In this section, we introduce sequences which are an important class of functions whose domains are, more or less, the set of natural numbers. ^[1] (Recall that this is the set \(\mathbb {N} = \{1, 2, 3, \ldots \}\).) Before we get to far ahead of ourselves, let’s look at what the term ‘sequence’ means mathematically. Informally, we can think of a sequence as an infinite list of numbers. For example, consider the sequence

As usual, the periods of ellipsis, \(\ldots \), indicate that the proposed pattern continues forever. Each of the numbers in the list is called a term, and we call \(\frac {1}{2}\) the ‘first term’, \(-\frac {3}{4}\) the ‘second term’, \(\frac {9}{8}\) the ‘third term’ and so forth. In numbering them this way, we are setting up a function, which we’ll call ‘\(a\)’ per tradition, between the natural numbers and the terms in the sequence.

In other words, \(a(n)\) is the \(n^{\text {th}}\) term in the sequence. We formalize these ideas in our definition of a sequence and introduce some accompanying notation.

Applying the notation provided in Definition sequencedefn to the sequence given (seq1), we have \(a_1 = \frac {1}{2}\), \(a_2 = -\frac {3}{4}\), \(a_3 = \frac {9}{8}\).

Suppose we wanted to know \(a_117\), that is, the \(117^{\text {th}}\) term in the sequence. While the pattern of the sequence is apparent, it would benefit us greatly to have an explicit formula for \(a_{n}\). Unfortunately, there is no general algorithm that will produce a formula for every sequence, so any formulas we do develop will come from that greatest of teachers, experience. In other words, it is time for an example.

Some remarks about Example seqex1 are in order. We first note that since sequences are functions, we can graph them in the same way we graph functions. For example, if we wish to graph the sequence \(\left \{ b_{k} \right \}_{k=0}^{\infty }\) from Example seqex1, we graph the equation \(y=b(k)\) for the values \(k \geq 0\). That is, we plot the points \((k, b(k))\) for the values of \(k\) in the domain, \(k = 0, 1, 2, \ldots \). The resulting collection of points is the graph of the sequence.

Note that we do not connect the dots in a pleasing fashion as we are used to doing, because the domain is just the whole numbers in this case, not a collection of intervals of real numbers. ^[2] (If you feel a sense of nostalgia, you should see Section FunctionsandtheirRepresentations.)

Speaking of \(\left \{ b_{k} \right \}_{k=0}^{\infty }\), the astute and mathematically minded reader will correctly note that this technically isn’t a sequence, since according to Definition sequencedefn, sequences are functions whose domains are the natural numbers, not the whole numbers, as is the case with \(\left \{ b_{k} \right \}_{k=0}^{\infty }\). In other words, to satisfy Definition sequencedefn, we need to shift the variable \(k\) so it starts at \(k=1\) instead of \(k=0\).

To see how we can do this, it helps to think of the problem graphically. What we want is to shift the graph of \(y=b(k)\) to the right one unit, and thinking back to Section Transformations, we can accomplish this by replacing \(k\) with \(k-1\) in the definition of \(\left \{ b_{k} \right \}_{k=0}^{\infty }\).

Specifically, let \(c_{k} = b_{k-1}\) where \(k - 1 \geq 0\). We get \(c_{k} = \frac {(-1)^{k-1}}{2(k-1)+1} = \frac {(-1)^{k-1}}{2k-1}\), where now \(k \geq 1\). We leave to the reader to verify that \(\left \{ c_{k} \right \}_{k=1}^{\infty }\) generates the same list of numbers as does \(\left \{ b_{k} \right \}_{k=0}^{\infty }\), but the former satisfies Definition sequencedefn, while the latter does not.

Like so many things in this text, we acknowledge that this point is pedantic and join the vast majority of authors who adopt a more relaxed view of Definition sequencedefn to include any function which generates a list of numbers which can then be matched up with the natural numbers. ^[3] (We’re basically talking about the ‘countably infinite’ subsets of the real number line when we do this.)

One last note about Example seqex1 concerns the manner in which the sequences in numbers introrecursiveex01 and factorialintroex are defined. We say these two sequences are described ‘recursively.’ In each instance, an initial value of the sequence is given which is then followed by a recursion equation \(-\) a formula which enables us to use known terms of the sequence to determine other terms.

