Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

9.1: Sequences

( \newcommand{\kernel}{\mathrm{null}\,}\)

When we first introduced a function as a special type of relation in Section 1.3, we did not put any restrictions on the domain of the function. All we said was that the set of x-coordinates of the points in the function F is called the domain, and it turns out that any subset of the real numbers, regardless of how weird that subset may be, can be the domain of a function. As our exploration of functions continued beyond Section 1.3, we saw fewer and fewer functions with ‘weird’ domains. It is worth your time to go back through the text to see that the domains of the polynomial, rational, exponential, logarithmic and algebraic functions discussed thus far have fairly predictable domains which almost always consist of just a collection of intervals on the real line. This may lead some readers to believe that the only important functions in a College Algebra text have domains which consist of intervals and everything else was just introductory nonsense. In this section, we introduce sequences which are an important class of functions whose domains are the set of natural numbers.1 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

12,34,98,2716,

As usual, the periods of ellipsis, , indicate that the proposed pattern continues forever. Each of the numbers in the list is called a term, and we call 12 the ‘first term’, 34 the ‘second term’, 98 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.

na(n)11223439842716

In other words, a(n) is the nth term in the sequence. We formalize these ideas in our definition of a sequence and introduce some accompanying notation.

Definition 9.1.

A sequence is a function a whose domain is the natural numbers. The value a(n) is often written as an and is called the nth term of the sequence. The sequence itself is usually denoted using the notation: an, n1 or the notation: {an}n=1.

Applying the notation provided in Definition 9.1 to the sequence given (1), we have a1=12, a2=34, a3=98 and so forth. Now suppose we wanted to know a117, that is, the 117th term in the sequence. While the pattern of the sequence is apparent, it would benefit us greatly to have an explicit formula for an. 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.

Example 9.1.1.

Write the first four terms of the following sequences.

  1. an=5n13n, n1
  2. bk=(1)k2k+1, k0
  3. {2n1}n=1{1+(1)ii}i=2
  4. {1+(1)ii}i=2
  5. a1=7, an+1=2an, n1
  6. [factorialintroex] f0=1, fn=nfn1, n1

Solution

  1. Since we are given n1, the first four terms of the sequence are a1, a2, a3 and a4. Since the notation a1 means the same thing as a(1), we obtain our first term by replacing every occurrence of n in the formula for an with n=1 to get a1=51131=13. Proceeding similarly, we get a2=52132=59, a3=53133=2527 and a4=54134=12581.
  2. For this sequence we have k0, so the first four terms are b0, b1, b2 and b3. Proceeding as before, replacing in this case the variable k with the appropriate whole number, beginning with 0, we get b0=(1)02(0)+1=1, b1=(1)12(1)+1=13, b2=(1)22(2)+1=15 and b3=(1)32(3)+1=17. (This sequence is called an alternating sequence since the signs alternate between + and . The reader is encouraged to think what component of the formula is producing this effect.)
  3. From {2n1}n=1, we have that an=2n1, n1. We get a1=1, a2=3, a3=5 and a4=7. (The first four terms are the first four odd natural numbers. The reader is encouraged to examine whether or not this pattern continues indefinitely.)
  4. Here, we are using the letter i as a counter, not as the imaginary unit we saw in Section 3.4. Proceeding as before, we set ai=1+(1)ii, i2. We find a2=1, a3=0, a4=12 and a5=0.
  5. To obtain the terms of this sequence, we start with a1=7 and use the equation an+1=2an for n1 to generate successive terms. When n=1, this equation becomes a1+1=2a1 which simplifies to a2=2a1=27=5. When n=2, the equation becomes a2+1=2a2 so we get a3=2a2=2(5)=7. Finally, when n=3, we get a3+1=2a3 so a4=2a3=27=5.
  6. As with the problem above, we are given a place to start with f0=1 and given a formula to build other terms of the sequence. Substituting n=1 into the equation fn=nfn1, we get f1=1f0=11=1. Advancing to n=2, we get f2=2f1=21=2. Finally, f3=3f2=32=6.

Some remarks about Example 9.1.1 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 {bk}k=0 from Example 9.1.1, we graph the equation y=b(k) for the values k0. That is, we plot the points (k,b(k)) for the values of k in the domain, k=0,1,2,. 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. If you feel a sense of nostalgia, you should see Section 1.2.

