# 5.3. Monotone Sequences, Bounded Sequences

This section deals with two properties that apply to sequences.

First we study sequences that represent a certain 'running direction' which means that e.g. the sequence members are steadily increasing in value. Second we study sequences whose 'running range' is bounded to the left and to the right, i.e. sequences in an interval.

Definition:  A sequence $\left({a}_{n}\right)$ is called

 increasing,  if [5.3.1] decreasing,  if [5.3.2]

In either case we call $\left({a}_{n}\right)$ a monotone sequence. If we replace in [5.3.1]  $\le$  by  $<$$\left({a}_{n}\right)$  is called strictly increasing. In an analogue way the terms strictly decreasing and strictly monotone are set.

Consider:

• $\left({a}_{n}\right)$ is not increasing, iff there is an  $n\in {ℕ}^{\ast }$ such that  ${a}_{n}>{a}_{n+1}\text{.}$

• $\left({a}_{n}\right)$ is not decreasing, iff there is an  $n\in {ℕ}^{\ast }$ such that  ${a}_{n}<{a}_{n+1}\text{.}$

 Example:   $\left(3n+1\right)$ is increasing, because for all  $n\in {ℕ}^{\ast }$ the following equivalence is valid and the final assertion is true: $\begin{array}{ll}\hfill & 3n+1\le 3\left(n+1\right)+1\hfill \\ ⇔\hfill & 3n+1\le 3n+4\hfill \\ ⇔\hfill & 1\le 4\hfill \end{array}$   $\left(\frac{1}{n}\right)$ is decreasing. Again we equivalently reduce the assertion to a true one: $\begin{array}{ll}\hfill & \frac{1}{n}\ge \frac{1}{n+1}\hfill \\ ⇔\hfill & n+1\ge n\hfill \\ ⇔\hfill & 1\ge 0\hfill \end{array}$   $\left({\left(-1\right)}^{n}\right)$ is neither decreasing nor increasing, as: $\begin{array}{l}{\left(-1\right)}^{1}=-1<1={\left(-1\right)}^{2}\hfill \\ {\left(-1\right)}^{2}=1>-1={\left(-1\right)}^{3}\hfill \end{array}$   Each constant sequence $\left(c\right)$ is decreasing and increasing as well, because both of the inequalities hold:

The subsequent proposition gathers some properties of monotony. For simplicity we only note them for increasing sequences, but of cause the corresponding results for decreasing ones are also valid.

Proposition:

 $\left({a}_{n}\right)$ is increasing$⇔{a}_{n+1}-{a}_{n}\ge 0$  for all n. [5.3.3]
 $\left({a}_{n}\right)$ is increasing$⇒{a}_{1}\le {a}_{n}$  for all n. [5.3.4]
 $\left({a}_{n}\right)$ is increasing$⇒\left({a}_{n+k}\right)$  is increasing for all k. [5.3.5]
 $\left({a}_{n}\right)$ is increasing$⇔\left(-{a}_{n}\right)$  is decreasing. [5.3.6]

Proof:  For 1. and 4. we only have to rearrange the condition ${a}_{n}\le {a}_{n+1}\text{.}$ 3. is trivial.
 2. ► We prove by induction: ${1\in A}:{a}_{1}\le {a}_{1}\text{.}$  ${n\in A⇒n+1\in A}:$Let ${a}_{1}\le {a}_{n}\text{.}$ By monotony we have ${a}_{n}\le {a}_{n+1}\text{,}$ but then ${a}_{1}\le {a}_{n+1}$ is also true.

The last example showed that constant sequences are monotone in both directions. By this property they are uniquely determined.

Proposition:

 $\left({a}_{n}\right)$ is decreasing and increasing$⇔\left({a}_{n}\right)$ is constant. [5.3.7]

Proof:
"$⇒$":  From [5.3.4] we have ${a}_{n}\le {a}_{1}\le {a}_{n}\text{.}$ Thus ${a}_{n}={a}_{1}$ for all n, which means that $\left({a}_{n}\right)$ is constant.

"$⇐$":  This is exactly the last example.

The cooperation between monotony and calculating with sequences is only poor. Take e.g. the increasing sequences $\left({n}^{2}\right)$ and $\left(4n-4\right)$. Their difference is no longer monotone in which direction ever.

$\left({n}^{2}\right)-\left(4n-4\right)=\left(1,0,1,\dots \right)$

Addition and, under certain circumstances, multiplication are a bit better in that way. Again we confine ourselves on increasing sequences.

