# Examples and properties of sequences – Serlo

## Examples

### Constant sequence

A sequence is called constant, if all of its elements are equal. An example is:

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }=(2,\,2,\,2,\,2,\,2,\,\ldots )}$

With ${\displaystyle c\in \mathbb {R} }$ , the general form of a constant sequence is ${\displaystyle a_{n}:=c}$ for all ${\displaystyle n\in \mathbb {N} }$.

### Arithmetic sequences

Arithmetic sequences have constant differences between two elements. For instance, the sequence of odd numbers is arithmetic, since any two neighbouring elements have difference ${\displaystyle 2}$:

${\displaystyle \left(a_{n}\right)=(1,\,3,\,5,\,7,\,9,\,11,\,\ldots )}$

A further example is sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ with ${\displaystyle a_{n}:=n}$ for all ${\displaystyle n\in \mathbb {N} }$:

${\displaystyle \left(a_{n}\right)=(1,\,2,\,3,\,4,\,5,\,\ldots )}$

Question: What is the recursive rule for a general arithmetic sequence?

The first element ${\displaystyle a_{1}}$ may be imposed arbitrarily. The next one has any constant distance from ${\displaystyle a_{1}}$.Let's call this difference ${\displaystyle d}$. Then, ${\displaystyle a_{2}-a_{1}=d}$ and hence ${\displaystyle a_{2}=a_{1}+d}$ . Analogously, since ${\displaystyle a_{3}-a_{2}=d}$ there is ${\displaystyle a_{3}=a_{2}+d}$ and so on. So we have the recursive definition:

{\displaystyle {\begin{aligned}a_{1}&\in \mathbb {R} \ {\text{arbitrary}}\\a_{n+1}&:=a_{n}+d\ {\text{for all }}n\in \mathbb {N} \\\end{aligned}}}

Question: What is the explicit rule for a general arithmetic sequence?

We already know the recursive rule ${\displaystyle a_{n+1}=a_{n}+d}$ for all ${\displaystyle n\in \mathbb {N} }$, where ${\displaystyle a_{1}\in \mathbb {R} }$ Is given. That means ${\displaystyle a_{2}=a_{1}+d}$ and ${\displaystyle a_{3}=a_{2}+d=(a_{1}+d)+d=a_{1}+2\cdot d}$. Analogously ${\displaystyle a_{4}=a_{1}+3\cdot d}$. So we get an explicit rule for all ${\displaystyle n\in \mathbb {N} }$:

${\displaystyle a_{n}=a_{1}+(n-1)\cdot d}$

### Geometric sequence

For the geometric sequence we have a constant ratio between two subsequent elements. No element is allowed to be 0, since else we would get into trouble dividing by 0 when computing ratios. An example for a geometric sequence is ${\displaystyle a_{n}=2^{n}}$ where the constant ratio is given by ${\displaystyle 2}$:

${\displaystyle \left(a_{n}\right)=(2,\,4,\,8,\,16,\,32,\,64,\,\ldots )}$

Question: What is the explicit rule for a general geometric sequence?

The first element ${\displaystyle a_{1}\neq 0}$ for a geometric series is arbitrary. The second element must have a fixed ratio to ${\displaystyle a_{1}}$. Let us call this ratio ${\displaystyle q\neq 0}$. This means ${\displaystyle {\tfrac {a_{2}}{a_{1}}}=q}$, or equivalently ${\displaystyle a_{2}=a_{1}\cdot q}$ . Now, as the ratio is always fixed, all other elements are given at this point. We have ${\displaystyle {\tfrac {a_{3}}{a_{2}}}=q}$ or equivalently ${\displaystyle a_{3}=a_{2}\cdot q}$ and so on. Hence, the recursive definition reads:

{\displaystyle {\begin{aligned}a_{1}&\in \mathbb {R} \setminus \{0\}{\text{ arbitrary}}\\a_{n+1}&:=a_{n}\cdot q\end{aligned}}}

Question: What is the explicit rule for a general geometric sequence?

