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# 7.5 Theorems About Convergent Sequences

7.29   Remark. Let be a complex sequence, and let . Then the following three statements are equivalent.
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 , .

By definition 7.11,  is a null sequence" means

 (7.30)

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.

7.31   Theorem (Decomposition theorem.) Let be a convergent complex sequence. Then we can write

where is a null sequence, and is a constant sequence. If , then .

Proof: .

7.32   Theorem (Sum theorems for convergent sequences.) Let and let be convergent complex sequences. Say and . Then , and are convergent and

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 .

7.33   Exercise. Prove the last statement in theorem 7.32; i.e., show that if then for all .

7.34   Theorem (Product theorem for convergent sequences.) Let be convergent complex sequences. Suppose and . Then is convergent and .

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,

7.35   Theorem (Uniqueness theorem for convergent sequences.) Let
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., .

7.36   Definition (Limit of a sequence.) Let be a convergent sequence. Then the unique complex number such that is denoted by or .

7.37   Remark. It follows from the sum and product theorems that if and are convergent sequences, then

and

and

7.38   Warning. We have only defined when is a convergent sequence. Hence is ungrammatical and should not be written down. (We showed in theorem 7.7 that diverges.) However, it is a standard usage to say  does not exist'' or  does not exist'' to mean that the sequence has no limit. Hence we may say  does not exist''.

7.39   Theorem. Let be a complex sequence. Then is convergent if and only if both and are convergent. Moreover,
 (7.40)

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 .

7.41   Definition (Bounded sequence.) A sequence in is bounded, if there is a disc such that for all ; i.e., is bounded if there is a number such that
 (7.42)

Any number satisfying condition (7.42) is called a bound for .

7.43   Examples. is bounded since for all . The sequence is not bounded since the statement for all contradicts the Archimedean property of . Every constant sequence is bounded. In fact, is a bound for .

7.44   Exercise (Null-times-bounded theorem.) A Show that if is a null sequence in , and is a bounded sequence in then is a null sequence.

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

7.45   Theorem (Reverse triangle inequality.) Let , then

Proof: By the triangle inequality.

Hence,

7.46   Lemma. Let be a convergent sequence that is not a null sequence; i.e., where . Suppose for all . Then is a bounded sequence.

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.

7.47   Theorem (Reciprocal theorem for convergent sequences.) Let be a complex sequence. Suppose that where , and that for all . Then is convergent, and .

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., .

7.48   Exercise (Quotient theorem for convergent sequences.) The following statement isn't quite true. Supply the missing hypotheses and prove the corrected statement.

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

7.49   Exercise. A
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.

7.50   Exercise. A Let be a divergent complex sequence. Show that if , then is divergent.

7.51   Example. Let be defined by
 (7.52)

Then
 (7.53)

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

7.54   Example. Let be the sequence
 (7.55)

Then for all ,

Since , I know and so

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.

7.56   Exercise. A Let be a sequence of non-negative numbers and suppose where . Prove that . (NOTE: The case follows from the root theorem for null sequences.

7.57   Exercise. A Investigate the sequences below, and find their limits if they have any.
a)
b)
c)
d)
e)

7.58   Exercise. A Show that the sum of two bounded sequences is a bounded sequence.

7.59   Theorem (Convergent sequences are bounded.) Let be a convergent complex sequence. Then is bounded.

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 .

7.60   Exercise. A Complete the proof of theorem 7.59; i.e., show that if is a convergent complex sequence, then is bounded.

7.61   Example. It follows from the fact that convergent sequences are bounded, that is not a convergent sequence.

7.62   Exercise. Give an example of a bounded sequence that is not convergent.

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