# 5.1. Sequences as Special Functions

In this part we study functions with a special domain, called sequences. We use them to mimic infinitely fine closeness, so we cannot take a finite domain. Our intention is a step by step approximation to which end a countable domain would be most useful. So we take the most simple of all countable and infinite sets, the positive intergers  ${ℕ}^{\ast }=\left\{1,2,3,4,\dots \right\}$.

Definition:   Let A be an arbitrary set. Any function

 $a:{ℕ}^{\ast }\to A$ [5.1.1]
is called a sequence in A.

We omit the adding "in A" if $A=ℝ$. Thus a sequence is a sequence in $ℝ$. Furthermore we often understand sequences in subsets of $ℝ$, e.g. in $ℤ$, in $ℚ$ or in ${ℝ}^{>0}$ as sequences in $ℝ$.

For historical reasons the common notions known from the functions are not used with sequences. We write

${a}_{n}$ instead of a(n) and
$\left({a}_{n}\right)$ for a.

Also in this context the value ${a}_{n}$ is called n-th sequence member of $\left({a}_{n}\right)$.

It is useful to name the constant sequences clearly: For every $c\in A$ the sequence

$\left(c\right)≔\left({c}_{{ℕ}^{\ast }}\right)$  is called the constant sequence c (in A).

As an example, the function

$a:{ℕ}^{\ast }\to ℝ$,  given by  $a\left(n\right)≔3{n}^{2}-1$ ,

is a sequence (in $ℝ$). The n-th sequence member calculates to  ${a}_{n}=3{n}^{2}-1$  and the sequence herself is noted as $\left({a}_{n}\right)=\left(3{n}^{2}-1\right)$.

Consider:

• Although choosing ${ℕ}^{\ast }$ as the domain of a sequence is certainly useful, it is nevertheless quite arbitrary. The crucial point is to have an infinite, countable domain with a first point and to that end every segment  ${ℤ}^{\ge k}$ of  $ℤ$ is suitable. Therefore we will occasionally call any function

$a:{ℤ}^{\ge k}\to A$

a sequence in A as well and take  ${\left({a}_{n}\right)}_{n\ge k}$ as notation.

There is no difference in calculating value tables for a sequence and for a common function. So with our example we get:

 n 1 2 3 4 5 6 7 ${a}_{n}$ 2 11 26 47 74 107 146

Normally the integers n are set in one after the other. If one fixes on this principle the header row is actually needless and we can take the advantage to write down the value table in a rather compact way: $\left({a}_{n}\right)=\left({a}_{1},{a}_{2},{a}_{3},\dots \right)$. Our example now looks like this:

$\left(3{n}^{2}-1\right)=\left(2,11,26,47,74,107,146,\dots \right)$

With the following examples value tables are produced on demand:

Example:
 $\left(n\left(1+{\left(-1\right)}^{n}\right)\right)$  is a sequence in $ℤ$.

 $\left(\frac{2n+1}{n+1}\right)$  is a sequence in $ℚ$.

 $\left(\sqrt{{n}^{2}+4}\right)$  is a sequence in $ℝ$.

 $\left(\left(n+{n}^{2},n-{n}^{2}\right)\right)$  is a sequence in ${ℤ}^{2}$.

 $\left(\left\{{i}^{3}|i\in {ℕ}^{\ast }\wedge {i}^{3}\le n\right\}\cup \left\{n\right\}\right)$  is a sequence in the power set $\mathcal{P}\left({ℕ}^{\ast }\right)$.

Calculating value tables is rarely a challenge, in most cases it is just boring. More fascinating but definitely more difficult is the reverse task:
Find a suitable sequence matching a given value table! The examples to follow show that we cannot expect a stiff scheme for solving such problems. We need to proceed intuitively using our creative mind.

 Example:   $\left({a}_{n}\right)=\left(4,7,10,13,16,19,\dots \right)$ This sequence is characterized by permanently adding  3, so it has a distinctive affinity to the multiples of  3:  $3,6,9,12,15,18,\dots$ Every value however lacks one unit, so it seems a sound idea to try   $\left({a}_{n}\right)=\left(3n+1\right)$.   $\left({a}_{n}\right)=\left(0,3,8,15,24,35,48,63,\dots \right)$ This sequence shows a similar situation: If we increase every value by 1 we gain the square numbers. Thus it should be rewarding to set   $\left({a}_{n}\right)=\left({n}^{2}-1\right)$.   $\left({a}_{n}\right)=\left(3,2,1,0,1,2,3,4,5,6,\dots \right)$ If there weren't the first four sequence members $\left({a}_{n}\right)=\left(n\right)$ would be a good solution. So we try to shift all members by 4 units to the right: $\left({a}_{n}\right)=\left(n-4\right)$. Then we have:   $-3,-2,-1,0,1,2,3,4,\dots$ Now all that's left is to adjust the algebraic sign of the first three members. We use the absolute value to achieve this:   $\left({a}_{n}\right)=\left(|n-4|\right)$.   $\left({a}_{n}\right)=\left(\frac{3}{2},\frac{5}{4},\frac{7}{8},\frac{9}{16},\frac{11}{32},\dots \right)$ It is easy to see that the numerators are the odd intergers starting with three and that every new denominator is twice the old one. Starting with two we get the following figures for the denominators:   $2,2\cdot 2,2\cdot 2\cdot 2,2\cdot 2\cdot 2\cdot 2,\dots$ . But in that way we have produced the powers of  2, so we set   $\left({a}_{n}\right)=\left(\frac{2n+1}{{2}^{n}}\right)$.