Let us take the recursive rule ${\displaystyle a_{n+1}=a_{n}\cdot q}$ and try to find an explicit formulation. There is ${\displaystyle a_{2}=a_{1}\cdot q}$ and ${\displaystyle a_{3}=a_{2}\cdot q=a_{1}\cdot q^{2}}$. Analogously,${\displaystyle a_{4}=a_{1}\cdot q^{3}}$. This suggests:

${\displaystyle a_{n}=a_{1}\cdot q^{n-1}}$

which can easily be checked by induction.

### Harmonic sequence

The sequence ${\displaystyle a_{n}={\tfrac {1}{n}}}$ is called harmonic sequence. The name originates from the fact that intervals in music theory can be defined by it: It describes octaves, fifths and thirds. Mathematicians like it, because it is one of the smallest sequences where the sum over all elements gives infinity (we will com to this later, when concerning series).The first elements of this sequence are:

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }=\left(1,\,{\tfrac {1}{2}},\,{\tfrac {1}{3}},\,{\tfrac {1}{4}},\,{\tfrac {1}{5}},\,\ldots \right)}$

The similar sequence ${\displaystyle a_{n}=(-1)^{n}\cdot {\tfrac {1}{n}}}$ or ${\displaystyle b_{n}=(-1)^{n+1}\cdot {\tfrac {1}{n}}}$ is called alternating harmonic sequence . Explicitly, the first elements are

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }=\left(-1,\,{\tfrac {1}{2}},\,-{\tfrac {1}{3}},\,{\tfrac {1}{4}},\,-{\tfrac {1}{5}},\,\ldots \right)}$

or

${\displaystyle \left(b_{n}\right)_{n\in \mathbb {N} }=\left(1,\,-{\tfrac {1}{2}},\,{\tfrac {1}{3}},\,-{\tfrac {1}{4}},\,{\tfrac {1}{5}},\,\ldots \right)}$

For ${\displaystyle k\in \mathbb {N} }$ the generalized harmonic sequence is given by

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }=\left({\frac {1}{n^{k}}}\right)_{n\in \mathbb {N} }=\left(1,\,{\tfrac {1}{2^{k}}},\,{\tfrac {1}{3^{k}}},\,{\tfrac {1}{4^{k}}},\,{\tfrac {1}{5^{k}}},\,\ldots \right)}$

### Alternating sequences

An alternating sequence is characterized yb a change of sign between any two sequence elements. The term "alternating" just means that the presign is "constantly changing". For instance, the sequence ${\displaystyle a_{n}=(-1)^{n}}$ alternates between the values ${\displaystyle 1}$ and ${\displaystyle -1}$, so we have an alternating sequence . A further example is ${\displaystyle a_{n}=(-1)^{n+1}\cdot n}$ with ${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }=(1,\,-2,\,3,\,-4,\,5,\,-6,\,\ldots )}$.

More generalyl, any alternating sequence can be put into the form:

1. ${\displaystyle a_{n}=(-1)^{n}\cdot b_{n}}$
2. ${\displaystyle a_{n}=(-1)^{n+1}\cdot b_{n}}$

Here, ${\displaystyle b_{n}=|a_{n}|}$ is a sequence of non-negative numbers.

Question: Which alternating sequences can be brought into which of the above two forms?

This is answered by taking a look at the first sequence element. If the first element with index 0 is positive(${\displaystyle a_{0}>0}$), then we have the form ${\displaystyle a_{n}=(-1)^{n}\cdot b_{n}}$. Conversely, if ${\displaystyle a_{0}<0}$ the we have ${\displaystyle a_{n}=(-1)^{n+1}\cdot b_{n}}$.

If the first index is 1 (the sequence starts with ${\displaystyle a_{1}}$, it is exactly the other way round.

If the first element is ${\displaystyle 0}$ , one needs to check the subsequent elements until one is strictly greater or smaller than 0. Only the sequence which is constantly 0 takes both forms at once.

