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1.

INFINITE SEQUENCES AND SERIES

2.

INFINITE SEQUENCES AND SERIES
Infinite sequences and series were
introduced briefly in A Preview of Calculus
in connection with Zeno’s paradoxes and
the decimal representation of numbers.

3.

INFINITE SEQUENCES AND SERIES
Their importance in calculus stems from
Newton’s idea of representing functions as
sums of infinite series.
▪ For instance, in finding areas, he often integrated
a function by first expressing it as a series and then
integrating each term of the series.

4.

INFINITE SEQUENCES AND SERIES
We will pursue his idea in Section 11.10
2
-x
in order to integrate such functions as e .
▪ Recall that we have previously been unable
to do this.

5.

INFINITE SEQUENCES AND SERIES
Many of the functions that arise in
mathematical physics and chemistry,
such as Bessel functions, are defined
as sums of series.
▪ It is important to be familiar with
the basic concepts of convergence
of infinite sequences and series.

6.

INFINITE SEQUENCES AND SERIES
Physicists also use series in another
way, as we will see in Section 11.11
▪ In studying fields as diverse as optics, special relativity,
and electromagnetism, they analyze phenomena by
replacing a function with the first few terms in the series
that represents it.

7.

INFINITE SEQUENCES AND SERIES
11.1
Sequences
In this section, we will learn about:
Various concepts related to sequences.

8. SEQUENCE

A sequence can be thought of as a list
of numbers written in a definite order:
a1, a2, a3, a4, …, an, …
▪ The number a1 is called the first term,
a2 is the second term, and in general
an is the nth term.

9. SEQUENCES

We will deal exclusively with infinite
sequences.
▪ So, each term an will have a successor an+1.

10. SEQUENCES

Notice that, for every positive integer n,
there is a corresponding number an.
So, a sequence can be defined as:
▪ A function whose domain is the set of positive
integers

11. SEQUENCES

However, we usually write an instead
of the function notation f(n) for the value
of the function at the number n.

12. SEQUENCES

Notation
SEQUENCES
The sequence { a1, a2, a3, . . .} is also
denoted by:
an
or
an n=1

13. SEQUENCES

Example 1
Some sequences can be
defined by giving a formula
for the nth term.

14. SEQUENCES

Example 1
In the following examples, we give three
descriptions of the sequence:
1.
Using the preceding notation
2.
Using the defining formula
3.
Writing out the terms of the sequence

15. SEQUENCES

Example 1 a
Preceding Defining
Terms of
Notation
Sequence
n
n + 1 n =1
Formula
n
n
1 2 3 4
an =
,...
, , , ,...,
n +1 2 3 4 5
n +1
▪ In this and the subsequent examples, notice
that n doesn’t have to start at 1.

16. SEQUENCES

Example 1 b
SEQUENCES
Preceding
Defining
Terms of
Notation
Formula
Sequence
(−1)n (n + 1)
(−1) n (n + 1)
an =
n
n
3
3
4 5
2 3
− 3 , 9 , − 27 , 81 ,...,
n
(−1) (n + 1) ,...
3n

17. SEQUENCES

Example 1 c
Preceding Defining
Terms of
Notation
Formula
Sequence
n − 3
an = n − 3
n =3
0,1, 2, 3,..., n − 3,...
(n 3)

18. SEQUENCES

Example 1 d
Preceding Defining
Terms of
Notation
Sequence
Formula
n
n
an = cos
6
cos
6 n=0
(n 0)
3 1
n
, , 0,..., cos
,...
1,
6
2 2

19. SEQUENCES

Example 2
Find a formula for the general term an
of the sequence
4 5
6
7
3
,−
,
,...
,− ,
5 25 125 625 3125
assuming the pattern of the first few terms
continues.

20. SEQUENCES

Example 2
SEQUENCES
We are given that:
3
a1 =
5
4
a2 = −
25
6
a4 = −
625
5
a3 =
125
7
a5 =
3125

21. SEQUENCES

Example 2
Notice that the numerators of these fractions
start with 3 and increase by 1 whenever we
go to the next term.
▪ The second term has numerator 4 and the third
term has numerator 5.
▪ In general, the nth term will have numerator n+2.

