Graphics Reference
In-Depth Information
Convergent sequences in closed intervals
a
0
aa a
3 2 1
75 We investigate the question 'Can a sequence of positive terms have
a negative limit?'
'Obviously not,' you think. Here is a more reasoned response.
Let (
a
)
a
,0
a
and
a
0. Usetheinequality 0
a
a
a
to
provethat
a
0.
Deduce that, if (
a
)
a
and 0
a
, then necessarily 0
a
.
76
The inequality rule
Let (
a
)
a
,(
b
)
b
and
a
b
for all
n
. By considering the
), and using the difference rule, prove that
a
b
.
77 By considering the sequences defined by
a
sequence (
b
a
1/
n
and
b
1/
n
,
show that even if
a
b
does
not
imply
a
b
. This gives a
counter
-
example
to theproposition that if
a
b
for all
n
,(
a
)
a
and (
b
)
b
for all
n
,(
a
)
a
and (
b
)
b
implies
a
b
.
78
The closed interval property
Let (
a
B
for all
n
. Provethat
A
a
B
.
The reason for calling the interval
x
A
x
B
a
closed
interval
is that no convergent sequence in the interval can have a limit
which escapes from the interval.
Givean exampleto show that if (
a
)
a
and
A
a
a
and
A
a
B
for all
n
,
)
it does not necessarily follow that
A
a
B
.
Theintrval
is called an
open
interval, partly
because a convergent sequence in the interval may have a limit
outsidetheintrval.
79 Let (
a
x
A
x
B
) be a sequence of positive terms such that
a
a
2,
and (
a
)
a
. Provethat
a
2. Usethesum rule(qn 54(iii)), the
product rule (qn 54(vi)) and the closed interval property.
(
Optional
) Provethat (
a
) is monotonic.
80 Let (
a
) be a monotonic increasing sequence with a convergent
subsequence (
a
)
A
.
a
n
?
a
n
k
A
n
