Serlo: EN: Monotony criterion

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In this chapter, we will prove that monotonous and bounded functions converge. So if you have a bounded sequence, which in addition is monotonous, you instantly know that it converges. There is no need ot make complicated bounds within the $ϵ$-definition. You do not even have to know what the limit is!

This especially turns out to be useful for recursive sequences. For those sequences, there is often no explicit form available which makes it hard to guess, what a limit may be or what the difference of a sequence element to a potential limit is.

Convergence of monotonous and bounded sequences

Datei:Konvergenz beschränkter monotoner Folgen - Quatematik.webm Mathe für Nicht-Freaks: Vorlage:Satz

Nested intervals

There is a useful implication for nested intervals.

Recap: A sequence of nested intervals is a sequence $(I_n{)}_{n\in \mathbb{N}}=([a_n,b_n]{)}_{n\in \mathbb{N}}$ with the following properties:

1. All intervals are subsets of their precursors:

2. For all $ϵ>0$ there is an interval $I_N$ smaller than $ϵ$:

In that case, there is always exactly one real number being included in all intervals. This statement can be shown using the monotony criterion:

We take a look at the two sequences $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$, i.e. upper and lower bounds of the intervals.

• Since $[a_1,b_1]\supseteq [a_2,b_2]\supseteq [a_3,b_3]\supseteq \dots \supseteq [a_n,b_n]$ there is

Vorlage:Einrücken So $(a_n)$ is monotonously increasing and $(b_n)$ monotonously decreasing.

• Since $a_n\le b_1$ and $b_n\ge a_1$ , the sequence $(a_n)$ is bounded from above by $b_1$ and conversely, $(b_n)$ is bounded from below by $a_1$ .

The monotony criterion hence implies that $(a_n)$ and $(b_n)$ converge. The limit of both is even equal and exactly this one number being in all intervals:

Since $(I_n{)}_{n\in \mathbb{N}}$ is a sequence of nested intervals, there is Vorlage:Einrücken

As $I_{n+1}\subseteq I_n$ we also have $b_n-a_n<ϵ$ for all $n\ge N$.

So the difference of $(a_n)$ and $(b_n)$ converges to0: Vorlage:Einrücken Using the limit theorems, we hence get that $(a_n)$ and $(b_n)$ have the same limit. This limit is Vorlage:Einrücken So $a_n\le a\le b_n$ , i.e. $a\in [a_n,b_n]$ and $a$ is included in all intervals. All other real numbers $a+ϵ$ or $a-ϵ$ with $ϵ>0$ can not be inside all intervals, since they either exceed $a$ as an upper bound or as a lower bound. More precisely, for $a+ϵ$, there is a $b_n<a+ϵ$ and $a+ϵ$ is not inside the interval $[a_n,b_n]$. An analogous problem appears for $a_n>a-ϵ$. So there is exactly one real number $a$ included in all intervals.

We recap those considerations in a theorem:

Application: Euler's number

Computing Euler's number e by nested intervals

We consider the sequence of intervals $(I_n{)}_{n\in \mathbb{N}}=([a_n,b_n]{)}_{n\in \mathbb{N}}$ with $a_n={(1+\frac{1}{n})}^n$ and $b_n={(1+\frac{1}{n})}^{n+1}$.

These can be shown to be nested intervals, which we will do in the following. In that case, both $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ converge. At first, Vorlage:Einrücken So $(I_n)=([a_n,b_n])$ is well-defined.

For a sequence of nested intervals, we need to establish two properties. At first, for all $n\in \mathbb{N}$: $I_{n+1}\subseteq I_n$ (intervals are included in each other). This is done in two steps:

• The lower bounds $(a_n)$are monotonously increasing, i.e. $a_{n+1}\ge a_n$. This is done by showing $\frac{a_{n+1}}{a_n}\ge 1$ , using Bernoulli's inequality:

• The upper bounds $(b_n)$ are monotonously increasing, i.e. $b_{n+1}\le b_n$.

Mathe für Nicht-Freaks: Vorlage:Aufgabe Since $(a_n)$is monotonously increasing and $(b_n)$ decreasing, the intervals are included in each other, i.e. $I_{n+1}=[a_{n+1},b_{n+1}]\subseteq [a_n,b_n]=I_n$. So we have established the first property for nested intervals.

The second property is that interval sizes go to 0: Vorlage:Einrücken This works by bounding $b_N-a_N$ from above. Vorlage:Einrücken But now, $a_N\le b_N\le b_1={(1+\frac{1}{1})}^2=4$, so Vorlage:Einrücken There is $\frac{4}{N}<ϵ\iff N>\frac{4}{ϵ}$. If we are given any $ϵ>0$ and choose a corresponding $N\in \mathbb{N}$ with $N>\frac{4}{ϵ}$, then Vorlage:Einrücken So the second property is also established and $(I_n)$ is indeed a sequence of nested intervals. The number included in all intervals is called Euler's number $e$. The sequence of nested intervals can be used for making estimates, what $e$ is. For instance, $e\in [a_{10},b_{10}]=[2.59374246,2.85311671]$. For greater $n$, one would get even more digits, e.g. $e\approx 2,718281828459045$.

With the theorem above, there is Vorlage:Einrücken In the series chapter, we will show that $e=\exp (1)$ with $\exp$ denoting the exponential series.

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