Here are some sequences to test our skills:

The use of  real-valued sequences, i.e. sequences in $ℝ$, is one basic concept of calculus. We can compute with them in the very same way we did with the ordinary real functions. The following propositon reveals if the real-valued sequences are compatible with the four rules.

Proposition:   Let $\left({a}_{n}\right)$ and $\left({b}_{n}\right)$ be two sequences, then we have:

 1. $\left({a}_{n}\right)+\left({b}_{n}\right)$ is a sequence and $\left({a}_{n}\right)+\left({b}_{n}\right)=\left({a}_{n}+{b}_{n}\right)$. [5.1.2]
 2. $\left({a}_{n}\right)-\left({b}_{n}\right)$ is a sequence and $\left({a}_{n}\right)-\left({b}_{n}\right)=\left({a}_{n}-{b}_{n}\right)$. [5.1.3]
 3. $\left({a}_{n}\right)\cdot \left({b}_{n}\right)$ is a sequence and $\left({a}_{n}\right)\cdot \left({b}_{n}\right)=\left({a}_{n}\cdot {b}_{n}\right)$. [5.1.4]
 4.Most often $\frac{\left({a}_{n}\right)}{\left({b}_{n}\right)}$ is no longer a sequence. [5.1.5]

Proof:   The only characteristic of a sequence is the domain ${ℕ}^{\ast }$. The sequences in 1., 2., and 3. have ${ℕ}^{\ast }\cap {ℕ}^{\ast }={ℕ}^{\ast }$ as their domain and thus are sequences again. The noted representations are nothing else than the rule for e.g. adding two functions. To point this out we once again write down the old notion:

$a+b\left(n\right)=a\left(n\right)+b\left(n\right)$.

To proof 4. we only need a counter example: Take for instance the function $\frac{\left(n+2\right)}{\left(n-2\right)}$.Her domain is no longer ${ℕ}^{\ast }$.

Consider:

• For a quotient of two sequences the sequence property isn't lost as a rule, but only if the value zero occurs in the denominator sequence. So we have:

 $\frac{\left({a}_{n}\right)}{\left({b}_{n}\right)}$ is a sequence$⇔\left({b}_{n}\right)$ has no zeros. [5.1.6]

• Denominator sequences with only finitely many zeros allow a kind of sound quotient using the generalized sequence concept as mentioned at the beginning. The counter example from the proof of 4. for instance could be saved the following way:

$\frac{{\left(n+2\right)}_{n\ge 3}}{{\left(n-2\right)}_{n\ge 3}}=\left(\frac{n+2}{n-2}{\right)}_{n\ge 3}$

Besides the possibility to produce new sequences from given ones the above proposition allows to separate sequences into simpler compounds. This is demonstrated with the last two sequences in the following example. Although separations of this kind represent sequences in a much more complicated way than necessary, they will prove to be very effective in our discussion to come. This applies especially to the very last example where we cancelled the fraction by ${n}^{2}$.

 Example:   $\left({n}^{2}+5\right)+\left(3n-7\right)-\left(4\right)\left(n+2\right)=\left({n}^{2}-n-10\right)$   $\frac{\left(n-2\right)}{\left(n+2\right)}=\left(\frac{n-2}{n+2}\right)$   $\left(3{n}^{2}-8n\right)=\left(3\right){\left(n\right)}^{2}-\left(8\right)\left(n\right)$   $\left(\frac{2{n}^{2}+n}{{n}^{2}+3}\right)=\frac{\left(2\right)+\left(\frac{1}{n}\right)}{\left(1\right)+\left(\frac{3}{{n}^{2}}\right)}$

The parts to follow mainly deal with real-valued sequences. All the results however can be stated and proved for generalized real-valued sequences  ${\left({a}_{n}\right)}_{n\ge k}$ as well.