# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 8

*Rules for sequences*

## Lemma

Let and be

convergent sequences. Then the following statements hold.- The sequence is convergent, and
holds.

- The sequence is convergent, and
holds.

- For
,
we have
- Suppose that
and
for all
.
Then is also convergent, and
holds.

- Suppose that
and that
for all
.
Then is also convergent, and
holds.

### Proof

(1). Denote the limits of the sequences by and , respectively. Let be given. Due to the convergence of the first sequence, there exists for

some such that for all the estimate

holds. In the same way there exists due to the convergence of the second sequence for some such that for all the estimate

holds. Set

Then for all the estimate

holds.

(2). Let
be given. The convergent sequence is
bounded,
due to
Lemma 5.
,
and therefore there exists a
such that
for all
.
Set
and .
We put
.
Because of the convergence, there are natural numbers
and
such that

These estimates hold also for all . For these numbers, the estimates

hold.

For the other parts, see Exercise 8.1 , Exercise 8.2 and Exercise 8.3 .

We give a typical application of this statement.

## Example

We consider the sequence given by

and want to know whether it converges and if so, what the limit is. We can not use Lemma 8.1 immediately, as neither the numerator nor the denominator converges. However, we can use the following trick. We write

In this form, the numerator and the denominator converges, and the limits are and respectively. Therefore, the sequence converges to .

*Cauchy sequences*

A problem with the concept of convergence is that in its very formulation already the limit is used, which in many cases is not known in advance. The Babylonian method to construct a sequence (for the computation of say) starting with a rational number gives a sequence of rational numbers. If we consider this sequence inside the real numbers where exists, this sequence converges. However, within the rational numbers, this sequence does not converge. We would like to formulate within the rational numbers alone the property that the members of the sequence are getting closer and closer without referring to a limit point. This purpose fulfills the notion of Cauchy sequence.

## Definition

A
real sequence
is called a *Cauchy sequence*, if the following condition holds.

For every , there exists an , such that for all , the estimate

## Lemma

Every convergent sequence is a Cauchy sequence.

### Proof

Let be a convergent sequence with limit . Let be given. We apply the convergence property for . Therefore there exists an with

For arbitrary we then have due to the triangle inequality

## Definition

Let be a real sequence. For any strictly increasing mapping , the sequence

*subsequence*of the sequence.

## Definition

A
real sequence
is called *increasing*, if
holds for all
,
and *strictly increasing*, if
holds for all
.

A sequence is called *decreasing* if
holds for all
,
and *strictly decreasing*, if
holds for all

## Lemma

Let be a real increasing sequence which is bounded from above. Then is a Cauchy sequence.

### Proof

Let denote a bound from above, so that holds for all . We assume that is not a Cauchy sequence. Then there exists some such that for every , there exist indices fulfilling . Because of the monotonicity, there is also for every an with . Hence, we can define inductively an increasing sequence of natural numbers satisfying

and so on. On the other hand, there exists, due to the axiom of Archimedes, some with

The sum of the first differences of the subsequence , , is

This implies , contradicting the condition that is an upper bound for the sequence.

*The completeness of the real numbers*

Within the rational numbers there are Cauchy sequences which do not converge, like the Heron sequence for the computation of . One might say that a nonconvergent Cauchy-sequence addresses a gap. Within the real numbers, all these gaps are filled.

## Definition

An
ordered field
is called *complete* or *completely ordered*, if every
Cauchy sequence
in

The rational numbers are not complete. We require the completeness for the real numbers as the final axiom.

## Axiom

The real numbers form a complete

Archimedian ordered field.Now we have gathered together all axioms of the real numbers: the field axioms, the ordering axiom and the completeness axiom. These properties determine the real numbers uniquely, i.e., if there are two models and , both fulfilling these axioms, then there exists a bijective mapping from to which respects all mathematical structures (such a thing is called an "isomorphism“).

The existence of the real numbers is not trivial. We will take the naive viewpoint that the idea of a "continuous number line“ gives the existence. In a strict set based construction, one starts with and constructs the real numbers as the set of all Cauchy sequences in with a suitable identification.

*Implications of completeness*

## Corollary

### Proof

Due to the condition, the sequence is increasing and bounded from above or decreasing and bounded from below. Because of Lemma 8.7 , we have a Cauchy sequence which converges in .

This statement is also the reason that any decimal expansion defines a real number. An (infinite) decimal expansion

with (we restrict to nonnegative numbers) and is just the sequence of rational numbers

This sequence is increasing. It is also bounded, e.g. by , so that it defines a Cauchy sequence and thus a real number.

*Nested intervals*

## Definition

A sequence of closed intervals

in is called
(a sequence of)
*nested intervals*, if
holds for all
,
and if the sequence of the lengths of the intervals, i.e.

In a family of nested intervals, the length of the intervals are a decreasing null sequence. However, we do not require a certain velocity of the convergence. An *interval bisection* is a special kind of nested intervals, where the next interval is either the lower or the upper half of the preceding interval.

## Theorem

Suppose that , , is a sequence of nested intervals in . Then the intersection

contains exactly one point . Nested intervals determine a unique real number.

### Proof

## Theorem

For every nonnegative real number and every there exists a unique nonnegative real number fulfilling

### Proof

We define recursively nested intervals . We set

and we take for an arbitrary real number with . Suppose that the interval bounds are defined up to index , the intervals fulfil the containment condition and that

holds. We set

and

Hence one bound remains and one bound is replaced by the arithmetic mean of the bounds of the previous interval. In particular, the stated properties hold for all intervals and we have a sequence of nested intervals. Let denote the real number defined by this nested intervals according to Theorem 8.12 . Because of Exercise 8.21 , we have

Due to Lemma 8.1 , we get

Because of the construction of the interval bounds and due to Lemma 7.12 , this is but also , hence .

This uniquely determined number is denoted by or by .

*Tending to infinity*

## Definition

A real
sequence
is said to *tend* to , if for every
,
there exists some
,
such that

The sequence is said to *tend* to , if for every
.
there exists some
.
such that

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