Examples and properties of sequences – Serlo

Examples

Constant sequence

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

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

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

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 $2$:

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

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

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

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

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

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

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

Geometric sequence

Example for a geometric sequence: $a_{n}=2^{n}$ for all $n\in \mathbb {N}$

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 $a_{n}=2^{n}$ where the constant ratio is given by $2$:

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

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

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

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

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

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

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

which can easily be checked by induction.

Harmonic sequence

The first ten element of the harmonic sequence

The sequence $a_{n}={\tfrac {1}{n}}$ 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:

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

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

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

or

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

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

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

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 $a_{n}=(-1)^{n}$ alternates between the values $1$ and $-1$, so we have an alternating sequence . A further example is $a_{n}=(-1)^{n+1}\cdot n$ with $\left(a_{n}\right)_{n\in \mathbb {N} }=(1,\,-2,\,3,\,-4,\,5,\,-6,\,\ldots )$.

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

Here, $b_{n}=|a_{n}|$ 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($a_{0}>0$), then we have the form $a_{n}=(-1)^{n}\cdot b_{n}$. Conversely, if $a_{0}<0$ the we have $a_{n}=(-1)^{n+1}\cdot b_{n}$.

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

If the first element is $0$ , 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 $100\,\%$ (oh my gosh, what a bank!) Then, after one year, you will get paid back $1+1=2$ "moneys" (2 units of money). Is there a way to get more money, if you are allowed to spread $100\,\%$ of interests over a year? You could ask the bank to pay you an interest rate of $50\,\%$, but twice a year. Then, after one year where multiplying your money twice, you get back

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

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

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

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

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

units of money.

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

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

units of money. This can be interpreted as a sequence in $n$: the sequence $\left(\left(1+{\frac {1}{n}}\right)^{n}\right)_{n\in \mathbb {N} }$ is also called "exponential sequence". Now, can you make infinitely much money within one year, just by splitting the $100\,\%$ 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 $e=2.71828\dots$. So you do not get above $2.72$ units of money. The proof why this sequence $\left(\left(1+{\frac {1}{n}}\right)^{n}\right)_{n\in \mathbb {N} }$ converges to $e$ can be found within the article "monotony criterion".

Sequence of Fibonacci numbers

The first elements of the Fibonacci sequence

How to determine the Fibonacci sequence

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

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

Or after an index shift:

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

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

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

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

Mixed sequences

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

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

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

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

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

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

Example (mixed sequence)

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

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

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

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

Yes, but only constant sequences.

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

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

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

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

Properties and important terms

Bounded sequence

An example for a bounded sequence ($a_{n}=(-1)^{n+1}\cdot {\tfrac {1}{n}}$) with some bounds.

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:

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

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:

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

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

We can formally invert the above statements:

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

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

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

So a sequence is unbounded from below, if for any $s$ there is a sequence element smaller than $s$. 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 $S$ is an upper bound of $(a_{n})_{n\in \mathbb {N} }$, if and only if $a_{n}\leq S$ for all $n\in \mathbb {N}$.

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 $S$ is a lower bound of $(a_{n})_{n\in \mathbb {N} }$, if and only if $a_{n}\geq S$ for all $n\in \mathbb {N}$.

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 $1$ , then $2$, $44$, $123$ and $502$ are upper bounds, as well. Boundedness from above can be shown by just stating any upper bound.

Theorem (alternative definition of boundedness)

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

Proof (alternative definition of boundedness)

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

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

or explicitly:

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

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 $S,s\in \mathbb {R}$, such that for all sequence elements $s\leq a_{n}\leq S$ holds. Then for all sequence elements there is also $|a_{n}|\leq \mathrm {max} \{|s|,\,|S|\}$. So the existence of some $S'$ within the alternative definition is established ($S'$ can be any positive, real number greater than or equal to $\mathrm {max} \{s|,\,|S|\}$).

And what about the other direction? Let now be $S'\in \mathbb {R} _{>0}$ given, with $|a_{n}|<S'$ for all sequence elements. Then the inequality $-S'\leq a_{n}\leq S'$ holds for all sequence elements. Thus $-S'$ represents a lower bound and $S'$ an upper bound for the sequence $\left(a_{n}\right)_{n\in \mathbb {N} }$, 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 $d>0$ it is bounded from below by $a_{0}$ and unbounded from above. For $d<0$ it is bounded from above by $a_{0}$ and unbounded from below. For $d=0$ we have a constant sequence, which is bounded.
3. geometric sequence: For $q=1$ we have a constant sequence, which is again bounded. For $q>1$ and $a_{0}>0$ the sequence is bounded from below (by $a_{0}$) and unbounded from above. For $q>1$ and $a_{0}<0$ it is the other way round: The sequence is bounded from above by the negative number $a_{0}$but unbounded from below. For $q<-1$ 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 $0<q<1$ and $a_{0}>0$ , it is bounded. The upper bound is $a_{0}$ and the lower one $0$. For $0<q<1$ and $a_{0}<0$ the lower bound is $a_{0}$ and the upper bound is $0$. So the sequence is bounded. And for $-1<q<0$ we also have boundedness by $a_{1}$ and $a_{0}$.
4. harmonic sequence: The harmonic sequence $\left(a_{n}\right)_{n\in \mathbb {N} }={\tfrac {1}{n}}$is bounded from above by $a_{1}=1$ and from below by $0$.
5. alternating harmonic sequence: This sequence is also bounded. An upper bound is $a_{2}$ and a lower bound is $a_{1}$ for $a_{n}=(-1)^{n}\cdot {\tfrac {1}{n}}$.
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 $a_{n+1}$ is larger than $a_{n}$), 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 $\left(a_{n}\right)_{n\in \mathbb {N} }$ we define:

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

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 $d>0$ the sequence is strictly monotonously increasing. For $d<0$ the sequence is strictly monotonously decreasing. For $d=0$ we have a constant sequence.
3. geometric sequence: For $q>1$ and $a_{0}>0$ it is strictly monotonously increasing and for $a_{0}<0$ it is strictly monotonously decreasing. For $0<q<1$ and $a_{0}>0$ it is strictly monotonously decreasing and for $a_{0}<0$ strictly monotonously increasing. For $q<0$ we have an alternating sequence, which is neither monotonously increasing, nor decreasing. For $q=1$ 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".