Proposition:  If $\left({a}_{n}\right)$ and $\left({b}_{n}\right)$ are increasing sequences then

 $\left({a}_{n}\right)+\left({b}_{n}\right)$  is increasing [5.3.8]
 $\left({a}_{n}\right)\cdot \left({b}_{n}\right)$  is increasing provided  ${a}_{n},{b}_{n}\ge 0$ [5.3.9]

Proof:
 1. ► According to our premise two inequalities hold for every n: $\begin{array}{l}{a}_{n}\le {a}_{n+1}\hfill \\ {b}_{n}\le {b}_{n+1}\hfill \end{array}$ Adding them on the left and on the right results in our assertion:  ${a}_{n}+{b}_{n}\le {a}_{n+1}+{b}_{n+1}\text{.}$ 2. ► We can proceed as before, as inequalities in ${ℝ}^{\ge 0}$ could be multiplied the same way.

The compatiblity of monotony with calculating is thus not very particular. We get remarkably better results with the next property, the boundedness.

Definition:  A sequence $\left({a}_{n}\right)$ is called bounded, if there are two numbers ("bounds")  $u,v\in ℝ$ such that

 [5.3.10]

$\left({a}_{n}\right)$ is called boundless, if there are no bounds for $\left({a}_{n}\right)$ in at least one direction.

Consider:

• $\left({a}_{n}\right)$ is bounded if and only if there is an interval $\left[u,v\right]$ such that $\left({a}_{n}\right)$ is a sequence in $\left[u,v\right]$.

• $\left({a}_{n}\right)$ is boundless if and only if at least one sequence member is missing in each interval.

When testing sequences for boundedness we often use a another version of [5.3.10] using the absolute value. The advantage is that we need only one bound instead of two.

Proposition:

 $\left({a}_{n}\right)$ is bounded$⇔$there is an $s\in {ℝ}^{\ge 0}$ such that [5.3.11]

Proof:
"$⇒$":  We just extend the equation $u\le {a}_{n}\le v$ on both ends:

$-\mathrm{max}\left\{|u|,|v|\right\}=\mathrm{min}\left\{-|u|,-|v|\right\}\le -|u|\le u\le {a}_{n}\le v\le |v|\le \mathrm{max}\left\{|u|,|v|\right\}$.

Setting  $s≔\mathrm{max}\left\{|u|,|v|\right\}$,  we have $-s\le {a}_{n}\le s$ and thus: .

"$⇐$":  $|{a}_{n}|\le s$ means: $-s\le {a}_{n}\le s$.

With the following examples we often use a certain technique, called 'estimating', which means that a given term could be increased (or decreased) by some legal methods.

The following tricks are standard for increasing a term:

• Increasing an addend within a sum, e.g.:

$n+1\le n+3\le 2n+3$

• Subtracting less than before within a difference, e.g.:

$7-{n}^{2}\le 7-n\le 7-1$

• Increasing a factor within a product of positive factors, e.g.:

$4n\le 10n\le 10\left(n+3\right)$

• Increasing the numerator or decreasing the denominator within a fraction of positive factors, e.g.:

$\frac{3n+1}{2n+3}\le \frac{3n+1}{2n}\le \frac{3n+n}{2n}$

There is no fault in estimating if it is done properly, even if terms change, as we do not manipulate = but $\le$ instead! However the options for transforming inequalities are really abundant compared to those for equalities. Sometimes this affects the clarity and we cannot always avoid the impression that we choose our manipulations arbitrarily, but always they are designed for a definite target.

Example:

 $\left({\left(-1\right)}^{n}\right)$ is bounded, because:  $|{\left(-1\right)}^{n}|=1\le 1$ for all n. [5.3.12]

 $\left(\frac{1}{n}\right)$ is bounded, because:  $|\frac{1}{n}|=\frac{1}{n}\le \frac{1}{1}=1$ for all n. [5.3.13]

 Every constant sequence $\left(c\right)$ is bounded, because if we set  $s≔|c|$ then $|c|\le s$ holds for any sequence member. [5.3.14]

 $\left(\frac{2{n}^{3}+n-2}{{n}^{3}+1}\right)$ is bounded, because for every n we have: $|\frac{2{n}^{3}+n-2}{{n}^{3}+1}|=\frac{2{n}^{3}+n-2}{{n}^{3}+1}\le \frac{2{n}^{3}+n-2}{{n}^{3}}\le \frac{2{n}^{3}+n-2}{{n}^{3}}\le \frac{2{n}^{3}+{n}^{3}}{{n}^{3}}=3$

As already mentioned the hereditary properties of basic arithmetics are more advanced with boundedness than with monotony.

