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

Rules for sequences

Lemma

Let ${\left(x_n\right)}_{n\in \mathbb{N}}$ and ${\left(y_n\right)}_{n\in \mathbb{N}}$ be

convergent sequences. Then the following statements hold.

1. The sequence ${(x_n+y_n)}_{n\in \mathbb{N}}$ is convergent, and

   $\underset{n\to \mathrm{\infty }}{lim}(x_n+y_n)=\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)+\left(\underset{n\to \mathrm{\infty }}{lim}{y}_n\right)$

   holds.

2. The sequence ${(x_n\cdot y_n)}_{n\in \mathbb{N}}$ is convergent, and

   $\underset{n\to \mathrm{\infty }}{lim}(x_n\cdot y_n)=\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)\cdot \left(\underset{n\to \mathrm{\infty }}{lim}{y}_n\right)$

   holds.

3. For $c\in ℝ$, we have

   $\underset{n\to \mathrm{\infty }}{lim}c{x}_n=c\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right).$

4. Suppose that $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x\ne 0$ and $x_n\ne 0$ for all $n\in \mathbb{N}$. Then ${\left(\frac{1}{x_n}\right)}_{n\in \mathbb{N}}$ is also convergent, and

   $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{x_n}=\frac{1}{x}$

   holds.

5. Suppose that $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x\ne 0$ and that $x_n\ne 0$ for all $n\in \mathbb{N}$. Then ${\left(\frac{y_n}{x_n}\right)}_{n\in \mathbb{N}}$ is also convergent, and

   $\underset{n\to \mathrm{\infty }}{lim}\frac{y_n}{x_n}=\frac{\underset{n\to \mathrm{\infty }}{lim}{y}_n}{x}$

   holds.

Proof

(1). Denote the limits of the sequences by $x$ and $y$, respectively. Let $ϵ>0$ be given. Due to the convergence of the first sequence, there exists for

${ϵ}^{\prime }=\frac{ϵ}{2}$

some $n_0$ such that for all $n\ge n_0$ the estimate

$|x_n-x|\le {ϵ}^{\prime }$

holds. In the same way there exists due to the convergence of the second sequence for ${ϵ}^{\prime }=\frac{ϵ}{2}$ some ${n_0}^{\prime }$ such that for all $n\ge {n_0}^{\prime }$ the estimate

$|y_n-y|\le {ϵ}^{\prime }$

holds. Set

$N=max(n_0,{n_0}^{\prime }).$

Then for all $n\ge N$ the estimate

$\begin{array}{cc}|x_n+y_n-(x+y)|& =|x_n+y_n-x-y|\\ & =|x_n-x+y_n-y|\\ & \le |x_n-x|+|y_n-y|\\ & \le {ϵ}^{\prime }+{ϵ}^{\prime }\\ & =ϵ\end{array}$

holds.

(2). Let $ϵ>0$ be given. The convergent sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ is bounded, due to Lemma 5. , and therefore there exists a $D>0$ such that $|x_n|\le D$ for all $n\in \mathbb{N}$. Set $x:=\underset{n\to \mathrm{\infty }}{lim}{x}_n$ and $y:=\underset{n\to \mathrm{\infty }}{lim}{y}_n$. We put $C:=max(D,|y|)$. Because of the convergence, there are natural numbers $N_1$ and $N_2$ such that

$|x_n-x|\le \frac{ϵ}{2C}\text{ for }n\ge N_1\text{ and }|y_n-y|\le \frac{ϵ}{2C}\text{ for }n\ge N_2.$

These estimates hold also for all $n\ge N:=max(N_1,N_2)$. For these numbers, the estimates

$\begin{array}{cc}|x_n{y}_n-xy|& =|x_n{y}_n-x_{n}y+x_{n}y-xy|\\ & \le |x_n{y}_n-x_{n}y|+|x_{n}y-xy|\\ & =|x_n||y_n-y|+|y||x_n-x|\\ & \le C\frac{ϵ}{2C}+C\frac{ϵ}{2C}\\ & =ϵ\end{array}$

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

$x_n=\frac{-5n^3+6n^2-n+8}{11n^3+7n^2+3n-1},$

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

$x_n=\frac{-5n^3+6n^2-n+8}{11n^3+7n^2+3n-1}=\frac{(-5n^3+6n^2-n+8)\frac{1}{n^3}}{(11n^3+7n^2+3n-1)\frac{1}{n^3}}=\frac{-5+\frac{6}{n}-\frac{1}{n^2}+\frac{8}{n^3}}{11+\frac{7}{n}+\frac{3}{n^2}-\frac{1}{n^3}}.$

In this form, the numerator and the denominator converges, and the limits are $-5$ and $11$ respectively. Therefore, the sequence converges to $-\frac{5}{11}$.

