5.7. Monotone and Bounded Sequences

Theorem:  For every sequence $(a_n)$ in $ℝ$ we have:

Proof:  We assume $(a_n)$ to be increasing. Due to boundedness we find an $s\in ℝ$ such that

Thus the set of all sequence members $\{a_n|n\in {\mathbb{N}}^{\ast }\}$ is a nonempty, bounded above subset of $ℝ$. According to the completeness axiom - and this is why this theorem is subject to the reals - this set has a least upper bound, its supremum. We set  $g≔\sup \{a_n|n\in {\mathbb{N}}^{\ast }\}$  and prove:

Let $\epsilon >0$ be given. As g is the least upper bound of $\{a_n|n\in {\mathbb{N}}^{\ast }\}$ it is impossible for $g-\epsilon$ to be an upper bound at all. Thus there is at least one sequence member above $g-\epsilon$, i.e. we have an $n_0\in {\mathbb{N}}^{\ast }$ such that $a_{n_0}>g-\epsilon$. Considering that g is a normal upper bound as well and that $(a_n)$ is increasing we get for all $n\ge n_0$ :

i.e.  $a_n\in ]g-\epsilon ,g+\epsilon [$. This proves our assertion according to [5.4.2].

Proposition:  With $q\in ℝ$ we have:

Proof:  

1. ►

Due to [5.5.6] it is sufficient to prove $|q^n|=|q{|}^n\to 0$, so we may assume $0\le q<1$. We multiply this inequality by $q^n$ and get $0\le q^{n+1}\le q^n<1\text{ for all }n\in {\mathbb{N}}^{\ast }$. Thus $(q^n)$ is decreasing and bounded, consequently convergent, say to g. For the remainder task, to prove $g=0$, we consider in addition the sequence $(q^{n+1})$, also converging to g. As a little trick we calculate the limit of $(q^{n+1})$ anew using the third limit theorem: $\begin{array}{l}{q}^{n+1}\to g\hfill \\ q^{n+1}=q\cdot q^n\to q\cdot g\hfill \end{array}$ From the uniqueness of g we know that  $g=q\cdot g\iff g\cdot (1-q)=0$  holds. So we have $g=0$ as $q\ne 1$ is our premise.

2. ►

If $|q|>1$, say $|q|=1+x$  with an $x>0$ we deduce from the Bernoulli inequality: $|q^n|=|q{|}^n={(1+x)}^n\ge 1+nx$. $(1+nx)$ is unbounded and so is $(q^n)$ and thus divergent.

Proposition:  For $r,s\in {\mathbb{N}}^{\ast }$ and  ${\displaystyle q≔\frac{r}{s}}$  the following holds:

Proof:  

1. ►

As the root operator is monotone we have for all $n\in {\mathbb{N}}^{\ast }$: ${\displaystyle \begin{array}{ll}\hfill & 1\le n\le n+1\hfill \\ \Rightarrow \hfill & 1\le \sqrt[s]{n}\le \sqrt[s]{n+1}\hfill \\ \Rightarrow \hfill & 1\ge \frac{1}{\sqrt[s]{n}}\ge \frac{1}{\sqrt[s]{n+1}}\ge 0\text{.}\hfill \end{array}}$ From that we see that ${\displaystyle (\frac{1}{\sqrt[s]{n}})}$ is decreasing and bounded thus convergent with a limit, say g. Again we calculate g using the third limit theorem: ${\displaystyle \frac{1}{n}={\left(\frac{1}{\sqrt[s]{n}}\right)}^s\to g^s\Rightarrow g^s=0\Rightarrow g=0\text{.}}$

2. ►

With the third limit theorem the result is immediate from 1.: ${\displaystyle \frac{1}{n^q}=\frac{1}{n^{\frac{r}{s}}}={\left(\frac{1}{\sqrt[s]{n}}\right)}^r\to 0^r=0\text{.}}$

Example:  

Proof:  Both assertions are best proved simultaneously. We need three steps to show that each sequence is monotone and bounded.

