Sequences of Real Numbers
Notes for Math 1000 by Gerald Beer and Matthew Stevenson June 2016
1. Sequences defined
Recall that the counting numbers, also called the positive integers, is the set {1, 2, 3, 4, . . .}. By a sequence
a, we mean a rule that assigns to each counting number n a real number an (read: a sub n). Let us consider
the sequence defined by an = n2 . This means that the number assigned to n is the square of n. In particular,
1
a1 = 1, a2 = 4, a3 = 9 and so forth. As another example, consider an = n+1
. Here, the number assigned to
1
1
1
1
1
= 41 , and so forth.
n is the reciprocal of one more than n. Then a1 = 1+1 = 2 , a2 = 2+1 = 3 , a3 = 3+1
We call an the nth term of the sequence. Thus, for an = n2 , the twelfth term is 122 = 144, and for
1
1
1
an = n+1
, the twenty-fifth term is 25+1
= 26
.
Example 1.1. For the sequence an = 7 + 3n, what term is 64?
Solution: We put 7 + 3n = 64 and solve for n. Subtracting 7 from both sides, 3n = 57 and upon division by
3, n = 57
3 = 19. Thus, 64 is the 19th term.
A sequence is often described by simply listing its terms rather than explicitly giving the rule of assignment. Thus the sequence described by an = 3n + 2 can be described by
5, 8, 11, 14, 17, 20, . . . ,
n
while an = 2 can be described by
2, 4, 8, 16, 32, 64, . . . .
There are certain letters that are typically used along with n to denote integer values: i, j, k, l, and m are
the most common. On the other, any letter can be used in place of a to denote the rule of assignment. All
of the following are equally valid descriptions of the sequence an = n2 :
bn = 2n , ci = 2i , xj = 2j , Pk = 2k
all describe the same rule of assignment.
If we look at the sequence, 5, 8, 11, 14, 17, 20, . . . , a certain pattern is clear: given a particular term, to
get the next term, simply add 3. Symbolically and more formally, this can be written as follows: for each
n ≥ 1, an+1 = an + 3. Of course we would need to specify a1 to get the process started. A description of a
sequence that tells how to get an+1 from previous terms is called a recursive description of the sequence.
Example 1.2. Describe the sequence 24, 12, 6, 3, 23 , 34 , . . . . . . recursively.
Solution: Here, to get the next term, multiply the last term by
recursive description is a1 = 24 and an+1 = 21 an = a2n for n ≥ 1.
1
2,
starting with a1 = 24. Thus, a full
Example 1.3. Describe the sequence 3, 8, 23, 68, . . . recursively.
Solution: Note 8 = 3 · 3 − 1, 23 = 3 · 8 − 1, 68 = 3 · 23 − 1, and so forth. Here, to get the next term, triple the
last term and subtract one. A complete recursive description is an+1 = 3an − 1 for n ≥ 1 along with a1 = 3.
Example 1.4. Describe the sequence 1, 2, 6, 24, 120, 720, . . . recursively.
Solution: Note that a2 = 2a1 , a3 = 3a2 , a4 = 4a3 , and so forth. Here, to get the next term, multiply the last
term by its subscript plus one, starting with a1 = 1. Specifically, an+1 = (n + 1)an . The nth term of this
sequence is denoted by n! (read : n factorial). Directly, n! = n(n − 1)(n − 2) · · · 2 · 1.
1
2
Example 1.5. List the first 6 terms of the recursively defined sequence xj+1 = xj − 7 where x1 = 98.
Solution: To get the next term, subtract 7 from from the previous term: 98, 91, 84, 77, 70, 63, . . ..
Example 1.6. List the first 6 terms of the recursively defined sequence ak+1 = (−2)ak + 4 where a1 = 6.
Solution: To get the next term, multiply the last term by -2 and add 4; to start, a2 = (−2)a1 +4 = −12+4 =
−8. The first six terms we obtain are 6, −8, 20, −36, 76, −148, . . ..
In the previous examples, to get the next term of our sequence, we needed only look at the immediately
preceding term. However, for some sequences, we might need to consider earlier terms as well. This is so in
the case of the celebrated Fibonnaci sequence which we now describe.