Screen Shot 2022-05-03 at 3.24.02 AM.png

Graphing y=bk=(1)k2k+1, k0

Speaking of {bk}k=0, the astute and mathematically minded reader will correctly note that this technically isn’t a sequence, since according to Definition 9.1, sequences are functions whose domains are the natural numbers, not the whole numbers, as is the case with {bk}k=0. In other words, to satisfy Definition 9.1, 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 1.7, we can accomplish this by replacing k with k1 in the definition of {bk}k=0. Specifically, let ck=bk1 where k10. We get ck=(1)k12(k1)+1=(1)k12k1, where now k1. We leave to the reader to verify that {ck}k=1 generates the same list of numbers as does {bk}k=0, but the former satisfies Definition 9.1, 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 9.1 to include any function which generates a list of numbers which can then be matched up with the natural numbers.2 Finally, we wish to note the sequences in parts 5 and 6 are examples of sequences 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 in part 6 are given a special name: fn=n! is called n-factorial. Using the ‘!’ notation, we can describe the factorial sequence as: 0!=1 and n!=n(n1)! for n1. After 0!=1 the next four terms, written out in detail, are 1!=10!=11=1, 2!=21!=21=2, 3!=32!=321=6 and 4!=43!=4321=24. From this, we see a more informal way of computing n!, which is n!=n(n1)(n2)21 with 0!=1 as a special case. (We will study factorials in greater detail in Section 9.4.) The world famous Fibonacci Numbers are defined recursively and are explored in the exercises. While none of the sequences worked out to be the sequence in (1), they do give us some insight into what kinds of patterns to look for. Two patterns in particular are given in the next definition.

Definition 9.2. Arithmetic and Geometric Sequences

Suppose {an}n=k is a sequencea

  • If there is a number d so that an+1=an+d for all nk, then {an}n=k is called an arithmetic sequence. The number d is called the common difference.
  • If there is a number r so that an+1=ran for all nk, then {an}n=k is called a geometric sequence. The number r is called the common ratio.

a Note that we have adjusted for the fact that not all ‘sequences’ begin at n=1.

Both arithmetic and geometric sequences are defined in terms of recursion equations. In English, an arithmetic sequence is one in which we proceed from one term to the next by always adding the fixed number d. The name ‘common difference’ comes from a slight rewrite of the recursion equation from an+1=an+d to an+1an=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 r0, we can rearrange the recursion equation to get an+1an=r, hence the name ‘common ratio.’ Some sequences are arithmetic, some are geometric and some are neither as the next example illustrates.3

Example 9.1.2.

Determine if the following sequences are arithmetic, geometric or neither. If arithmetic, find the common difference d; if geometric, find the common ratio r.

  1. an=5n13n, n1
  2. bk=(1)k2k+1, k0
  3. {2n1}n=112,34,98,2716,
  4. 12,34,98,2716,

Solution

A good rule of thumb to keep in mind when working with sequences is “When in doubt, write it out!” Writing out the first several terms can help you identify the pattern of the sequence should one exist.

  1. From Example 9.1.1, we know that the first four terms of this sequence are 13,59,2527 and 12581. To see if this is an arithmetic sequence, we look at the successive differences of terms. We find that a2a1=5913=29 and a3a2=252759=1027. Since we get different numbers, there is no ‘common difference’ and we have established that the sequence is not arithmetic. To investigate whether or not it is geometric, we compute the ratios of successive terms. The first three ratios a2a1=5913=53,a3a2=252759=53 and a4a3=125812527=53 suggest that the sequence is geometric. To prove it, we must show that an+1an=r for all n.

    an+1an=5(n+1)13n+15n13n=5n3n+13n5n1=53

    This sequence is geometric with common ratio r=53.

  2. Again, we have Example 9.1.1 to thank for providing the first four terms of this sequence: 1,13,15 and 17. We find b1b0=43 and b2b1=815. Hence, the sequence is not arithmetic. To see if it is geometric, we compute b1b0=13 and b2b1=35. Since there is no ‘common ratio,’ we conclude the sequence is not geometric, either.
  3. As we saw in Example 9.1.1, the sequence {2n1}n=1 generates the odd numbers: 1,3,5,7,. Computing the first few differences, we find a2a1=2, a3a2=2, and a4a3=2. This suggests that the sequence is arithmetic. To verify this, we find an+1an=(2(n+1)1)(2n1)=2n+212n+1=2 This establishes that the sequence is arithmetic with common difference d=2. To see if it is geometric, we compute a2a1=3 and a3a2=53. Since these ratios are different, we conclude the sequence is not geometric.
  4. We met our last sequence at the beginning of the section. Given that a2a1=54 and a3a2=158, the sequence is not arithmetic. Computing the first few ratios, however, gives us a2a1=32, a3a2=32 and a4a3=32. Since these are the only terms given to us, we assume that the pattern of ratios continue in this fashion and conclude that the sequence is geometric.