1. ►	${\displaystyle ({(1+\frac{1}{n})}^n)}$ is increasing, because from the Bernoulli inequality we get for $n\in {\mathbb{N}}^{\ast }$ ${\displaystyle {(1+\frac{1}{n+1})}^{n+1}\cdot {(1-\frac{1}{n+1})}^{n+1}={(1-\frac{1}{{(n+1)}^2})}^{n+1}\ge 1-\frac{1}{n+1}}$ and thus have: ${\displaystyle {(1+\frac{1}{n+1})}^{n+1}\ge \frac{1}{{(1-\frac{1}{n+1})}^n}={\left(\frac{n+1}{n}\right)}^n={(1+\frac{1}{n})}^n}$.
2. ►	${\displaystyle ({(1+\frac{1}{n})}^{n+1})}$ is decreasing: Again we employ the Bernoulli inequality for $n\in {\mathbb{N}}^{\ast }$ to get ${\displaystyle {\left(\frac{{(n+1)}^2}{{(n+1)}^2-1}\right)}^{n+1}={(1+\frac{1}{{(n+1)}^2-1})}^{n+1}\ge {(1+\frac{1}{{(n+1)}^2})}^{n+1}\ge 1+\frac{1}{n+1}\text{.}}$ This allows the following estimate: ${\displaystyle \begin{array}{ll}{(1+\frac{1}{n+1})}^{n+2}\hfill & \le {\left(\frac{{(n+1)}^2}{{(n+1)}^2-1}\right)}^{n+1}\cdot {(1+\frac{1}{n+1})}^{n+1}\hfill \\ \hfill & ={(\frac{{(n+1)}^2}{(n+2)n}\cdot \frac{n+2}{n+1})}^{n+1}\hfill \\ \hfill & ={\left(\frac{n+1}{n}\right)}^{n+1}\hfill \\ \hfill & ={(1+\frac{1}{n})}^{n+1}\text{.}\hfill \end{array}}$
3. ►	${\displaystyle ({(1+\frac{1}{n})}^n)}$ and  ${\displaystyle ({(1+\frac{1}{n})}^{n+1})}$ are bounded. As ${(1+\frac{1}{1})}^1=2$ and ${(1+\frac{1}{1})}^2=4$ the just proven monotony implies: ${\displaystyle 2\le {(1+\frac{1}{n})}^n\le {(1+\frac{1}{n})}^{n+1}\le 4}$.

Example:  

Proof:  As ${\displaystyle {(1+\frac{1}{n})}^n\le 4}$ the following assertions hold for each $n\ge 4$:

Therefor ${(\sqrt[n]{n})}_{n\ge 4}$ is decreasing and, because of $1\le \sqrt[n]{n}$, bounded as well thus convergent to a number $g\ge 1$. The case $g>1$ however is impossible: The ordering of $ℝ$ is archimedic, so for every $x>0$ there is an $n\in {\mathbb{N}}^{\ge 4}$ such that ${\displaystyle 1<(n-1)\cdot \frac{x^2}{2}}$. The generalized binomial theorem [5.2.5] now allows the following estimate for this n:

So we have $\sqrt[n]{n}<1+x$ and thus $1+x\ne g$, considering that  $\sqrt[n]{n}\ge g$.

Example:  

• Let ${\displaystyle a_1≔2\wedge a_{n+1}≔\frac{2a_n}{a_n+1}}$  generate the recursive sequence $(a_n)$. We have:  $a_n\to 1$.

Proof:  First we show by induction
 

which in fact means that $(a_n)$ is already bounded.

►   $1\le 2\le 2\text{, i.e. }1\le a_1\le 2\text{.}$

►   From $1\le a_n\le 2$ we can estimate

and thus have  $1\le a_{n+1}\le 2$.

Furthermore $(a_n)$ proves to be decreasing: Because $a_n\ge 1$ implies $a_n(a_n-1)\ge 0$, the assertion holds due to the following equivalence:

All in all $(a_n)$ has a limit, say $g\in [1,2]$. But g is the limit of $(a_{n+1})$ as well, so we can calculate g using the limit theorems:

As g is unique, g satisfies the equation  ${\displaystyle g=\frac{2g}{g+1}}$ . So we have (note that $g\in [1,2]$):

Proposition (Babylonian Square Root Algorithm):  Let $a>0$ be arbitrary. For each initial value $a_1>0$ the recursive sequence $(a_n)$ given by  ${\displaystyle a_{n+1}≔\frac{1}{2}(a_n+\frac{a}{a_n})}$  is convergent, more precise:

Proof:  A simple inductive consideration shows that $a_n>0$ for all n. As squares are always positive we can argue as follows
 

to get the estimate $a_{n+1}\ge \sqrt{a}$. Employing this result we get for $n\ge 2$ :

Thus $(a_{n+1})$ is decreasing and due to  $\sqrt{a}\le a_{n+1}\le a_2$  bounded as well and thus in fact convergent with a limit $g\ge \sqrt{a}>0$. As $a_n\to g$ is also true our 'standard trick' is applicable and from

we calculate g like this:  ${\displaystyle g=\frac{1}{2}(g+\frac{a}{g})\iff 2g=g+\frac{a}{g}\iff g^2=a\iff g=\sqrt{a}}$.