Example 1.7. List the first eight terms of the recursively defined sequence ak+1 = ak + ak−1 for k ≥ 2 where
a1 = a2 = 1.
Solution: To get the next term, add the two immediately preceding terms. Thus to get a3 , both a1 and a2
must be specified in advance and they are. We get a3 = a2 + a1 = 1 + 1 = 2. Continuing, a4 = a3 + a2 =
2 + 1 = 3. For our first eight terms, we get
1, 1, 2, 3, 5, 8, 13, 21, . . . .
Example 1.8. List the first six terms of the recursively defined sequence yi+1 = −yi · yi−1 for i ≥ 2, given
y1 = 3 and y2 = −2.
Solution: To get the next term, multiply the two immediately preceding terms and change the sign of the
product. Thus for y3 , change the sign of (−2)(3), that is, y3 = 6. Similarly, y4 = −(6)(−2) = 12. For our
first six terms, we get
3, −2, 6, 12, −72, 864, . . . .
2. Arithmetic sequences and Geometric sequences
In an arithmetic sequence, we go from one term to the next by adding a fixed number d, that is, an+1 =
an + d for n ≥ 1. Since d = an+1 − an , it is traditional to call d the common difference.
Example 2.1. Write the first 5 terms of the arithmetic sequence defined by a1 = 1.26 and an+1 = an + .08
for n ≥ 1.
Solution: We add .08 to each term to get the next; this produces 1.26, 1.34, 1.42, 1.50, 1.58, . . ..
Example 2.2. Find the first term and common difference for the arithmetic sequence 48, 40, 32, 24, 16, 8, 0, −8, . . ..
Solution: Clearly, the first term is 48. To get from one term to the next, we always subtract 8, which means
we add −8. Thus, d = −8 for this arithmetic sequence.
Suppose we have an arithmetic sequence with first term a and common difference d; then a1 = a, a2 =
a + d, a3 = (a + d) + d = a + 2d, and so forth. The emergent pattern is clear:
an = a + (n − 1)d
is the formula for the nth term.
3
Example 2.3. Suppose the second term of an arithmetic sequence is 18 and the common difference is 5. Find
the formula for an and the 24th term.
Solution: Since a1 + 5 = a2 = 18, we get a = a1 = 13. The formula for the nth term is an = 13 + (n − 1)5.
In particular, a24 = 13 + (23)5 = 128.
Example 2.4. The eleventh term of an arithmetic sequence is 20 and the seventeenth term is 29. Find the
formula an for the nth term.
Solution: Using the formula an = a + (n − 1)d we obtain two equations in the unknowns a and d:
20 = a + 10d
29 = a + 16d
Subtraction of the first equation from the second gives 9 = 6d so d = 32 . With this value of d the first
equation becomes 20 = a + 10( 23 ) = a + 15, so 20 − 15 = a. The desired formula is an = 5 + (n − 1) 32 .
In a geometric sequence, we generate successive terms by multiplying the last term on our list by a number
r that does not depend on the term. The number r can be positive or negative. Thus, if the nth term is
denoted by an , we have
an+1 = an · r
for n = 1, 2, 3, . . . .
It is customary to call r the common ratio for the geometric sequence.
As with arithmetic sequences, let us denote the first term by a, that is, a1 = a; from this, a2 = ar, a3 =
a2 r = (ar)r = ar2 , a4 = a3 r = (ar2 )r = ar3 , etc. Paralleling the pattern we saw for arithmetic sequences,
the nth term is given by
an = arn−1 for n ≥ 1.
Example 2.5. Write down the first six terms of a geometric sequence whose first term is 8 and whose common
ratio is 23 . Find the 12th term using the formula an = arn−1 .
Solution: To get the second term, we multiply 8 by the common ratio, i.e, a2 = 8 × 23 = 4 · 3 = 12. Similarly,
a3 = 12 × 32 = 18. Continuing we get for our first five terms 8, 12, 18, 27, 81
2 , . . ..
