- a)
- b)
- is a null sequence.
- c)
- is a null sequence.

Proof: By definition 7.10, `` " means

for every
there
is some
such that

for every , .

for every , .

By definition 7.11, `` is a null sequence" means

Both definitions say the same thing. If we write out the definition for `` is a null sequence" we get (7.30) with `` " replaced by

`` ." Since

conditions b) and c) are equivalent.

Proof:
.

Proof: Suppose and .
By the decomposition theorem, we can write

where and are null sequences. Then

By the sum theorem for null sequences, is a null sequence, so is a null sequence, and hence .

Proof: Suppose and . Write , where
are null sequences. Then

Now , and are null sequences by the product theorem and sum theorem for null sequences, and , so by several applications of the sum theorem for convergent sequences,

be a complex sequence, and let . If and , then .

Proof: Suppose and . Then and are null
sequences, so
is a null
sequence. Hence, by theorem 7.15, ; i.e., .

and

and

Proof: If
and
are convergent, then it follows from the sum
theorem for convergent sequences that is convergent and (7.40) is valid.

Suppose that . Then is a null sequence, so
is a null sequence (by Theorem 7.26). For all
,

so is a null sequence and it follows that converges to . A similar argument shows that .

Any number satisfying condition (7.42) is called a

The next theorem I want to prove is a quotient theorem for convergent sequences. To prove this, I will need some technical results.

Proof: By the triangle inequality.

Hence,

Proof: Since , we know that is a null sequence. Let
be a
precision function for . Then for all
,

i.e., if , then

Let

Then for and for , so

for all , and hence is bounded.

Proof: By the preceding lemma,
is a bounded sequence, and since
, we know that is a null sequence. Hence
is a null sequence, and it follows that
. Then we have

i.e., .

Let be convergent complex sequences. If and , then is convergent and .

- a)
- Let be complex sequences. Show that if converges and diverges, then diverges.
- b)
- Show that if converges and diverges, then does not necessarily diverge.

Then

Hence can be written as a quotient of two sequences:

and

where for all . Since

and

it follows from numerous applications of product and sum rules that and and hence . Once I have expressed in the final form in (7.53), I can see what the final result is, and I will usually just write

In the last two examples, I was motivated by the following considerations. I think:
In the numerator and denominator for (7.52), for large the `` "
term overwhelms the other terms - so that's the term I factored out. In the
numerator of (7.55), the overwhelming term is , and in the denominator,
the overwhelming term is so those are the terms I factored out.

- a)
- b)
- c)
- d)
- e)

Proof: I will show that null sequences are bounded and leave the general case to
you. Let be a null sequence and let be a precision function for .

Let

I claim that is a bound for . If , then

If , then , so . Hence

i.e., for all .