22. SEQUENCES

Example 2
SEQUENCES
The denominators are the powers
of 5.
n
▪ Thus, an has denominator 5 .

23. SEQUENCES

Example 2
The signs of the terms are alternately
positive and negative.
▪ Hence, we need to multiply by a power of –1.

24. SEQUENCES

Example 2
In Example 1 b, the factor (–1)n meant
we started with a negative term.
Here, we want to start with a positive term.

25. SEQUENCES

Example 2
SEQUENCES
Thus, we use (–1)n–1 or (–1)n+1.
▪ Therefore,
an = (−1)
n −1
n+2
n
5

26. SEQUENCES

Example 3
We now look at some sequences
that don’t have a simple defining
equation.

27. SEQUENCES

Example 3 a
The sequence {pn}, where pn
is the population of the world as
of January 1 in the year n

28. SEQUENCES

Example 3 b
If we let an be the digit in the nth decimal place
of the number e, then {an} is a well-defined
sequence whose first few terms are:
{7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, …}

29. FIBONACCI SEQUENCE

Example 3 c
The Fibonacci sequence {fn} is defined
recursively by the conditions
f1 = 1
f2 = 1
fn = fn–1 + fn–2
n≥3
▪ Each is the sum of the two preceding term terms.
▪ The first few terms are:
{1, 1, 2, 3, 5, 8, 13, 21, …}

30. FIBONACCI SEQUENCE

Example 3
This sequence arose when the 13th-century
Italian mathematician Fibonacci solved a
problem concerning the breeding of rabbits.
▪ See Exercise 71.

31. SEQUENCES

A sequence such as
that in Example 1 a
[an = n/(n + 1)] can
be pictured either
by:
▪ Plotting its terms on
a number line
▪ Plotting its graph

32. SEQUENCES

Note that, since
a sequence is
a function whose
domain is the set
of positive integers,
its graph consists of isolated points with
coordinates
(1, a1) (2, a2) (3, a3) … (n, an) …

33. SEQUENCES

From either figure,
it appears that
the terms of
the sequence
an = n/(n + 1) are
approaching 1 as n
becomes large.

34. SEQUENCES

In fact, the difference
n
1
1−
=
n +1 n +1
can be made as small as we like by taking n
sufficiently large.
▪ We indicate this by writing
n
lim
=1
n → n + 1

35. SEQUENCES

In general, the notation
lim an = L
n →
means that the terms of the sequence {an}
approach L as n becomes large.

36. SEQUENCES

Notice that the following definition of
the limit of a sequence is very similar to
the definition of a limit of a function at infinity
given in Section 2.6

37. LIMIT OF A SEQUENCE

Definition 1
A sequence {an} has the limit L, and we write
lim an = L
n →
or
an → L as n →
if we can make the terms an as close to L
as we like, by taking n sufficiently large.
an exists, the sequence converges
▪ If lim
n →
(or is convergent).
▪ Otherwise, it diverges (or is divergent).

38. LIMIT OF A SEQUENCE

Here, Definition 1 is illustrated by showing
the graphs of two sequences that have
the limit L.

39. LIMIT OF A SEQUENCE

A more precise version of
Definition 1 is as follows.

40. LIMIT OF A SEQUENCE

Definition 2
LIMIT OF A SEQUENCE
A sequence {an} has the limit L, and we write
lim an = L
n →
or
an → L as n →
if for every ε > 0 there is a corresponding
integer N such that:
if
n>N
then
|an – L| < ε

41. LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure,
in which the terms a1, a2, a3, . . . are plotted
on a number line.

42. LIMIT OF A SEQUENCE

No matter how small an interval (L – ε, L + ε)
is chosen, there exists an N such that all
terms of the sequence from aN+1 onward must
lie in that interval.

43. LIMIT OF A SEQUENCE

Another illustration of Definition 2
is given here.
▪ The points on the graph of {an} must lie between
the horizontal lines y = L + ε and y = L – ε if n > N.

44. LIMIT OF A SEQUENCE

This picture must be valid no matter how
small ε is chosen.
▪ Usually, however, a smaller ε requires a larger N.

45. LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7
in Section 2.6, you will see that the only
difference between lim an = L and lim f ( x) = L
n →
x →
is that n is required to be an integer.
▪ Thus, we have the following theorem.

46. LIMITS OF SEQUENCES

Theorem 3
If lim f ( x) = L and f(n) = an when n
x →
is an integer, then
lim an = L
n →

47. LIMITS OF SEQUENCES

Theorem 3 is illustrated here.

48. LIMITS OF SEQUENCES

Equation 4
In particular, since we know that
lim(1/ x ) = 0 when r > 0 (Theorem 5
r
x →
in Section 2.6), we have:
1
lim r = 0
n → n
if r 0

49. LIMITS OF SEQUENCES

If an becomes large as n becomes large,
we use the notation
lim an =
n →
▪ The following precise definition is similar to
Definition 9 in Section 2.6

50. LIMIT OF A SEQUENCE

Definition 5
LIMIT OF A SEQUENCE
lim an = means that, for every positive
n →
number M, there is an integer N such that
if
n>N
then
an > M

51. LIMITS OF SEQUENCES

If lim an = , then the sequence {an}
n →
is divergent, but in a special way.
▪ We say that {an} diverges to ∞.

52. LIMITS OF SEQUENCES

The Limit Laws given in Section 2.3
also hold for the limits of sequences
and their proofs are similar.

53. LIMIT LAWS FOR SEQUENCES

Suppose {an} and {bn} are
convergent sequences and
c is a constant.

54. LIMIT LAWS FOR SEQUENCES

Then,
lim(an + bn ) = lim an + lim bn
n →
n →
n →
lim(an − bn ) = lim an − lim bn
n →
n →
lim can = c lim an
n →
n →
n →
lim c = c
n →

55. LIMIT LAWS FOR SEQUENCES

Also,
lim(anbn ) = lim an lim bn
n →
n →
an
an lim
lim = n→
n → b
lim bn
n
n →
if lim bn 0
n →
n →
lim a = lim an
n →
n→
P
n
P
if p 0 and an 0

56. LIMITS OF SEQUENCES

The Squeeze Theorem
can also be adapted for
sequences, as follows.

57. SQUEEZE THEOREM FOR SEQUENCES

If an ≤ bn ≤ cn for n ≥ n0
and lim an = lim cn = L ,
then
n →
n →
lim bn = L
n →

58. LIMITS OF SEQUENCES

Another useful fact about limits of
sequences is given by the following
theorem.
▪ The proof is left as Exercise 75.

59. LIMITS OF SEQUENCES

If lim an = 0
n→
then lim an = 0
n →
Theorem 6

60. LIMITS OF SEQUENCES

n
Find lim
n → n + 1
Example 4
▪ The method is similar to the one we used
in Section 2.6
▪ We divide the numerator and denominator by the
highest power of n and then use the Limit Laws.

61. LIMITS OF SEQUENCES

Example 4
Thus,
lim1
n
1
n →
lim
= lim
=
n → n + 1
n →
1
1
1+
lim1 + lim
n n→ n→ n
1
=
=1
1+ 0
▪ Here, we used Equation 4 with r = 1.

62. LIMITS OF SEQUENCES

Calculate lim ln n
n →
Example 5
n
▪ Notice that both the numerator and denominator
approach infinity as n → ∞.

63. LIMITS OF SEQUENCES

Example 5
Here, we can’t apply l’Hospital’s Rule
directly.
▪ It applies not to sequences but to functions
of a real variable.

64. LIMITS OF SEQUENCES

Example 5
However, we can apply l’Hospital’s Rule
to the related function f(x) = (ln x)/x and
obtain:
ln x
1/ x
lim
= lim
=0
x → x
x → 1

65. LIMITS OF SEQUENCES

Therefore, by Theorem 3,
we have:
ln n
lim
=0
n → n
Example 5

66. LIMITS OF SEQUENCES

Example 6
Determine whether
the sequence an = (–1)n
is convergent or divergent.

67. LIMITS OF SEQUENCES

Example 6
If we write out the terms of the sequence,
we obtain:
{–1, 1, –1, 1, –1, 1, –1, …}

68. LIMITS OF SEQUENCES

Example 6
The graph of the sequence is shown.
▪ The terms oscillate between 1 and –1 infinitely often.
▪ Thus, an does not approach any number.