We are now one step away from determining an explicit formula for the sequence given in (1). We know that it is a geometric sequence and our next result gives us the explicit formula we require.

Equation 9.1. Formulas for Arithmetic and Geometric Sequences
  • An arithmetic sequence with first term a and common difference d is given by an=a+(n1)d,n1
  • A geometric sequence with first term a and common ratio r0 is given by an=arn1,n1

While the formal proofs of the formulas in Equation 9.1 require the techniques set forth in Section 9.3, we attempt to motivate them here. According to Definition 9.2, 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 nth term, we add d to a exactly (n1) times, which is what the formula says. The derivation of the formula for geometric series follows similarly. Here, we start with a and go from one term to the next by multiplying by r. We get a,ar,ar2,ar3 and so forth. The nth term results from multiplying a by r exactly (n1) times. We note here that the reason r=0 is excluded from Equation 9.1 is to avoid an instance of 00 which is an indeterminant form.4 With Equation 9.1 in place, we finally have the tools required to find an explicit formula for the nth term of the sequence given in (1). We know from Example 9.1.2 that it is geometric with common ratio r=32. The first term is a=12 so by Equation 9.1 we get an=arn1=12(32)n1 for n1. After a touch of simplifying, we get an=(3)n12n for n1. Note that we can easily check our answer by substituting in values of n and seeing that the formula generates the sequence given in (1). We leave this to the reader. Our next example gives us more practice finding patterns.

Example 9.1.3.

Find an explicit formula for the nth term of the following sequences.

  1. 0.9,0.09,0.009,0.0009,
  2. 25,2,23,27,
  3. 1,27,413,819,

Solution

  1. Although this sequence may seem strange, the reader can verify it is actually a geometric sequence with common ratio r=0.1=110. With a=0.9=910, we get an=910(110)n1 for n0. Simplifying, we get an=910n, n1. There is more to this sequence than meets the eye and we shall return to this example in the next section.
  2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt to find a pattern, we rewrite the second term with a denominator to make all the terms appear as fractions. We have 25,21,23,27,. If we associate the negative ‘’ of the last two terms with the denominators we get 25,21,23,27,. This tells us that we can tentatively sketch out the formula for the sequence as an=2dn where dn is the sequence of denominators. Looking at the denominators 5,1,3,7,, we find that they go from one term to the next by subtracting 4 which is the same as adding 4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a=5 and d=4, we get the nth denominator by the formula dn=5+(n1)(4)=94n for n1. Our final answer is an=294n, n1.
  3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem to try to get patterns individually for the numerator and denominator. Letting cn and dn denote the sequence of numerators and denominators, respectively, we have an=cndn. After some experimentation,5 we choose to write the first term as a fraction and associate the negatives ‘’ with the numerators. This yields 11,27,413,819,. The numerators form the sequence 1,2,4,8, which is geometric with a=1 and r=2, so we get cn=(2)n1, for n1. The denominators 1,7,13,19, form an arithmetic sequence with a=1 and d=6. Hence, we get dn=1+6(n1)=6n5, for n1. We obtain our formula for an=cndn=(2)n16n5, for n1. We leave it to the reader to show that this checks out.

While the last problem in Example 9.1.3 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,, we would be hard-pressed to find a formula for an 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 nth 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, is rather recognizable as the squares, so the formula an=n2, n1 may not be too hard to determine. With this in mind, it’s possible to see 2,5,10,17, as the sequence 1+1,4+1,9+1,16+1,, so that an=n2+1, n1. 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 bn=14n4+52n3314n2+252n5, n1. 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.6 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 6.2 in Section 6.5. Recall that if we invest P dollars at an annual percentage rate r and compound the interest n times per year, the formula for Ak, the amount in the account after k compounding periods, is Ak=P(1+rn)k=[P(1+rn)](1+rn)k1, k1. We now spot this as a geometric sequence with first term P(1+rn) and common ratio (1+rn). In retirement planning, it is seldom the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called an annuity and will be discussed in the next section once we have developed more mathematical machinery.

9.1.1 Exercises

In Exercises 1 - 3, write out the first four terms of the given sequence.