### The exponential sequence

To-Do:

Sketch the first sequence elements

A common example for a sequence is the exponential sequence. For instance, it appears when you invest money and get a return (e.g. in terms of interests). For instance, imagine you invest one "money" of any currency (dollar or pound or whatever) at a bank with a rate of interest of ${\displaystyle 100\,\%}$ (oh my gosh, what a bank!) Then, after one year, you will get paid back ${\displaystyle 1+1=2}$ "moneys" (2 units of money). Is there a way to get more money, if you are allowed to spread ${\displaystyle 100\,\%}$ of interests over a year? You could ask the bank to pay you an interest rate of ${\displaystyle 50\,\%}$, but twice a year. Then, after one year where multiplying your money twice, you get back

${\displaystyle \left(1+{\frac {1}{2}}\right)\cdot \left(1+{\frac {1}{2}}\right)=\left(1+{\frac {1}{2}}\right)^{2}=1{.}5^{2}=2.25}$

units of money. Those are ${\displaystyle 0.25}$ units more! If you split the interest rate in even smaller parts, you get even more: for 4 times ${\displaystyle 25\,\%}$, you get back ${\displaystyle 2.44}$ units of money.

Question: Why do you get back ${\displaystyle 2.44}$ units if you split the interest rate of ${\displaystyle 100\,\%}$ in 4 equal parts of ${\displaystyle 25\,\%}$?

After the first 3 months, the amount of money you have will be ${\displaystyle 1+{\tfrac {1}{4}}=1.25}$ . Another 3 months later, it will have increased by the same factor and you get ${\displaystyle (1+{\tfrac {1}{4}})\cdot (1+{\tfrac {1}{4}})=1.25^{2}=1{.}56}$ (approximately). Doing this two more times, you get

${\displaystyle \left(1+{\frac {1}{4}}\right)\cdot \left(1+{\frac {1}{4}}\right)\cdot \left(1+{\frac {1}{4}}\right)\cdot \left(1+{\frac {1}{4}}\right)=\left(1+{\frac {1}{4}}\right)^{4}=1{.}25^{4}=2.44}$

units of money.

In general, if you split the ${\displaystyle 100\,\%}$ into ${\displaystyle n}$ parts, then in the end you will receive

${\displaystyle \left(1+{\frac {1}{n}}\right)\cdot \ldots \cdot \left(1+{\frac {1}{n}}\right)=\left(1+{\frac {1}{n}}\right)^{n}}$

units of money. This can be interpreted as a sequence in ${\displaystyle n}$: the sequence ${\displaystyle \left(\left(1+{\frac {1}{n}}\right)^{n}\right)_{n\in \mathbb {N} }}$ is also called "exponential sequence". Now, can you make infinitely much money within one year, just by splitting the ${\displaystyle 100\,\%}$ infinitely often? The answer is: unfortunately no. There is an upper bound to how much money you can make that way. It is called Euler's number ${\displaystyle e=2.71828\dots }$. So you do not get above ${\displaystyle 2.72}$ units of money. The proof why this sequence ${\displaystyle \left(\left(1+{\frac {1}{n}}\right)^{n}\right)_{n\in \mathbb {N} }}$ converges to ${\displaystyle e}$ can be found within the article "monotony criterion".

### Sequence of Fibonacci numbers

The Fibonacci sequence has been discovered already in 1202 by Leonardo Fibonacci . He investigated populations of rabbits, which approximately spread by the following rule:

1. At first, there is one pair of rabbits being able to mate.
2. A pair of rabbits being able to mate gives birth to another pair of rabbits every month.
3. A newborn pair takes one month where it cannot give birth to rabbits until it is finally able to do so.
4. We consider an ideal world, with no rabbits leaving, no predators, infinitely much food and no rabbits dying.

Question: How many rabbits will be there in each month?

Let ${\displaystyle f_{n}}$ be the number of rabbit pairs being able to mate within month ${\displaystyle n}$. What is then ${\displaystyle f_{n+1}}$ ? Within this month, an additional amount of rabbit pairs ${\displaystyle x}$ will become able to mate. So there is ${\displaystyle f_{n+1}=f_{n}+x}$. But now, the newly mating rabbit pairs in month ${\displaystyle (n+1)}$ are exactly those born in month ${\displaystyle (n-1)}$ (because rabbits are born, then they take a pause of one turn and after 2 months, they start to mate). I.e. ${\displaystyle x=f_{n-1}}$. Plugged into the above equation, we get:

