By comparing this definition with definition 7.10, you see that

Definition 7.11 is important. You should memorize it.

The definitions of null sequence and dull sequence use the same words, but they are not in the same order, and the definitions are not equivalent.

If satisfies condition (7.12), then whenever ,

If , this condition would say , which is false. Hence if , then ; i.e., if , then . Hence a dull sequence has the property that there is some such that for all . Thus every dull sequence is a null sequence. The sequence

is a dull sequence, but

is not a dull sequence. In the next theorem we show that is a null sequence, so null sequences are not necessarily dull.

Proof:
Let
. By the Archimedean property for
, there is an
such that
.
Then for all
,

so for all .

The difference between a null sequence and a dull sequence is that the `` " in the definition of null sequence can (and usually does) depend on , while the ``" in the definition of dull sequence depends only on . To emphasize that depends on (and also on ), I will often write or instead of .

Here is another reformulation of the definition of null sequence.

I will call such a function a

This formulation shows that in order to show that a sequence is a null sequence, you need to find a *function*
such that

In the proof of theorem 7.13, for the sequence we had

This description for could be made more precise, but it is good enough for our purposes.

Proof: If , then
. Suppose, to get a
contradiction, that is a null sequence. Then there is a number
such that for all
. Then for all
,

If then (7.16) is false and this shows that is not a null sequence.

complex sequences. Suppose that is a null sequence and that

Proof: Since is a null sequence, there is a function
such
that for all
,

Then

Hence, we can let .

Scratchwork: Let
. I want to find so that for all
and all
,

i.e.

i.e.

This suggests that I should take .

Proof: Let be a null sequence in and let be a precision
function for . Define
by
for all
. Then for all
,

Hence is a precision function for .

Consider the sequence . For all ,

Hence , so it follows from the comparison theorem that is a null sequence.

Since
is a null sequence, it follows
from the root theorem that
is a null sequence. Now
,
so
so

for all , and by another comparison test, is a null sequence. Since , it follows that is a null sequence for all with .

You probably suspect that is a null sequence for all with . This is correct, but we will not prove it yet.

and

that are sketched above 7.1 are in fact
null sequences.

- a)
- b)
- c)

is a null sequence. (If you succeed, you will probably find a proof that is a null sequence whenever .) NOTE: If you use calculator operations, then is not a null sequence on most calculators.

It follows from remark 5.38 that we can add, subtract and multiply complex
sequences, and that the usual associative, commutative, and distributive laws hold.
If and then
and
. If
then the constant sequences
satisfy

Proof: All four results follow by the comparison theorem. We have, for all :