Our discussion begins with a very basic concept in mathematics: sequences. Probably most people have at least a general intuitive idea of what a sequence is. One simple example of a sequence is
{1,2,3,4,5,…},
while another more irregular sequence is
{3,7,π,2/5,−6,…}.
In the second case, it is not possible to predict what the next entry will be after −6, but nonetheless, both examples seem to satisfy the essential sense of what a sequence is. What is needed now is an exact mathematical definition:
Definition.
Asequence is a functiona:A→R where eitherA=Z+:={1,2,3,…} orA=N:={0,1,2,3,…} (thenatural numbers) or perhapsA is any countable set.1 Thus, either
a:{an}={a1,a2,a3,…}
or
a:{an}={a0,a1,a2,a3,…},
where in both casesan∈R. We refer toan as then-thelement orentry of the sequence.
Remarks.
- 1.
Notice that a sequence differs from a set in that a sequence has an order, whereas a set is simply a collection of elements. Technically, there is no first element in a set, but there is in a sequence.
- 2.
This is not the most general definition of the word “sequence,” but it is easily general enough for our purposes. It contains all the key ideas to discuss sequences. In particular, we will not discuss finite sequences.
- 3.
Notice that for our purposes, sequences begin either withn=0 or withn=1. Beginning with a zeroth element is less common, but it occurs sometimes when a zeroth element makes sense. Among other places, this happens in computer science. Still, unless there is some specific reason to do otherwise, letA=Z+.
A sequence can be depicted graphically on the real line as inFigure 1.1.
Figure 1.1 A sequence on the real line. Here several elements of the sequence are shown in blue. Notice that there is no particular pattern to this sequence. Which of the elements are positive, and which are negative?
1.1 Convergent sequences
Now that we have a definition for sequences, we can introduce the central concept of convergence, starting with a basic example.
Example 1.1.1.
Suppose thatan:=1/n (i. e.,an isdefined equal to1/n), so that
{an}={1/n}={1,12,13,14,…}.
Notice that this sequence must start withn=1 sincea0 is undefinable using the given formula. Also notice thatan decreases toward zero asn increases toward infinity, even thoughan≠0 for anyn∈Z+. In symbols, that isan↘0 asn↗∞.
All of this may seem clear intuitively, but as we will see, intuition is not always so clear. So again an exact mathematical definition is needed.
Definition.
A sequence{an}⊂Rconverges to a limitL∈R iff givenε>0,∃N∈Z+ such that|an−L|<ε whenevern>N. Written with a minimal amount of symbols, this definition is “a sequence{an} of real numbersconverges to a limitL in the reals if and only if givenε>0, there existsN in the positive integers such that|an−L|<ε whenevern>N.” The sequence{an}diverges iff it does not converge.
Notation.
The notation to indicate that a sequence{an} converges to a limitL is
limn→∞an=L,
or more briefly
{an}→Loran→L.
Remark.
In this definition,ε can be any positive real number, but it is generally thought of as asmall positive real number. The definition works becauseε can be an arbitrary small positive real number. A schematic diagram of the relationship betweenε,L and the elements of the sequenceak is shown inFigure 1.2.
Figure 1.2 A sequence approaching a limitL on the real line. The early elements of the sequence throughaN can be outside the interval(L−ε,L+ε), but starting withaN+1, all of the elements must lie inside the interval. In particular, whenn>N, thenan must lie in the interval. Where mightaN+5 be on this number line? Ifε is made smaller, then likelyN will have to be made larger to keep the elements of the sequence fromaN+1 onward in the new smaller interval.
Example 1.1.1.
Returning to our first example, suppose again thatan:=1/n. Does the sequence{an} converge, and if so, to where? For a givenε, what value ofN guarantees that the convergence definition is satisfied?
Answer. Notice that there are two questions here: The first is a calculus question (Where does the sequence converge to? What is the value ofL?), while the second is an analysis question (How does one show that the definition is satisfied?). In this case, the calculus question is relatively easy, and in fact we have already mentioned the answer:L=0. The second question is more interesting: Givenε>0, what value ofN is needed to guarantee that|an−L|<ε whenn>N? Notice that here
|an−L|=|an−0|=|an|=1/n<ε⟸n>1/ε.
This may seem obvious, but notice that in fact, it gives us the answer we are looking for: defineN:=⌈1/ε⌉, where⌈x⌉ is the ceiling function or next greater integer function forx (the least integer greater than or equal tox). It is important that this ceiling function be used becauseN must be an integer, but for most values ofε, its reciprocal1/ε is not itself an integer. Then for this example,|an−L|<ε whenevern>N:=⌈1/ε⌉≥1/ε.
In summary, the limit here isL=0 and settingN:=⌈1/ε⌉ allows the definition of convergence to be satisfied.
Remarks.
- 1.
Notice that the value ofN increases asε decreases; this is almost always the case. Expectε to be in the denominator in the expression forN.
- 2.
Notice the direction of the double arrow above. Although we started on the left working on|an−L|, we need to keep in mind that this is the conclusion to be reached provided thatn is large enough. So a backward implication (a backward double arrow) is what is needed. Indeed, the symbol “⟸” is perhaps best read as “whenever”. Thus, the implication above becomes|an−L|<ε whenevern>1/ε.
The first example above is relatively straightforward; the next two examples are more interesting (“more interesting” means “harder”)....