${\displaystyle f_{n+1}=f_{n}+f_{n-1}}$

Or after an index shift:

${\displaystyle f_{n+2}=f_{n+1}+f_{n}}$

In the beginning, there is ${\displaystyle f_{1}=1}$ and ${\displaystyle f_{2}=1}$ (rabbits born in month 1 only start to mate in month 3). So we have a recursively defined sequence, where each ${\displaystyle f_{n+2}}$ can be determined if ${\displaystyle f_{n+1}}$ and ${\displaystyle f_{n}}$ are known. This sequence is also called Fibonacci sequence. The shorthand definition reads:

{\displaystyle {\begin{aligned}f_{0}&=0\\f_{1}&=1\\f_{n+2}&=f_{n+1}+f_{n}\end{aligned}}}

Its first elements are ${\displaystyle 1,1,2,3,5,8,13,21,34,55,89,...}$

### Mixed sequences

Mixed sequences are a generalization of alternating sequence. We merge two sequences ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ into a new one which consists alternately of elements of ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$, i.e.

${\displaystyle (a_{n})_{n\in \mathbb {N} }=(b_{1},c_{1},b_{2},c_{2},b_{3},c_{3},\ldots )}$

An element with odd index, e.g. ${\displaystyle a_{2k-1}}$ for ${\displaystyle k\in \mathbb {N} }$ will be equal to ${\displaystyle b_{k}}$ from the sequence ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ . And an element with even index, e.g. ${\displaystyle a_{2k}}$ for ${\displaystyle k\in \mathbb {N} }$ agrees with ${\displaystyle c_{k}}$ from the sequence ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ .

In order to get a general formula for ${\displaystyle a_{n}}$ with ${\displaystyle n\in \mathbb {N} }$ , we just have to distinguish the cases of even and odd ${\displaystyle n}$ . For odd ${\displaystyle n}$ , there is ${\displaystyle k={\tfrac {n+1}{2}}}$ or equivalently ${\displaystyle n=2k-1}$ , so we get ${\displaystyle a_{n}=a_{2k-1}=b_{k}=b_{\frac {n+1}{2}}}$. For an even ${\displaystyle n}$ there is ${\displaystyle a_{n}=c_{\frac {n}{2}}}$. Together, we have

${\displaystyle a_{n}={\begin{cases}b_{\frac {n+1}{2}}&{\text{ for }}n{\text{ odd}}\\c_{\frac {n}{2}}&{\text{ for }}n{\text{ even}}\end{cases}}}$

${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is then said to be a mixed sequence composed by ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$.

Example (mixed sequence)

The alternating sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ given by ${\displaystyle a_{n}=(-1)^{n}}$ (${\displaystyle n\in \mathbb {N} }$) is a merger of the sequences ${\displaystyle b_{n}=-1}$ and ${\displaystyle c_{n}=1}$ , since for ${\displaystyle n\in \mathbb {N} }$ there is

${\displaystyle a_{n}={\begin{cases}-1&{\text{ for }}n{\text{ odd}}\\1&{\text{ for }}n{\text{ even}}\end{cases}}}$

If you encounter an exercise where a sequence is defined with a distinction between even and odd ${\displaystyle n}$ , then it is just a mixed sequence. Basically, any sequence can be interpreted as a mixed sequence: Any ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is composed by ${\displaystyle (a_{2n-1})_{n\in \mathbb {N} }}$ and ${\displaystyle (a_{2n})_{n\in \mathbb {N} }}$. For instance ${\displaystyle (1,2,3,\ldots )}$can be seen as a merger of ${\displaystyle (1,3,5,\ldots )}$ and ${\displaystyle (2,4,6,\ldots )}$.

Question: Are there sequences which remain invariant, if they are merged with itself?

Yes, but only constant sequences.

For ${\displaystyle c\in \mathbb {R} }$ , if we mix the constant sequence ${\displaystyle b_{n}=c}$ with itself ${\displaystyle c_{n}=c}$, we again get the constant sequence ${\displaystyle a_{n}=c}$.