69. LIMITS OF SEQUENCES

Example 6
Thus, lim(−1) does not exist.
n
n→
That is, the sequence {(–1)n} is divergent.

70. LIMITS OF SEQUENCES

Example 7
(
−
1)
Evaluate lim
if it exists.
n →
n
n
(−1)n
1
= lim = 0
▪ lim
n →
n → n
n
(−1) n
▪ Thus, by Theorem 6, lim
=0
n →
n

71. LIMITS OF SEQUENCES

The following theorem says that, if we
apply a continuous function to the terms
of a convergent sequence, the result is also
convergent.

72. LIMITS OF SEQUENCES

Theorem 7
If lim an = L and the function f is
n →
continuous at L, then
lim f (an ) = f ( L)
n →
▪ The proof is left as Exercise 76.

73. LIMITS OF SEQUENCES

Example 8
LIMITS OF SEQUENCES
Find lim sin( / n)
n →
▪ The sine function is continuous at 0.
▪ Thus, Theorem 7 enables us to write:
(
lim sin( / n) = sin lim( / n)
n →
n →
= sin 0 = 0
)

74. LIMITS OF SEQUENCES

Example 9
Discuss the convergence
of the sequence an = n!/nn,
where n! = 1 . 2 . 3 . … . n

75. LIMITS OF SEQUENCES

Example 9
Both the numerator and denominator
approach infinity as n → ∞.
However, here, we have no corresponding
function for use with l’Hospital’s Rule.
▪ x! is not defined when x is not an integer.

76. LIMITS OF SEQUENCES

Example 9
Let’s write out a few terms to get a feeling
for what happens to an as n gets large:
a1 = 1
1 2
a2 =
2 2
1 2 3
a3 =
3 3 3

77. LIMITS OF SEQUENCES

E. g. 9—Equation 8
Therefore,
1 2 3 n
an =
n n n n

78. LIMITS OF SEQUENCES

Example 9
From these expressions and the graph here,
it appears that the terms are decreasing and
perhaps approach 0.

79. LIMITS OF SEQUENCES

Example 9
To confirm this, observe from Equation 8
that
1 2 3 n
an =
n n n n
▪ Notice that the expression in parentheses is
at most 1 because the numerator is less than
(or equal to) the denominator.

80. LIMITS OF SEQUENCES

1
Thus, 0 < an ≤
n
Example 9
▪ We know that 1/n → 0 as n → ∞.
▪ Therefore an → 0 as n → ∞ by the Squeeze
Theorem.

81. LIMITS OF SEQUENCES

Example 10
For what values of r is the sequence {r n}
convergent?
▪ From Section 2.6 and the graphs of the exponential
functions in Section 1.5, we know that lim a x =
for a > 1 and lim a x = 0 for 0 < a < 1. x→
x →

82. LIMITS OF SEQUENCES

Example 10
LIMITS OF SEQUENCES
Thus, putting a = r and using Theorem 3,
we have:
lim r =
n →
0
n
if r 1
if 0 r 1
▪ It is obvious that
lim1 = 1
n
n→
and
lim0 = 0
n
n→

83. LIMITS OF SEQUENCES

Example 10
LIMITS OF SEQUENCES
If –1 < r < 0, then 0 < |r | < 1.
▪ Thus, lim | r
n→
n
|= lim | r | = 0
n
n→
▪ Therefore, by Theorem 6, lim r n = 0
n →

84. LIMITS OF SEQUENCES

Example 10
If r ≤ –1, then {r n} diverges as in
Example 6.

85. LIMITS OF SEQUENCES

Example 10
The figure shows the graphs for various
values of r.

86. LIMITS OF SEQUENCES

Example 10
The case r = –1 was shown earlier.

87. LIMITS OF SEQUENCES

The results of Example 10
are summarized for future use,
as follows.