3 n−1
The formula for our nth term is an = 8( 2 )
. In particular, a12 = 8(1.5)12−1 = 8(1.5)11 . To compute
x
this we must use our y button on our calculator: Rounding off, (1.5)11 ≈ 86.498 so that a12 ≈ 8(86.498) ≈
691.98.
Example 2.6. Write the first five terms of a geometric sequence with third term 20 and common ratio −4.
Solution: Moving forward, the fourth term is 20 × −4 = −80 and the fifth term is −80 × −4 = 320. Moving
backwards, we divide by −4 to get the immediately preceding term. Thus the second term is 20 ÷ −4 = −5
and the first term is −5 ÷ −4 = 45 . The first five terms are 54 , −5, 20, −80, 320, . . ..
Example 2.7. A national park hopes to grow a population of an endangered species of deer living in the park
by 2 percent each year. Currently, there are 150 deer in the park. How many deer would it like to have in
the park after n years? After 7 years?
Solution: If P is the population at the end of a given year, then after another year has elapsed, the hoped-for
population would be
4
P + 2 percent of P = P + (.02)P = P (1 + .02) = P (1.02).
Let Pn denote the target population after n years. As the initial population is 150 deer, after 1 year we
calculate P1 = 150(1.02) = 153 deer. More generally, Pn = 153(1.02)n−1 is the target deer population after
n years. After 7 years, the park would like to have P7 = 153(1.02)6 ≈ 153(1.12616) ≈ 172 deer.
Example 2.8. A midwestern city had 600 violent crimes this year. The police department sets a goal to
reduce the rate by 3 percent each year.
(a) What is the target crime rate figure after one year?
(b) What is the target crime rate figure after n years?
(c) Estimate the violent crime rate after 10 years if their goal is realized.
Solution: (a) If C is the violent crime rate in a given year, in the next year, the goal is to reduce it by 3
percent of C, that is, to make it C − (.03)C = C(1 − .03) = C(.97) in the next year. Thus, after one year,
the target violent crime rate is 600(.97) = 582 violent crimes per year.
(b) Let Cn be the target violent crime rate after n years; we have computed C1 = 582, and since
Cn+1 = Cn (.97)
n−1
we obtain the formula Cn = 582(.97)
.
(c) C10 = 582(.97)9 ≈ 582(.760) ≈ 442 violent crimes per year.
3. Partial Sums of a Sequence
Given a sequence with nth term an , we denote the sum of its first n terms by Sn , that is,
Sn = a1 + a2 + a3 + · · · + an .
For each positive number n, Sn is called the nth partial sum of the original sequence.
Example 3.1. Determine the first five partial sums for each sequence below.
(a) −2, 4, 10, 16, 22, 28, . . .;
(b) cj = (−1)j .
Solution: (a) For this arithmetic sequence, S1 = −2, S2 = −2 + 4 = 2, S3 = −2 + 4 + 10 = 12,
S4 = −2 + 4 + 10 + 16 = 28, and S5 = −2 + 4 + 10 + 16 + 22 = 50.
(b) S1 = −1, S2 = −1 + 1 = 0, S3 = −1 + 1 + (−1) = −1, S4 = −1 + 1 + (−1) + 1 = 0, and
S5 = −1 + 1 + (−1) + 1 + (−1) = −1. If we continued to compute partial sums, it is clear that for n odd,
Sn = −1 while for n even, Sn = 0.
Notice that for an arbitrary sequence with nth term an , S1 = a1 and for each n the grouping
Sn+1 = (a1 + a2 + a3 + · · · + an ) + an+1
says that Sn+1 = Sn + an+1 . In words, to get the next partial sum after a particular Sn has been computed,
simply add the next term of the original sequence to Sn ; there is no need to re-add everything from the
start! We can list the partial sums as a sequence associated with the original sequence. We of course have
just provided a recursive description of this sequence of partial sums.
5
Example 3.2. Using the recursive point of view, list the first six terms of the sequence of partial sums of
each sequence below.
(a) an = n2 ;
(b) x1 = 2 and xk+1 = 2xk + 1 for k ≥ 1.
