# MATH 409 Lecture 7

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## Review

- Limits of Sequences
- Properties of Convergent sequences
- Limit is unique
- Convergent sequence is bounded
- Subsequence of convergent sequence converges to same ilmit
- Modifying a finite number of elements can't affect convergence or change its limit
- Rearranging elements of a sequence can't affect convergence or change its limit

- Limit Theorems

### Examples

**Theorem.**

*Proof.* for all since for all . As shown last time, as . Then as . By the Squeeze theorem, as .

**Theorem.**

*Proof.* The sequence is a subsequence of . Hence it converges to the same limit.

**Theoreom.**

*Proof.* Divide both numerator and denominator by :

Since as , it follows that

- as ,
- as
- as
- as

Finally, as .

## Monotone Sequences

A sequence is called **increasing** (or **nondecreasing**) if , or, to be precise, for all .

It is called **strictly increasing** if for all .

**Note:**These definitions are according to the textbooks. Other people prefer to use

**nondecreasing**for ≤ and

**increasing**for strictly increasing.

Similarly, is called **decreasing** (or **nonincreasing**) if for all .

It is called **strictly decreasing** if for all .

### Examples

- is strictly decreasing
- The sequence 1, 1, 2, 2, 3, 3, … is increasing, but not strictly increasing
- The sequence -1, 1, -1, 1, … is neither increasing nor decreasing
- A constant sequence , , … is both increasing and decreasing

### Theorems

**Theorem.** Any increasing sequence converges to a limit if it is bounded, and diverges to otherwise.

*Proof.* Let be an increasing sequence.

First consider the case when is bounded. In this case, the set of all elements occurring in the sequence is bounded. By the completeless axiom, any bounded set has a supremum; thus exists.

We claim that as . Take any . Then is not an upper bound of . Hence there exists such that . Since the sequence is increasing, it follows that for . At the same time, for all . Therefore for all , which proves the claim.

Now consider the case when is not bounded. Note that the set is bounded below (as is a lower bound). Hence is not bounded above. Then for any , there exists such that . It follows that for all . Thus diverges to .

**Theorem.** Any decreasing sequence converges to a limit if it is bounded, and diverges to otherwise.

*Proof.* Let be a decreasing sequence. Then the sequence is increasing since is equivalent to for all . By the previous theorem, either for some as , or else diverges to . In the former case, as . In the latter case, diverges to .

**Corollary.** Any monotone sequence converges to a limit if it is bounded, and diverges to infinity otherwise.

### Nested Intervals Property

A sequence of sets is called **nested** if . That is, for all .

**Theorem.** If is a nested sequence of nonempty closed bounded intervals, then the intersection is nonempty. Moreover, if lengths of the intervals satisfy as , then the intersetion consists of a single point.

*Proof.* Let for . Since the sequence is nested, it follows that the sequence is increasing while the sequence is decreasing. Besides, both sequences are bonded (since both are contained in the bounded interval ). Hence both are convergent: and as . Since for all , the Comparison Theorem implies that .

We claim that . Indeed, we have for all (by Comparison Theorem applied to and constant sequence ). Similarly, for all . Therefore is contained in the intersection.

On the other hand, if , then for some so that . Similarly, if then for some so that . This proves the claim.

Clearly, the length of cannot exceed for any . Therefore as implies that is a degenerate interval: .

A few remarks:

- The theorem may not hold if the intervals are open. Counterexample: , The intervals are nested and bounded, but their intersection is empty since as .
- The theorem may not hold if the intervals are not bounded. Counterexample: , . The intervals are nested and closed, but their intersection is empty since the sequence diverges to .
- This is another form of the completeness axiom.
- This theorem can be used to prove the existence of a decimal expansion

## Bolzano-Weierstrass Theorem

**Theorem.** Every bounded sequence of real numbers has a convergent subsequence.

*Proof.* Let be a bounded sequence of real numbers. We are going to build a nested sequence of intervals for , such that each contains infinitely many elements of and for all . The sequence is built inductively.

*Basis.* First we set to be any closed bounded interval that contains all elements of (such an interval exists because the sequence is bounded).

*Induction.* Now assume that for some the interval is already chosen and it contains infinitely many elements of . Then at least one of the subintervals and also contains infinitely many elements of . We set to be such an interval. By construction, and .

Since for all , it follows by induction that for all . As a consequence, as . By the Nested Intervals Property, the intersection of the intervals consists of a single number .

Next we are going to build a strictly increasing sequence of natural numbers such that for all . The sequence is built inductively:

*Basis.* First let .

*Induction.* Now assume that for some the number is already chosen. Since the interval contains infinitely many elements of the sequence , there exists such that . We set .

Now we claim that the subsequence of the sequence converges to . Indeed, for any , the points and both belong to the interval . Hence . Since as , it follows that as .