88. LIMITS OF SEQUENCES

Equation 9
LIMITS OF SEQUENCES
The sequence {r n} is convergent
if –1 < r ≤ 1 and divergent for all other
values of r.
0
lim r =
n →
1
n
if − 1 r 1
if r = 1

89. LIMITS OF SEQUENCES

Definition 10
A sequence {an} is called:
▪ Increasing, if an < an+1 for all n ≥ 1,
that is, a1 < a2 < a3 < · · ·
▪ Decreasing, if an > an+1 for all n ≥ 1
▪ Monotonic, if it is either increasing or decreasing

90. DECREASING SEQUENCES

Example 11
The sequence 3 is decreasing
n + 5
because
3
3
3
=
n + 5 (n + 1) + 5 n + 6
and so an > an+1 for all n ≥ 1.

91. DECREASING SEQUENCES

Show that the sequence
n
an = 2
n +1
is decreasing.
Example 12

92. DECREASING SEQUENCES

E. g. 12—Solution 1
We must show that an+1 < an,
that is,
n +1
n
2
2
(n + 1) + 1 n + 1

93. DECREASING SEQUENCES

E. g. 12—Solution 1
DECREASING SEQUENCES
This inequality is equivalent to the one
we get by cross-multiplication:
n +1
n
2
2
(
n
+
1)(
n
+
1)
n
[(
n
+
1)
+ 1]
2
2
(n + 1) + 1 n + 1
n 3 + n 2 + n + 1 n 3 + 2n 2 + 2n
1 n + n
2

94. DECREASING SEQUENCES

E. g. 12—Solution 1
Since n ≥ 1, we know that the inequality
n2 + n > 1 is true.
▪ Therefore, an+1 < an.
▪ Hence, {an} is decreasing.

95. DECREASING SEQUENCES

E. g. 12—Solution 2
x
Consider the function f ( x ) =
:
2
x +1
x +1 − 2x
f '( x) =
2
2
( x + 1)
2
2
1− x
= 2
0
2
( x + 1)
2
whenever x 1
2

96. DECREASING SEQUENCES

E. g. 12—Solution 2
Thus, f is decreasing on (1, ∞).
▪ Hence, f(n) > f(n + 1).
▪ Therefore, {an} is decreasing.

97. BOUNDED SEQUENCES

Definition 11
A sequence {an} is bounded:
▪ Above, if there is a number M such that an ≤ M
for all n ≥ 1
▪ Below, if there is a number m such that m ≤ an
for all n ≥ 1
▪ If it is bounded above and below

98. BOUNDED SEQUENCES

For instance,
▪ The sequence an = n is bounded below (an > 0)
but not above.
▪ The sequence an = n/(n+1) is bounded
because 0 < an < 1 for all n.

99. BOUNDED SEQUENCES

We know that not every bounded
sequence is convergent.
▪ For instance, the sequence an = (–1)n satisfies
–1 ≤ an ≤ 1 but is divergent from Example 6.

100. BOUNDED SEQUENCES

Similarly, not every
monotonic sequence is
convergent (an = n → ∞).

101. BOUNDED SEQUENCES

However, if a sequence is both bounded
and monotonic, then it must be convergent.
▪ This fact is proved as Theorem 12.
▪ However, intuitively, you can understand why
it is true by looking at the following figure.

102. BOUNDED SEQUENCES

If {an} is increasing and an ≤ M for all n,
then the terms are forced to crowd together
and approach some number L.

103. BOUNDED SEQUENCES

The proof of Theorem 12 is based on
the Completeness Axiom for the set
of real numbers.
▪ This states that, if S is a nonempty set of real numbers that has an
upper bound M (x ≤ M for all x in S), then
S has a least upper bound b.

104. BOUNDED SEQUENCES

This means:
▪ b is an upper bound for S.
▪ However, if M is any other upper bound,
then b ≤ M.

105. BOUNDED SEQUENCES

The Completeness Axiom is
an expression of the fact that there is
no gap or hole in the real number line.

106. MONOTONIC SEQ. THEOREM

Theorem 12
Every bounded, monotonic
sequence is convergent.

107. MONOTONIC SEQ. THEOREM

Theorem 12—Proof
Suppose {an} is an increasing
sequence.
▪ Since {an} is bounded, the set S = {an|n ≥ 1}
has an upper bound.
▪ By the Completeness Axiom, it has a least
upper bound L.