Solution: (a) The terms of our sequence listed out are 1, 4, 9, 16, 25, 36, 49, 64, 81, . . . . From this we get
S1 = 1, S2 = 1 + 4 = 5, S3 = 5 + 9 = 14, S4 = 14 + 16 = 30, S5 = 30 + 25 = 55 and S6 = 55 + 36 = 91.
This produces the following sequence of partial sums as a list:
1, 5, 14, 30, 55, 91, . . .
.
(b) Successive terms of our recursively defined sequence are determined by doubling the last term and
then adding 1. This produces
2, 4 + 1, 10 + 1, 22 + 1, 66 + 1, 134 + 1, . . . or 2, 5, 11, 23, 67, 135, . . . .
The associated sequence of partial sums we obtain is
2, 2 + 5, 7 + 11, 18 + 23, 41 + 67, 108 + 135 . . . or 2, 7, 18, 41, 108, 243, . . . .
There is an easy way to find the nth partial sum of an arithmetic sequence that was observed by the
famous mathematician Gauss as a young child. Let us return to the initial arithmetic sequence of Example
3.1: −2, 4, 10, 16, 22, 28, . . .. As is easily seen, an = −2 + (n − 1)6 describes the nth term. Let us compute
S20 using the method of Gauss. Since a20 = 112, we can write S20 in the following way:
S20 = −2 + 4 + 10 + · · · + 100 + 106 + 112
Since the sum is not changed if we reverse the order of addition, we can write
S20 = 112 + 106 + 100 + · · · + 10 + 4 + −2.
Now if we add vertically, we get
2S20 = (−2 + 112) + (4 + 106) + (10 + 100) + · · · + (100 + 10) + (106 + 4) + (112 + −2).
The right-hand side consists of 20 identical summands of 110 = a1 + a20 , and so
S20 =
20
20
(a1 + a20 ) =
(110) = 1100.
2
2
The same trick works for any arithmetic sequence: the sum of the first n terms is the number of terms
times the sum of the first and last terms, all divided by 2. We record this as a theorem.
Theorem 3.3. Let a1 , a2 , a3 , a4 , . . . be an arithmetic sequence. Then its nth partial sum is
n
Sn = (a1 + an ) = n
2
a1 + an
2
.
6
The last formula for Sn is the more intuitive of the two, as it says that Sn is the number of terms times
the average of the first and last terms. Remember, however, that this formula only applies to arithmetic
sequences!
Example 3.4. Find the sum of the first 23 terms of this arithmetic sequence: 4.07, 4.20, 4.33, 4.46, . . ..
Solution: The first term is a = 4.07 and the common difference d = .13. As the nth term is
an = 4.07 + (n − 1)(.13), we get a23 = 4.07 + (22)(.13) = 4.07 + 2.86 = 6.93. By Theorem 3.3,
S23 = 23
4.07 + 6.93
2
= 23(5.5) = 126.5.
For a geometric sequence with an = arn−1 , there is also a simple formula for Sn . We now assume that
r 6= 1, where it is clear that Sn = a + a + · · · + a + a = na in that case. We compute
(1 − r)Sn = Sn − rSn = a + ar + ar2 + · · · + arn−2 + arn−1 − ar − ar2 − ar3 − · · · − arn−1 − arn
= a − arn = a(1 − rn ),
and dividing both sides by 1 − r which is nonzero by assumption gives the formula we are after:
Sn = a ·
1 − rn
.
1−r
Theorem 3.5. If r 6= 1, then a + ar + ar2 + ar3 + · · · + arn−1 = a ·
1−r n
1−r
=a·
r n −1
r−1 .
It is convenient to use the first formula for Sn when r < 1 and the second when r > 1 to keep numerator
and denominator both positive.
Example 3.6. Compute 100 + 100(1.05)2 + 100(1.05)3 + · · · + 100(1.05)10 .
Solution: Here, a = 100, r = 1.05, and n − 1 = 10, or n = 11. The second formula yields
S11 = 100
(1.05)11 − 1
1.05 − 1
≈ 100
1.7103 − 1
.05
= 100
.7103
.05
= 1420.6.
