=
∞
:
12. Let
epsilon1 >
0. Since
(
s
n
) is convergent, it is bounded;

s
n
 ≤
M
for all
n
.

s
2
n

s
2

=

s
n

s
 
s
n
+
s
 ≤ 
s
n

s

(
M
+

s

).
Since
s
n
→
s
there is a
positive integer
N
such that

s
n

s

<
epsilon1
M
+

s

for all
n > N
. Therefore
vextendsingle
vextendsingle
s
2
n

s
2
vextendsingle
vextendsingle
=

s
n

s
 
s
n
+
s
 ≤ 
s
n

s

(
M
+

s

)
<
(
M
+

s

)
epsilon1
M
+

s

=
epsilon1
for all
n > N
and it follows that
s
2
n
→
s
2
.
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 Fall '08
 Staff
 Tn, Natural number, Sn