Conversely, if ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is a mixture by ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$, then

${\displaystyle a_{n}={\begin{cases}a_{\frac {n+1}{2}}&{\text{ for }}n{\text{ odd}}\\a_{\frac {n}{2}}&{\text{ for }}n{\text{ even}}\end{cases}}}$

For any element ${\displaystyle a_{n}}$ we can apply this formula. Since for ${\displaystyle n\geq 2}$ there is ${\displaystyle {\tfrac {n+1}{2}} or ${\displaystyle {\tfrac {n}{2}} , we get smaller and smaller indices until we reach ${\displaystyle a_{1}}$ . So the sequence has to be constant with only value ${\displaystyle a_{1}}$.

## Properties and important terms

### Bounded sequence

a sequence is called bounded from above, if there is an upper bound, i.e. a large number, which is never exceeded by any sequence element. This number bounds the sequence from above. The mathematical definition of this expression reads:

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }{\text{ is bounded from above }}:\iff \exists S\in \mathbb {R} :\ \forall n\in \mathbb {N} :\ a_{n}\leq S}$

Or explicitly:

${\displaystyle {\begin{array}{c}\left(a_{n}\right)_{n\in \mathbb {N} }{\text{ is bounded from above }}\\[1em]\underbrace {:\iff } _{\mathrm {...\ is\ defined\ by\ ...} }\\[2em]\underbrace {\exists S\in \mathbb {R} :} _{\mathrm {there\ is\ an\ upper\ bound\ } S}\ \underbrace {\forall n\in \mathbb {N} } _{{\text{for all indices }}n}\ \underbrace {a_{n}\leq S} _{\mathrm {element\ with\ index\ } n\mathrm {\ is\ smaller\ or\ equal\ } S}\end{array}}}$

Analogously, a sequence is bounded from below if and only if there is a lower bound, i.e. a number for which all sequence elements are greater than this number. The mathematical definition hence reads:

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }{\text{ is bounded from below }}\ :\iff \ \exists s\in \mathbb {R} :\ \forall n\in \mathbb {N} :\ a_{n}\geq s}$

Question: What does it mean that a sequence in not bounded from below or above?

We can formally invert the above statements:

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }{\text{ is not bounded from above }}\ :\Longleftrightarrow \ \forall S\in \mathbb {R} :\ \exists n\in \mathbb {N} :\ a_{n}>S}$

So a sequence is unbounded from above, if for any ${\displaystyle S}$ there is some sequence element bigger than ${\displaystyle S}$. That means, parts of the sequence grow infinitely big. Conversely,:

${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }{\text{ is not bounded from above }}\ :\Longleftrightarrow \ \forall s\in \mathbb {R} :\ \exists n\in \mathbb {N} :\ a_{n}

So a sequence is unbounded from below, if for any ${\displaystyle s}$ there is a sequence element smaller than ${\displaystyle s}$. Intuitively, a part of the sequence gets infinitely small.

If a sequence is both bounded from above and from below, we just call it bounded. So we have the following definitions:

upper bound
An upper bound is a number, which is greater than any sequence element. So ${\displaystyle S}$ is an upper bound of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$, if and only if ${\displaystyle a_{n}\leq S}$ for all ${\displaystyle n\in \mathbb {N} }$.
sequence bounded from above
A sequence is bounded from above, if it has any upper bound.
lower bound
A lower bound is a number, which is smaller than any sequence element. So ${\displaystyle S}$ is a lower bound of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$, if and only if ${\displaystyle a_{n}\geq S}$ for all ${\displaystyle n\in \mathbb {N} }$.
sequence bounded from below
A sequence is bounded from below, if it has any lower bound.
bounded sequence
A sequence is bounded, if it has both an upper and a lower bound.

Hint

An upper bound does not need to be the smallest (best) upper bound. And a lower bound does not need to be the greatest lower bound, either. For instance, if a sequence is bounded from above by ${\displaystyle 1}$ , then ${\displaystyle 2}$, ${\displaystyle 44}$, ${\displaystyle 123}$ and ${\displaystyle 502}$ are upper bounds, as well. Boundedness from above can be shown by just stating any upper bound.