The terms of the sequence from number factorialintroex is given notation and name: \(f_{n}=n!\) is called \(n\)-factorial. Using the ‘!’ notation, we can describe the factorial sequence as: \(0! = 1\) and \(n! = n(n-1)!\) for \(n \geq 1\).

After \(0! = 1\) the next four terms, written out in detail, are \(1! = 1 \cdot 0! = 1 \cdot 1 = 1\), \(2! = 2 \cdot 1! = 2 \cdot 1 = 2\), \(3! = 3 \cdot 2! = 3 \cdot 2 \cdot 1 = 6\) and \(4! = 4 \cdot 3! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\). From this, we see a more informal way of computing \(n!\), which is \(n! = n\cdot (n -1)\cdot (n -2) \cdots 2 \cdot 1\) with \(0! = 1\) as a special case. (We will study factorials in greater detail in Section Binomial.) ^[4] (Another famous sequence, the Fibonacci Numbers are defined also recursively and are explored in the exercises.)

While none of the sequences in Example seqex1 worked out to be the sequence in (seq1), they do give us some insight into what kinds of patterns to look for. Two patterns in particular are given in the next definition.

In English, an arithmetic sequence is one in which we proceed from one term to the next by always adding the fixed number \(d\). If this sort of ‘constant change’ idea sounds familiar, it should. Indeed, arithmetic sequences are merely linear functions, something we will explore in more detail shortly. Note the name ‘common difference’ comes from a slight rewrite of the recursion equation from \(a_{n+1} = a_{n} + d\) to \(a_{n+1} - a_{n} = d\). That is, every pair of successive terms has the same or common difference, \(d\).

Analogously, a geometric sequence is one in which we proceed from one term to the next by always multiplying by the same fixed number \(r\). If this notion sounds familiar, it is because geometric sequences are, in fact, exponential functions. Again, we will explore this connection in more detail later. We note that if \(a_{n} \neq 0\), we can rearrange the recursion equation to get \(\frac {a_{n+1}}{a_{n}} = r\). Hence, every pair of successive terms has the same or common ratio, \(r\).

Some sequences are arithmetic, some are geometric and some are neither as the next example illustrates. ^[6] (Can a sequence be both arithmetic and geometric? See Exercise arithmeticandgeometricexercise.)

We are now one step away from determining an explicit formula for the sequence given in (seq1). We know that it is a geometric sequence and our next result gives us the explicit formula we require.

An intuitive way to arrive at Equation arithgeoformula appeals to Definition arithgeodefn directly. Given an arithmetic sequence with first term \(a\) and common difference \(d\), the way we get from one term to the next is by adding \(d\). Hence, the terms of the sequence are: \(a\), \(a+d\), \(a+2d\), \(a+3d\), …. We see that to reach the \(n\)th term, we add \(d\) to \(a\) exactly \((n-1)\) times, which is exactly what the formula says. ^[7] (We formalize this argument in Section Induction.)

Note if we rewrite the formula \(a_{n} = a_{1} + (n-1) d\) using traditional function notation as \(a(n) = a(1) + d (n-1)\) we can see arithmetic sequences are linear functions. ^[8] (Note here \(a\) is a function, so the expressions \(a(n)\) and \(a(1)\) here represent the outputs from \(a\). On the other hand, the expression \(d(n-1)\) indicates multiplication of the real numbers \(d\) and \((n-1)\).) Indeed, relabeling the function \(a\) as ‘\(f\)’ and the independent variable \(n\) as ‘\(x\),’ we can make the identifications \(x_{0} = 1\), and \(m=d\) so as to put the equation \(a(n) = a(1) + d (n-1)\) into the form of Equation linearfunctionpointslope:

Hence, arithmetic sequences are linear functions with slope \(d\) whose domains are the natural numbers.

