# MATH 409 Lecture 8

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## New Challenges

### Challenge 7

Build a sequence of real numbers such that every real number is a limit point of (a limit point f a sequence is, by definition, the limit of a convergent subsequence).

### Challenge 8

Let be the sequence of Fibbonacci numbers. Prove that

## Review

### Examples

**Theorem.** If , then as .

*Proof.* Since and , it follows that and for all . Hence the sequence is strictly decreasing and bounded. Therefore it converges to some .

Since for all , it follows that as . However, is a subsequence of , hence it converges to the same limit as . Thus , which implies that .

**Theorem.** If , then as .

*Proof.* Since , it follows that for all . Hence is strictly increasing. Then either diverges to or converges to a limit . In the latter case, we argue as above to obtain that . However, this contradicts with . Thus diverges to .

**Theorem.** If , then as .

Observe that by definition, is a unique positive number such that .

*Proof.* If , then for all , which implies that . Notice that and . Hence for all .

Similarly, in the case , we obtain that for all . In either case, the sequence is monotone and bounded. Therefore it converges to a limit . Then the sequence also converges to since it is a subsequence of . At the same time, , which implies that . Hence . However, the limit cannot be 0 since (the first element sets the lower bound of an increasing function). Thus the limit is 1.

**Theorem.** The sequence for is increasing and bounded, hence it is convergent.

**Note:**The limit is the number

*Proof.* First let us show that is increasing. For any ,

If , then similarly

Hence

To continue, we need the following lemma:

**Lemma.** If , then for all .

*Proof by induction.* *Basis.* For , we have .

*Induction.* Assume for some and all . Then

Thus the lemma holds by induction on .

*Remark.* According to the binomial formula,

Using this lemma, we obtain that

Thus the sequence is strictly increasing.

Now let us show that the sequence is bounded. Since is increasing, it is enough to show that it is bounded above. By the binomial formula,

Observe that for all because

Note there are terms in both the numerator and the denominator, and the denominator is obviously larger.

It follows that

Further observe that for all because . There are factors greater than 2 on the LHS, and factors equal to 2 on the RHS.

Therefore we obtain

So the sequence is bounded.

## Cauchy Sequences

A sequence of real numbers is called a **Cauchy sequence** if, for any , there exists such that whenever .

Note that the definition of cauchy sequences is ~~redundant with~~ equivalent to the theorem regarding convergent of sequences, only cauchy sequences do not need a limit to prove they converge.

**Theorem.** Any convergent sequence is Cauchy.

*Proof.* Let be a convergent sequence and be its limit. Then for any there exists such that whenever . Now for any natural numbers we have

Thus is a Cauchy sequence.

**Theorem.** Any Cauchy sequence is convergent.

*Proof.* (proved by Cauchy, but he called them *fundamental sequences*) Suppose is a Cauchy sequence. First let us show that the sequence is bounded. Since is Cauchy, there exists such that whenever . In particular, for all . Then

It follows that for any natural number , we have , where .

Now the Bolzano-Weierstrass theorem implies that has a subsequence converging to some . Given , there exists such that for all . Also, there exists such that whenever . Let . Then and . Therefore for any , we obtain

Thus the entire sequence converges to .

## Limit Points

A **limit point** of a sequence is the limit of any convergent subsequence of .

- A convergent sequence has only one limit point, its limit.
- Any bounded sequence has at least one limit point (by Bolzano-Wierstrass theorem)
- If a bounded sequence is not convergent, then it has at least two limit points.
- The sequence has 2 limit points, namely 1 and −1
- If all elements of a sequence belong to a closed interval , then all its limits belong to as well (by comparison theorem)
- The set of limit points of the sequence is the entire interval (key: π is irrational)
- If a sequence diverges to infinity, then it has no limit points
- If a sequence does not diverge to infinity, then it has a bounded subsequence and hence it has a limit point.