Example 3.7. Compute S7 for the sequence xn = 400(−.8)n−1 .
Solution: For this geometric sequence, a = 400 and r = −.8 < 1, so we use the first formula to compute the
partial sum S7 .
S7 = 400 ·
1 − (−.8)7
1 + .20972
≈ 400 ·
≈ 100(.6721) = 67.21.
1 − (−.8)
1.8
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Sequences of Real Numbers Exercises
Concepts
(1) What is a sequence?
(2) In a sequence, an is called the nth . . . . . . of the sequence.
(3) What does it mean for a sequence to be recursively defined ?
(4) Describe in words the difference between an arithmetic sequence and a geometric sequence.
(5) Given a sequence with general term an , write down the formula we use to find the nth partial sum
of the sequence.
In Exercises 6-11, list the first five terms of each sequence.
(6) an = n3 − 1
(9) yi = 32
(7) an =
n−1
n+6
(10) wk = (−1)k+1
(8) cj = |3 − 7j|
√
(11) bn = 2n2 + 3.
In Exercises 12-17, list the first five terms of each recursively defined sequence.
(12) a1 = 7; an+1 = an + 5, n ≥ 1
(13) y1 = 92 ; ym+1 = 3ym , m ≥ 1
(14) a1 = −2; an+1 = 3 − an , n ≥ 1
(15) b1 = b2 = 0; bn+1 = 1 − bn − bn−1 , n ≥ 2;
(16) c1 = 3; cj+1 = −cj · j 2 , j ≥ 1
(17) x1 = 4, x2 = −2; xk+1 = 2xk + xk−1 , k ≥ 2.
(18) Given the sequence an = 74 n − 3 what term is 25?
(19) Given the sequence bj =
2
j+1
what term is 18 ?
(20) Write a formula for the nth term of each sequence:
√ √
√
(b) 7, 8, 3, 10, . . . (c) −9, 16, −25, 36, −49, . . .
(a) 12 , 23 , 34 , 45 , . . .
(21) The 20th term of an arithmetic sequence is 88 and the common difference of
the sequence is 6. Find the formula for an and use it to find a100 .
(22) The 23rd term of an arithmetic sequence is −1; the 29th term is 11. Find a formula for an .
(23) Find a formula for nth term of each arithmetic sequence below:
(a) 31 , 1, 53 , 73 , 3, . . .
(b) 18,15,12,9,6, . . .
(c) 4.26, 4.49, 4.72, 4.95, . . .
(24) Find a formula for nth term of each geometric sequence below:
(a) 200, 200(1.04), 200(1.04)2 , 200(1.04)3 , . . .
(b) 12, 4, 43 , 49 , . . .
(25) List the first five terms of each geometric sequence:
(a) an = 5(−2)n−1
(b) bj = 2π j−1 .
(26) (a) Find the first five terms of a geometric sequence whose 4th term is 90 and common ration is 3.
(b) Write a formula for an and use it to find the 10th term of the sequence.
(27) A zombie apocalypse begins in Los Angeles with 26,411 zombies and grows at a rate
of 25.1 percent per month.
(a) If the zombie population is Z at a certain point of time, what is its population 1 month later?
(b) How many zombies will Los Angeles have after 16 months?
8
(28) List the first five partial sums of the sequences described in Exercises (6), (8), (9) and (10).
(29) List the first five partial sums for the recursively defined sequence whose first term is
2
z1 = 3 and whose general term is zm+1 = zm
− 1, where m ≥ 1.
(30) Find the 60th partial sum of each arithmetic sequence below using the formula of Theorem 3.3.
(a) an = 4 + (n − 1)(2)
(b) b1 = 6 and bn+1 = bn + 21 , n ≥ 1
(c) 120, 117, 114, . . ..
(31) Find the 12th partial sum of each geometric sequence below using the formula of Theorem 3.5.
(a) an = 30(.8)n−1
(b) a1 = 4 and ai+1 = 1.5ai , i ≥ 1;
(c) 500, 500(1.07), 500(1.07)2 , . . ..