The derivation of the formula for geometric series follows similarly. Here, we start with the first term \(a\) and go from one term to the next by multiplying by \(r\). We get \(a, ar, ar^2, ar^3\) and so forth. The \(n\)th term results from multiplying \(a\) by \(r\) exactly \((n-1)\) times. ^[9] (We note here that the reason \(r=0\) is excluded from Equation arithgeoformula is to avoid an instance of \(0^{0}\) which is an indeterminate form. (See the remarks following Definition monomialfunction in Section GraphsofPolynomials.))

In the same way arithmetic sequences are linear functions, geometric sequences are exponential functions. Writing \(a_{n} = a_{1} r^{n-1}\) as \(a(n) = a(1) r^{n-1}\), we can relabel \(a\) as \(f\) and \(n\) as \(x\) and make the identifications \(x_{0} = 1\) and \(b = r\) to put the equation into the form described in Definition expfcnpointbaseform:

So, geometric sequences are exponential functions with base \(r\) whose domains are the natural numbers.

With Equation arithgeoformula in place, we finally have the tools required to find an explicit formula for the \(n\)th term of the sequence given in (seq1). We know from Example arithgeoex1 that it is geometric with common ratio \(r = -\frac {3}{2}\). The first term is \(a = \frac {1}{2}\) so by Equation arithgeoformula we get \(a_{n} = ar^{n-1} = \frac {1}{2} \left ( - \frac {3}{2}\right )^{n-1}\) for \(n \geq 1\). After a touch of simplifying, we get \(a_{n} = \frac {(-3)^{n-1}}{2^{n}}\) for \(n \geq 1\). Note that we can easily check our answer by substituting in values of \(n\) and seeing that the formula generates the sequence given in (seq1). We leave this to the reader. In particular, the 117th term in the sequence is \(a_{117} = \frac {1}{2} \left (-\frac {3}{2} \right )^{117-1} = \frac {3^{116}}{2^{117}}\).

Our next example gives us more practice finding patterns.

While the last problem in Example seqex2 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence \(2, 5, 10, 17, \ldots \), we would be hard-pressed to find a formula for \(a_{n}\) if we restrict our attention to these two archetypes. We said before that there is no general algorithm for finding the explicit formula for the \(n\)th term of a given sequence, and it is only through experience gained from evaluating sequences from explicit formulas that we learn to begin to recognize number patterns.

The pattern \(1, 4, 9, 16, \ldots \) is rather recognizable as the squares, so the formula \(a_{n} = n^2\), \(n \geq 1\) may not be too hard to determine. With this in mind, it’s possible to see \(2, 5, 10, 17, \ldots \) as the sequence \(1 + 1, 4 + 1, 9 + 1, 16 + 1 , \ldots \), so that \(a_{n} = n^2 + 1\), \(n \geq 1\).

Of course, since we are given only a small sample of the sequence, we shouldn’t be too disappointed to find out this isn’t the only formula which generates this sequence. For example, consider the sequence defined by \(b_{n} = -\frac {1}{4} n^4 + \frac {5}{2} n^3 - \frac {31}{4} n^2 + \frac {25}{2} n - 5\), \(n \geq 1\). The reader is encouraged to verify that it also produces the terms \(2, 5, 10, 17\). In fact, it can be shown that given any finite sample of a sequence, there are infinitely many explicit formulas all of which generate those same finite points. This means that there will be infinitely many correct answers to some of the exercises in this section. ^[11] (For more on this, see When Every Answer is Correct: Why Sequences and Number Patterns Fail the Test.) Just because your answer doesn’t match ours doesn’t mean it’s wrong. As always, when in doubt, write your answer out. As long as it produces the same terms in the same order as what the problem wants, your answer is correct.

Sequences play a major role in the Mathematics of Finance, as we have already seen with Equation compoundinterest in Section ExpLogApplications. Recall that if we invest \(P\) dollars at an annual percentage rate \(r\) and compound the interest \(n\) times per year, the formula for \(A_{k}\), the amount in the account after \(k\) compounding periods, is \(A_{k} = P\left (1 + \frac {r}{n}\right )^{k} = \left [P\left (1 + \frac {r}{n}\right )\right ]\left (1 + \frac {r}{n}\right )^{k-1}\), \( k \geq 1\). We leave it to the reader to show this is a geometric sequence with first term \(P\left (1 + \frac {r}{n}\right )\) and common ratio \(\left (1 + \frac {r}{n}\right )\).