108. MONOTONIC SEQ. THEOREM

Theorem 12—Proof
Given ε > 0, L – ε is not an upper bound
for S (since L is the least upper bound).
▪ Therefore,
aN > L – ε
for some integer N

109. MONOTONIC SEQ. THEOREM

Theorem 12—Proof
However, the sequence is increasing.
So, an ≥ aN for every n > N.
▪ Thus, if n > N, we have an > L – ε
▪ Since an ≤ L, thus 0 ≤ L – an < ε

110. MONOTONIC SEQ. THEOREM

Theorem 12—Proof
Thus,
|L – an| < ε
Therefore,
whenever n > N
lim an = L
n →

111. MONOTONIC SEQ. THEOREM

Theorem 12—Proof
A similar proof (using
the greatest lower bound)
works if {an} is decreasing.

112. MONOTONIC SEQ. THEOREM

The proof of Theorem 12 shows that
a sequence that is increasing and bounded
above is convergent.
▪ Likewise, a decreasing sequence that is bounded
below is convergent.

113. MONOTONIC SEQ. THEOREM

This fact is used many times in
dealing with infinite series.

114. MONOTONIC SEQ. THEOREM

Example 13
Investigate the sequence {an} defined by
the recurrence relation
a1 = 2
an+1 = 12 (an + 6)
for n = 1, 2,3,...

115. MONOTONIC SEQ. THEOREM

Example 13
We begin by computing the first several
terms:
a1 = 2
a2 = 12 (2 + 6) = 4
a3 = 12 (4 + 6) = 5
a4 = 12 (5 + 6) = 5.5 a5 = 5.75
a6 = 5.875
a7 = 5.9375
a9 = 5.984375
a8 = 5.96875
▪ These initial terms suggest the sequence is
increasing and the terms are approaching 6.

116. MONOTONIC SEQ. THEOREM

Example 13
To confirm that the sequence is increasing,
we use mathematical induction to show
that an+1 > an for all n ≥ 1.
▪ Mathematical induction is often used
in dealing with recursive sequences.

117. MONOTONIC SEQ. THEOREM

Example 13
That is true for n = 1 because
a2 = 4 > a1.

118. MONOTONIC SEQ. THEOREM

Example 13
If we assume that it is true for n = k,
we have:
ak +1 ak
ak +1 + 6 ak + 6
▪ Hence,
and
▪ Thus,
1
2
(ak +1 + 6) 12 (ak + 6)
ak +2 ak +1

119. MONOTONIC SEQ. THEOREM

Example 13
We have deduced that an+1 > an is true
for n = k + 1.
Therefore, the inequality is true for all n
by induction.

120. MONOTONIC SEQ. THEOREM

Example 13
Next, we verify that {an} is bounded by
showing that an < 6 for all n.
▪ Since the sequence is increasing, we already know
that it has a lower bound: an ≥ a1 = 2 for all n

121. MONOTONIC SEQ. THEOREM

Example 13
We know that a1 < 6.
So, the assertion is true for n = 1.

122. MONOTONIC SEQ. THEOREM

Example 13
Suppose it is true for n = k.
▪ Then,
ak 6
▪ Thus,
ak + 6 12
and
▪ Hence,
1
2
(ak + 6) 12 (12) = 6
ak +1 6

123. MONOTONIC SEQ. THEOREM

Example 13
This shows, by mathematical
induction, that an < 6 for all n.

124. MONOTONIC SEQ. THEOREM

Example 13
Since the sequence {an} is increasing
and bounded, Theorem 12 guarantees
that it has a limit.
▪ However, the theorem doesn’t tell us
what the value of the limit is.

125. MONOTONIC SEQ. THEOREM

Example 13
Nevertheless, now that we know L = lim an
n →
exists, we can use the recurrence relation
to write:
lim an +1 = lim 12 (an + 6)
n →
n →
= 12 (lim an + 6)
n →
= 12 ( L + 6)

126. MONOTONIC SEQ. THEOREM

Example 13
Since an → L, it follows that an+1 → L, too
(as n → ∞, n + 1 → ∞ too).
▪ Thus, we have:
L = 12 ( L + 6)
▪ Solving this equation for L, we get L = 6,
as predicted.