There is an alternative definition of boundedness:

Theorem (alternative definition of boundedness)

A sequence is bounded if and only if there is a real number ${\displaystyle S'\geq 0}$ , such that for all elements ${\displaystyle a_{n}}$ there is ${\displaystyle |a_{n}|\leq S'}$ .

Proof (alternative definition of boundedness)

This is equivalent to the first definition: We can show that

${\displaystyle {\begin{array}{c}(\exists S\in \mathbb {R} :\ \forall n\in \mathbb {N} :\ a_{n}\leq S)\land \ (\exists s\in \mathbb {R} :\ \forall n\in \mathbb {N} :\ a_{n}\geq s)\\\Longleftrightarrow \\(\exists S'\in \mathbb {R} _{\geq 0}:\ \forall n\in \mathbb {N} :\ |a_{n}|\leq S')\end{array}}}$

or explicitly:

${\displaystyle {\begin{array}{c}\overbrace {\underbrace {(\exists S\in \mathbb {R} :\ \forall n\in \mathbb {N} :\ a_{n}\leq S)} _{a_{n}{\text{ is bounded from above}}}\ \land \underbrace {(\exists s\in \mathbb {R} :\ \forall n\in \mathbb {N} :\ a_{n}\geq s)} _{a_{n}{\text{ is bounded from below}}}} ^{\mathrm {first\ definition} }\\[2em]\underbrace {\Longleftrightarrow } _{\mathrm {...\ is\ equivalent\ to\ ...} }\\[2em]\underbrace {(\exists S'\in \mathbb {R} _{\geq 0}:\ \forall n\in \mathbb {N} :\ |a_{n}|\leq S')} _{\mathrm {alternative\ definition} }\end{array}}}$

So we have to prove equivalence, which means that we have to prove both directions of the above double arrow. Let us start with the first direction: We assume that the first definition is fulfilled by a sequence. Thus, there are real numbers ${\displaystyle S,s\in \mathbb {R} }$, such that for all sequence elements ${\displaystyle s\leq a_{n}\leq S}$ holds. Then for all sequence elements there is also ${\displaystyle |a_{n}|\leq \mathrm {max} \{|s|,\,|S|\}}$. So the existence of some ${\displaystyle S'}$ within the alternative definition is established (${\displaystyle S'}$ can be any positive, real number greater than or equal to ${\displaystyle \mathrm {max} \{s|,\,|S|\}}$).

And what about the other direction? Let now be ${\displaystyle S'\in \mathbb {R} _{>0}}$ given, with ${\displaystyle |a_{n}| for all sequence elements. Then the inequality ${\displaystyle -S'\leq a_{n}\leq S'}$ holds for all sequence elements. Thus ${\displaystyle -S'}$ represents a lower bound and ${\displaystyle S'}$ an upper bound for the sequence ${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }}$, so that the sequence is also bounded by the first definition.

Question: Which of the following sequences is bounded/unbounded from above/below?

1. constant sequence
2. arithmetic sequence
3. geometric sequence
4. harmonic sequence
5. alternating harmonic sequence
6. Fibonacci sequence

Lösung:

1. constant sequence: Bounded.
2. arithmetic sequence: For ${\displaystyle d>0}$ it is bounded from below by ${\displaystyle a_{0}}$ and unbounded from above. For ${\displaystyle d<0}$ it is bounded from above by ${\displaystyle a_{0}}$ and unbounded from below. For ${\displaystyle d=0}$ we have a constant sequence, which is bounded.
3. geometric sequence: For ${\displaystyle q=1}$ we have a constant sequence, which is again bounded. For ${\displaystyle q>1}$ and ${\displaystyle a_{0}>0}$ the sequence is bounded from below (by ${\displaystyle a_{0}}$) and unbounded from above. For ${\displaystyle q>1}$ and ${\displaystyle a_{0}<0}$ it is the other way round: The sequence is bounded from above by the negative number ${\displaystyle a_{0}}$but unbounded from below. For ${\displaystyle q<-1}$ the absolute values grow infinitely large and the sign is alternating. So the sequence is unbounded both from above and from below. If we choose ${\displaystyle 0 and ${\displaystyle a_{0}>0}$ , it is bounded. The upper bound is ${\displaystyle a_{0}}$ and the lower one ${\displaystyle 0}$. For ${\displaystyle 0 and ${\displaystyle a_{0}<0}$ the lower bound is ${\displaystyle a_{0}}$ and the upper bound is ${\displaystyle 0}$. So the sequence is bounded. And for ${\displaystyle -1 we also have boundedness by ${\displaystyle a_{1}}$ and ${\displaystyle a_{0}}$.
4. harmonic sequence: The harmonic sequence ${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }={\tfrac {1}{n}}}$is bounded from above by ${\displaystyle a_{1}=1}$ and from below by ${\displaystyle 0}$.
5. alternating harmonic sequence: This sequence is also bounded. An upper bound is ${\displaystyle a_{2}}$ and a lower bound is ${\displaystyle a_{1}}$ for ${\displaystyle a_{n}=(-1)^{n}\cdot {\tfrac {1}{n}}}$.
6. Fibonacci sequence: Bounded from below (by 0), unbounded from above.