In section ExpLogApplications, we showed \(\lim _{n \rightarrow \infty } P\left (1 + \frac {r}{n}\right )^{nt} = Pe^{rt}\), where we noted, at the time, that the limit here was taken on a discrete, rather than continuous variable. That didn’t stop us from using the limit properties listed in Theorem LimitProp01. We talk more on the limits of sequences next.

1 Limits of Sequences

Consider the sequence \(a_{n} = \frac {3n}{n+1}\), \(n \geq 1\). Suppose we wished to find \(\lim _{n \rightarrow \infty } a_{n}\).

We should first note that even though the domain of the sequence \(a_{n}\) is discrete, we can nonetheless discuss what happens as \(n \rightarrow \infty \) since for any real number \(M>0\), we can find a natural number \(n > M\). (We could, for instance take \(n = \lfloor M \rfloor + 1\).)

Next, note that we can visualize the sequence \(a_{n} = \frac {3n}{n+1}\), \(n \geq 1\) as being points on the graph of \(f(x) = \frac {3x}{x+1}\):

Comparing leading terms of numerator and denominator, as \(x \rightarrow \infty \), \(\frac {3x}{x+1} \approx \frac {3x}{x} = 3\). Hence, \(\lim _{x \rightarrow \infty } \frac {3x}{x+1} = 3\), which means all of the \(y\)-values on the graph of \(y = f(x)\), including the \(y\)-values of the graph of the sequence, approach \(3\) as \(x \rightarrow \infty \). It stands to reason, then that \(\lim _{n \rightarrow \infty } \frac {3n}{n+1} = 3\).

The long and short of the above argument is that since \(\lim _{x\rightarrow \infty } \frac {3x}{x+1}\) exists, \(\lim _{n \rightarrow \infty } \frac {3n}{n+1} = \lim _{x \rightarrow \infty } \frac {3x}{x+1}\). While the syntax of ‘\(\lim _{n \rightarrow \infty }\frac {3n}{n+1} = \lim _{x \rightarrow \infty } \frac {3x}{x+1}\)’ appears to be just a switch in a dummy variable, ^[12] (which it would be without context) the switch from ‘\(n\)’ to ‘\(x\)’ indicates switching from a discrete variable to a continuous one. This sort of maneuver is called passing to a continuous variable and is one of the primary ways we can use what we’ve already studied to analyze limits of sequences.

Note the phrase ‘\(f(n) = a_{n}\) for all \(n \geq k\) for some natural number \(k\)’ indicates the focus on what is happening as \(n \rightarrow \infty \). In other words, the first finitely many sequence values do not impact the value of \(\lim _{n \rightarrow \infty } a_{n}\). We are just concerned with the long-run or end behavior here. The reason Theorem 1 works, indeed why all of the limit properties discussed in Section ?? work with sequences as well as functions of continuous variables is because the formal definitions associated with limits of both sequences and functions share the same mathematical ‘bones.’ (See a Calculus instructor for more details.) We’ll explore these connections more deeply in the Exercises.

We have special words to describe sequences which have limits and those which do not.

The sequence in number 2 in Example 4 is an example of an alternating sequence, so-named because the terms alternate in sign. Alternating sequences play a large role in the study of infinite series (whatever those are) in Calculus, ^[14] (We’ll touch on these in the next section, too.) so it is worth pointing them out here.

Next, the reasoning we used to determine \(\lim _{n \rightarrow \infty } \frac {(-1)^{n}}{n^2 +1} = 0\) is sound and is codified in the following theorem. We state the result for both sequences (discrete functions) and (continuous) functions.

The Squeeze Theorem is so-named because the two sequences (functions) which bound the middle sequence (function) ‘squeeze’ the middle function to the common limit, \(L\).

Passing to a continuous variable in conjunction with the Squeeze Theorem can be used to prove certain classes of Geometric Sequences converge. We have the following:

We encourage the reader to think through each of the cases stated in Theorem 3 to make sure the conclusions seem reasonable. We’ll have occasion to cite Theorem 3 in the next section. For now, it’s time for some Exercises.