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 ${\left(x_n\right)}_{n\in \mathbb{N}}$ (for the computation of $\sqrt{5}$ say) starting with a rational number gives a sequence of rational numbers. If we consider this sequence inside the real numbers $ℝ$ where $\sqrt{5}$ 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.

Augustin Louis Cauchy (1789-1857)

Definition

A real sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ is called a Cauchy sequence, if the following condition holds.

For every $ϵ>0$, there exists an $n_0\in \mathbb{N}$, such that for all $n,m\ge n_0$, the estimate

$|x_n-x_m|\le ϵ$

holds.

Lemma

Every convergent sequence is a

Cauchy sequence.

Proof

Let ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a convergent sequence with limit $x$. Let $ϵ>0$ be given. We apply the convergence property for $ϵ/2$. Therefore there exists an $n_0$ with

$|x_n-x|\le ϵ/2\text{ for all }n\ge n_0.$

For arbitrary $n,m\ge n_0$ we then have due to the triangle inequality

$|x_n-x_m|\le |x_n-x|+|x-x_m|\le ϵ/2+ϵ/2=ϵ.$

  Hence we have a Cauchy sequence.

$◻$

Definition

Let ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a real sequence. For any strictly increasing mapping $\mathbb{N}\to \mathbb{N},i\mapsto n_i$, the sequence

$i\mapsto x_{n_i}$

is called a subsequence of the sequence.

Definition

A real sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ is called increasing, if $x_{n+1}\ge x_n$ holds for all $n\in \mathbb{N}$, and strictly increasing, if $x_{n+1}>x_n$ holds for all $n\in \mathbb{N}$.

A sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ is called decreasing if $x_{n+1}\le x_n$ holds for all $n\in \mathbb{N}$, and strictly decreasing, if $x_{n+1}<x_n$ holds for all

$n\in \mathbb{N}$.

Lemma

Let ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a real increasing sequence which is bounded from above. Then ${\left(x_n\right)}_{n\in \mathbb{N}}$ is a

Cauchy sequence.

Proof

Let $b\in ℝ$ denote a bound from above, so that $x_n\le b$ holds for all $x_n$. We assume that ${\left(x_n\right)}_{n\in \mathbb{N}}$ is not a Cauchy sequence. Then there exists some $ϵ>0$ such that for every $n_0$, there exist indices $n>m\ge n_0$ fulfilling $x_n-x_m\ge ϵ$. Because of the monotonicity, there is also for every $n_0$ an $n>n_0$ with $x_n-x_{n_0}\ge ϵ$. Hence, we can define inductively an increasing sequence of natural numbers satisfying

$n_1>n_0\text{ such that }{x}_{n_1}-x_{n_0}\ge ϵ,$

$n_2>n_1\text{ such that }{x}_{n_2}-x_{n_1}\ge ϵ,$

and so on. On the other hand, there exists, due to the axiom of Archimedes, some $k\in \mathbb{N}$ with

$kϵ>b-x_{n_0}.$

The sum of the first $k$ differences of the subsequence $x_{n_j}$, $j\in \mathbb{N}$, is

$\begin{array}{cc}{x}_{n_k}-x_{n_0}& =(x_{n_k}-x_{n_{k-1}})+(x_{n_{k-1}}-x_{n_{k-2}})+\cdots +(x_{n_2}-x_{n_1})+(x_{n_1}-x_{n_0})\\ & \ge kϵ\\ & >b-x_{n_0}.\end{array}$

This implies $x_{n_k}>b$, contradicting the condition that $b$ 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 $\sqrt{5}$. One might say that a nonconvergent Cauchy-sequence addresses a gap. Within the real numbers, all these gaps are filled.