Proposition:  If $\left({a}_{n}\right)$ and $\left({b}_{n}\right)$ are bounded sequences, so are

 $\left({a}_{n}\right)+\left({b}_{n}\right)$ [5.3.15]
 $\left({a}_{n}\right)-\left({b}_{n}\right)$ [5.3.16]
 $\left({a}_{n}\right)\cdot \left({b}_{n}\right)$ [5.3.17]

Proof:  In all cases we use boundedness in absolute value according to [5.3.11]. From the premise we have numbers  ${s}_{1},{s}_{2}\in {ℝ}^{\ge 0}$, such that

From that the triangle inequality allows for all $n\in {ℕ}^{\ast }$ the following estimates:

 1. ► $|{a}_{n}+{b}_{n}|\le |{a}_{n}|+|{b}_{n}|\le {s}_{1}+{s}_{2}$
 2. ► $|{a}_{n}-{b}_{n}|=|{a}_{n}+\left(-{b}_{n}\right)|\le |{a}_{n}|+|-{b}_{n}|=|{a}_{n}|+|{b}_{n}|\le {s}_{1}+{s}_{2}$
 3. ► $|{a}_{n}\cdot {b}_{n}|=|{a}_{n}|\cdot |{b}_{n}|\le {s}_{1}\cdot {s}_{2}$

So in each case all the sequence members are estimated in absolute value by a real number. This means boundedness due to [5.3.11].

A first result from [5.3.13] and [5.3.14] is:

 For each $c\in ℝ$ the sequence  $\left(\frac{c}{n}\right)=\left(c\right)\cdot \left(\frac{1}{n}\right)$  is bounded. [5.3.18]

The next proposition shows that boundedness is transferable to 'smaller' whereas unboundedness is transferable to 'bigger' sequences.

Proposition:  Let $\left({a}_{n}\right)$ and $\left({b}_{n}\right)$ be sequences such that

,

then the following holds:

 $\left({b}_{n}\right)$ bounded$⇒\left({a}_{n}\right)$ bounded  $\left({a}_{n}\right)$ unbounded$⇒\left({b}_{n}\right)$ unbounded [5.3.19]

Proof:
 1. ► We have a number $s\in {ℝ}^{\ge 0}$ such that . But in fact this is already the assertion.
 2. ► A bounded sequence $\left({b}_{n}\right)$ would yield a bounded sequence $\left({a}_{n}\right)$ as shown in 1.   Contradiction!

As we have got one bounded reference sequence in [5.3.18] we can use 1. to produce more of them. From the estimate  $|\frac{c}{{n}^{k}}|=\frac{|c|}{{n}^{k}}\le \frac{|c|}{n}=|\frac{c}{n}|$  e.g. we get:

 For each the sequence  $\left(\frac{c}{{n}^{k}}\right)$  is bounded. [5.3.20]

If we are to benefit from 2. in a similar way we need to establish an unbounded reference sequence first.

Proposition:

 $\left(n\right)$ is unbounded. [5.3.21]

Proof:  We refer to the completeness axiom

Each non-empty bounded subset of $ℝ$ has a greatest lower bound, its infimum, and a least upper bound, its supremum

and proceed indirectly:  If $\left(n\right)$ is bounded then ${ℕ}^{\ast }$ will be a non-empty bounded subset of $ℝ$. Let $s≔\mathrm{sup}{ℕ}^{\ast }$  be its supremum.

As s is the least upper bound of ${ℕ}^{\ast }$ the number $s-1$ can't be an upper bound. So there has to be an $n\in {ℕ}^{\ast }$ such that $s-1. From that we have for the positive natural number $n+1$ :

$s

But that means s is no upper bound for ${ℕ}^{\ast }$.   Contradiction!

Consider:

• [5.3.21] states again that the ordering of $ℝ$ is archimedic. So the proof is just a copy from chapter 3.

Taking  $\left(n\right)$ as a reference sequence now we get further examples of unbounded sequences.

Example:

 $\left(\frac{3{n}^{2}}{n+1}\right)$ is unbounded,
 because:

 $\left(cn\right)$ is unbounded for each $c\ne 0$: [5.3.22]
 $\left(cn\right)$ is bounded implies  $\left(\frac{1}{c}\right)\cdot \left(cn\right)=\left(n\right)$ is bounded.   Contradiction!

 $\left(c{n}^{k}\right)$ is unbounded for each , [5.3.23]
 because:

 $\left(\sqrt[m]{{n}^{k}}\right)$ is unbounded for $k,m\in {ℕ}^{\ast }$: [5.3.24]
 If $\left(\sqrt[m]{{n}^{k}}\right)$ is bounded so will be $\underset{m\text{-times}}{\underbrace{\left(\sqrt[m]{{n}^{k}}\right)\cdot \dots \cdot \left(\sqrt[m]{{n}^{k}}\right)}}=\left({n}^{k}\right)$ .   Contradiction!

We studied two properties in this part, monotony and boundedness. There is however no relationship between them:

We will find bounded and unbounded sequences among the monotone ones, and on the other hand non-monotone sequences could be bounded and unbounded as well.

 5.2. 5.4.