### Monotone sequences

Sequences are also distinguished according to their growth behaviour: If the sequence elements of become larger and larger (i.e. each subsequent sequence member ${\displaystyle a_{n+1}}$ is larger than ${\displaystyle a_{n}}$), this sequence is called a strictly monotonically growing/increasing sequence. Similarly, a sequence with ever smaller sequence elements is called a strictly monotonously falling/decreasing sequence. If you want to allow a sequence to be constant between two sequence elements, the sequence is called only monotonously growing/increasing sequence or monotonously falling/decreasing sequence (without the "strictly"). Remember: "strictly monotonous" means as much as "getting bigger and bigger" or "getting smaller and smaller". In contrast, "monotonous", without the "strict", means as much as "getting bigger and bigger or remaining constant" or "getting smaller and smaller or remaining constant". The mathematical definition is:

Definition (monotone sequences)

For a real sequence ${\displaystyle \left(a_{n}\right)_{n\in \mathbb {N} }}$ we define:

{\displaystyle {\begin{aligned}&\left(a_{n}\right)_{n\in \mathbb {N} }{\text{ grows strictly monotonously }}&:\iff \ \forall n\in \mathbb {N} :\ a_{n+1}>a_{n}\\&\left(a_{n}\right)_{n\in \mathbb {N} }{\text{ grows monotonously }}&:\iff \ \forall n\in \mathbb {N} :\ a_{n+1}\geq a_{n}\\&\left(a_{n}\right)_{n\in \mathbb {N} }{\text{ falls strictly monotonously }}&:\iff \ \forall n\in \mathbb {N} :\ a_{n+1}

Question: Which of the following sequences are monotonously increasing/decreasing? For which ones, the monotony is strict?

1. constant sequence
2. arithmetic sequence
3. geometric sequence
4. harmonic sequence
5. alternating harmonic sequence
6. Fibonacci sequence

Lösung:

1. constant sequence: Both monotonously increasing and decreasing, but not strictly.
2. arithmetic sequence: For ${\displaystyle d>0}$ the sequence is strictly monotonously increasing. For ${\displaystyle d<0}$ the sequence is strictly monotonously decreasing. For ${\displaystyle d=0}$ we have a constant sequence.
3. geometric sequence: For ${\displaystyle q>1}$ and ${\displaystyle a_{0}>0}$ it is strictly monotonously increasing and for ${\displaystyle a_{0}<0}$ it is strictly monotonously decreasing. For ${\displaystyle 0 and ${\displaystyle a_{0}>0}$ it is strictly monotonously decreasing and for ${\displaystyle a_{0}<0}$ strictly monotonously increasing. For ${\displaystyle q<0}$ we have an alternating sequence, which is neither monotonously increasing, nor decreasing. For ${\displaystyle q=1}$ the sequence is constant.
4. harmonic sequence: strictly monotonously decreasing.
5. alternating harmonic sequence: alternating and hence neither monotonously increasing, nor decreasing.
6. Fibonacci sequence: monotonously increasing.

### Remark: convergent sequences

Sequences are also distinguished by whether they have a limit or not. Sequences which have a limit are called convergent and all other ones are divergent. This property requires a bit more explanation. We will come back to it later within the article "convergence and divergence".