  1. an=2n1dj=(1)j(j+1)2, n0
  2. dj=(1)j(j+1)2, j1
  3. {5k2}k=1{n2+1n+1}n=0
  4. {n2+1n+1}n=0
  5. {xnn2}n=1
  6. {ln(n)n}n=1{xnn2}n=1
  7. a1=3, an+1=an1, n1dm=dm1100
  8. d0=12, dm=dm1100, m1
  9. b1=2,bk+1=3bk+1,k1
  10. c0=2,cj=cj1(j+1)(j+2),j1
  11. a1=117, an+1=1an, n1
  12. s0=1, sn+1=xn+1+sn, n0
  13. F0=1, F1=1, Fn=Fn1+Fn2, n2 (This is the famous Fibonacci Sequence)

In Exercises 14 - 21 determine if the given sequence is arithmetic, geometric or neither. If it is arithmetic, find the common difference d; if it is geometric, find the common ratio r.

  1. {3n5}n=1
  2. an=n2+3n+2, n1
  3. 13, 16, 112, 124{3(15)n1}n=1, …
  4. {3(15)n1}n=1
  5. 17, 5, 7, 19, …
  6. 2, 22, 222, 2222, …
  7. 0.9, 9, 90, 900an=n!2, …
  8. an=n!2, n0.

In Exercises 22 - 30, find an explicit formula for the nth term of the given sequence. Use the formulas in Equation 9.1 as needed.

  1. 3, 5, 7, 918, …
  2. 1, 12, 14, 18, …
  3. 1, 23, 45, 87, …
  4. 1, 23, 13, 427x77, …
  5. 1, 14, 19, 116x77, …
  6. x, x33, x55, x77, …
  7. 0.9,0.99,0.999,0.9999,
  8. 27,64,125,216,
  9. 1,0,1,0,
  10. Find a sequence which is both arithmetic and geometric. (Hint: Start with an=c for all n.)
  11. Show that a geometric sequence can be transformed into an arithmetic sequence by taking the natural logarithm of the terms.
  12. Thomas Robert Malthus is credited with saying, “The power of population is indefinitely greater than the power in the earth to produce subsistence for man. Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will show the immensity of the first power in comparison with the second.” (See this webpage for more information.) Discuss this quote with your classmates from a sequences point of view.
  13. This classic problem involving sequences shows the power of geometric sequences. Suppose that a wealthy benefactor agrees to give you one penny today and then double the amount she gives you each day for 30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the 30\scriptsize th day? What is the dollar value of the gift you have received?
  14. Research the terms ‘arithmetic mean’ and ‘geometric mean.’ With the help of your classmates, show that a given term of a arithmetic sequence ak, k2 is the arithmetic mean of the term immediately preceding, ak1 it and immediately following it, ak+1. State and prove an analogous result for geometric sequences.
  15. Discuss with your classmates how the results of this section might change if we were to examine sequences of other mathematical things like complex numbers or matrices. Find an explicit formula for the n\scriptsize th term of the sequence i,1,i,1,i,. List out the first four terms of the matrix sequences we discussed in Exercise 8.3.1 in Section 8.3.

9.1.2 Answers

  1. 0,1,3,7
  2. 1,1,1,1
  3. 3,8,13,18
  4. 1,1,53,52
  5. x,x24,x39,x416
  6. 0,ln(2)2,ln(3)3,ln(4)4
  7. 3,2,1,0
  8. 12,0.12,0.0012,0.000012
  9. 2,7,22,67
  10. 2,13,136,1720
  11. 117,1117,117,1117
  12. 1,x+1,x2+x+1,x3+x2+x+1
  13. 1,1,2,3
  14. arithmetic, d=3
  15. neither
  16. geometric, r=12
  17. geometric, r=15
  18. arithmetic, d=12
  19. neither
  20. geometric, r=10
  21. neither
  22. an=1+2n,n1
  23. an=(12)n1,n1
  24. an=2n12n1,n1
  25. an=n3n1,n1
  26. an=1n2,n1
  27. (1)n1x2n12n1,n1
  28. an=10n110n,n1
  29. an=(n+2)3,n1
  30. an=1+(1)n12,n1

Reference

1 Recall that this is the set {1, 2, 3, . . .}.

2 We’re basically talking about the ‘countably infinite’ subsets of the real number line when we do this.

3 Sequences which are both arithmetic and geometric are discussed in the Exercises.

4 See the footnotes on page 237 in Section 3.1 and page 418 of Section 6.1.

5 Here we take ‘experimentation’ to mean a frustrating guess-and-check session.

6 For more on this, see When Every Answer is Correct: Why Sequences and Number Patterns Fail the Test.


This page titled 9.1: Sequences is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Carl Stitz & Jeff Zeager via source content that was edited to the style and standards of the LibreTexts platform.

  • Was this article helpful?

Support Center

How can we help?