Definition

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

converges.

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

Axiom

The real numbers $ℝ$ 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 ${ℝ}_1$ and ${ℝ}_2$, both fulfilling these axioms, then there exists a bijective mapping from ${ℝ}_1$ to ${ℝ}_2$ 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 $\mathbb{Q}$ and constructs the real numbers as the set of all Cauchy sequences in $\mathbb{Q}$ with a suitable identification.

Implications of completeness

Corollary

A bounded and monotone sequence

in $ℝ$ converges.

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 $ℝ$.

$◻$

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

$a.a_{-1}{a}_{-2}{a}_{-3}\dots$

with $a\in \mathbb{N}$ (we restrict to nonnegative numbers) and $a_{-n}\in \{0,\dots ,9\}$ is just the sequence of rational numbers

$x_0:=a,x_1:=a+a_{-1}\cdot \frac{1}{10},x_2:=a+a_{-1}\cdot \frac{1}{10}+a_{-2}\cdot {\left(\frac{1}{10}\right)}^2,\mathrm{e}\mathrm{t}\mathrm{c}\mathrm{.}$

This sequence is increasing. It is also bounded, e.g. by $a+1$, so that it defines a Cauchy sequence and thus a real number.

Nested intervals

Definition

A sequence of closed intervals

$I_n=[a_n,b_n],n\in \mathbb{N},$

in $ℝ$ is called (a sequence of) nested intervals, if $I_{n+1}\subseteq I_n$ holds for all $n\in \mathbb{N}$, and if the sequence of the lengths of the intervals, i.e.

${(b_n-a_n)}_{n\in \mathbb{N}},$

converges

to $0$.

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 $I_n$, $n\in \mathbb{N}$, is a sequence of nested intervals in $ℝ$. Then the intersection

$\bigcap _{n\in \mathbb{N}}{I}_n$

contains exactly one point

$x\in ℝ$. Nested intervals determine a unique real number.

Proof

See Exercise 8.20 .

$◻$

Theorem

For every nonnegative real number $c\in {ℝ}_{\ge 0}$ and every $k\in {\mathbb{N}}_{+}$ there exists a unique nonnegative real number $x$ fulfilling

$x^k=c.$

Proof

We define recursively nested intervals $[a_n,b_n]$. We set

$a_0=0$

and we take for $b_0$ an arbitrary real number with $b_0^k\ge c$. Suppose that the interval bounds are defined up to index $n$, the intervals fulfil the containment condition and that

$a_n^k\le c\le b_n^k$

holds. We set

$a_{n+1}:=\{\begin{array}{c}\frac{a_n+b_n}{2},\text{ if }{\left(\frac{a_n+b_n}{2}\right)}^k\le c,\\ a_n\text{ else},\end{array}$

and

$b_{n+1}:=\{\begin{array}{c}\frac{a_n+b_n}{2},\text{ if }{\left(\frac{a_n+b_n}{2}\right)}^k>c,\\ b_n\text{ else}.\end{array}$

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 $x$ denote the real number defined by this nested intervals according to Theorem 8.12 . Because of Exercise 8.21 , we have

$x=\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{b}_n.$

Due to Lemma 8.1   (2), we get

$x^k=\underset{n\to \mathrm{\infty }}{lim}{a}_n^k=\underset{n\to \mathrm{\infty }}{lim}{b}_n^k.$

Because of the construction of the interval bounds and due to Lemma 7.12 , this is $\le c$ but also $\ge c$, hence $x^k=c$.

$◻$

This uniquely determined number is denoted by $\sqrt[k]{c}$ or by $c^{1/k}$.

Tending to infinity

Definition

A real sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ is said to tend to $+\mathrm{\infty }$, if for every $s\in ℝ$, there exists some $N\in \mathbb{N}$, such that

$x_n\ge s\text{ holds for all }n\ge N.$

The sequence is said to tend to $-\mathrm{\infty }$, if for every $s\in ℝ$. there exists some $N\in \mathbb{N}$. such that

$x_n\le s\text{ holds for all }n\ge N.$

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