1
Section 9.1 Sequences
Definition A sequence is an infinite list of numbers s1 , s2 , s3 , . . . , sn , . . . We call s1 the first term, s2
the second term, and so on. sn is the general term, say nth term.
Example 1 1, 4, 9, 16, · · · is a sequence. To be more precise, s1 = 1, s2 = 4, s3 = 9, and so
on. In this example, the general nth term is given by sn =
.
Example 2 Give the first six terms of the following sequences:
a. sn =
n(n+1)
2
b. tn =
n+(−1)n
n
Example 3 Consider the sequence 1, 1, 2, 3, 5, 8, 13, . . . Can you read off the pattern? What is the
next term?
Example 4 Find the general nth term sn when the first few terms of the sequence are
a. 2, 4, 6, 8, 10, 12, . . .
b. 1, 3, 5, 7, 9, 11, . . .
c. 1, 2, 4, 8, 16, 32, . . .
d. 1, −1, 1, −1, 1, −1, . . .
e.
1
, − 32 , 43 , − 45 , 56 , − 67 , . . .
2
Sequences can also be defined recursively, by giving an equation relating the nth term to the previous
terms (called recursion formula) and as many of the first terms as are needed to get started.
Example 5 Give the first few terms of the following recursively defined sequences.
a. sn = sn−1 + 3 for n > 1 and s1 = 4.
b. sn = 2sn−1 for n > 1 and s1 = 1.
c. sn = sn−1 + sn−2 for n > 2 and s1 = s2 = 1.
d. sn =
sn−1
for n > 1 and s1 = 1.
n
2
Example 6 Consider a recursive formula given by
( sn−1
sn =
if sn−1 is even
2
3sn−1 + 1 if sn−1 is odd
a. With s1 = 4, write the first few terms of the sequence.
b. With s1 = 5, write the first few terms of the sequence.
c. With s1 = 7, write the first few terms of the sequence.
d. Try other initial terms. Does your answer always end up with 4, 2, 1, 4, 2, 1, . . .?
We now consider the limit of a sequence.
Definition If sn approaches a fixed number L as n → ∞, then we say that sn converges to L
and L is the limit of sn . When there is no such L, we say that the sequence diverges.
Remark The concept of the limit of a sequence is not much different from that of a function as
x → ∞ and the usual limit rules on functions still hold.
Example 7 Do the following sequences converge or diverge? If a sequence converges, find its limit.
a. sn =
1
n
d. sn =
(−1)n
n
b. sn =
5n2 −100
2n2 −3n+5
e. sn =
sin n
n
c. sn = n2
f. sn = 1 + (−1)n
3
Example 8 Examine the limit of each of the following sequences
a. sn = ( 14 )n
c. sn = ( 43 )n
b. sn = (− 21 )n
d. sn = (−2)n
Remark In summary we get

∞



1
lim rn =
n→∞

0



does not exist
if
if
if
if
r>1
r=1
−1<r <1
r ≤ −1
Example 9 Examine the limit of each of the following sequences:
a. sn =
1+3n
2−5n
c. sn =
6n +32n
3n +5·9n
b. sn =
2+4n
3n
d. sn =
2n +4n
3n +2n
Definition Let sn be a sequence.
a. sn is said to be bounded from above if there is a constant M such that sn ≤ M for all n.
b. sn is said to be bounded from below if there is a constant K such that K ≤ sn for all n.
c. sn is said to be bounded if sn is both bounded from above and from below.
d. sn is said to be increasing (strictly increasing, respectively) if sn+1 ≥ sn (sn+1 > sn , respectively)
for all n.
e. sn is said to be decreasing (strictly decreasing, respectively) if sn+1 ≤ sn (sn+1 < sn , respectively) for all n.
f. sn is called monotone if it is either increasing or decreasing.
4
Example 10 Which of the following sequences are bounded from above? bounded from below?
increasing? decreasing?
a. 2, 4, 6, 8, 10, 12, . . .
b. 1, −1, 1, −1, 1, −1, . . .
c.
1
, − 32 , 43 , − 45 , 56 , − 67 , . . .
2
d. 1, 21 , 31 , 41 , 51 , 16 , . . .
e. cos 1, cos 2, cos 3, cos 4, cos 5, cos 6, . . .
Theorem 1 A convergent sequence is bounded.
Remark The converse of Theorem 1 is not true: consider sn = (−1)n .
Theorem 2 Let sn be a sequence.
a. If sn is increasing and bounded from above, then sn converges.
b. If sn is decreasing and bounded from below, then sn converges.
Example 11 Let sn = (1 + n1 )n . It is known that sn is increasing and sn < 4 for all n. Therefore,
lim sn exists. This limit is called
.
n→∞
5
Section 9.2 Geometric Series
Definition A finite series is the sum of finitely many terms of a sequence. For a given sequence
an , the sum of the first k terms, called the k th partial sum and denoted by Sk , is defined by
Sk =
k
X
ai = a1 + a2 + a3 + · · · + ak−1 + ak .
i=1
Example 1 Let an = 2n + 1. Then
5
X
ai = a1 + a2 + a3 + a4 + a5 = 3 + 5 + 7 + 9 + 11 = 35.
i=1
Remark
a. i is a dummy variable in the expression above, that is, we could have used
k
X
aj or
j=1
k
X
an to
n=1
express the same finite series.
b. The summation notation can be used in a different way. For example,
5
X
i2 = 32 + 42 + 52 = 50 and
i=3
X
p = 2 + 3 + 5 + 7 = 17.
p:prime
p≤10
c. Sn = Sn−1 + an , so an = Sn − Sn−1 .
Definition A geometric sequence or geometric progression (GP) is a sequence of the form
a, ar, ar2 , ar3 , . . .
Here a is called the initial term and r is called the common ratio.
Example 2 Consider the sequence 4, 2, 1, 12 , 41 , 18 , . . .. Check that it is a GP. Find the initial term
and the common ratio.
Remark
a. The general nth term of a GP with initial term a and common ratio r is given by an = arn−1 .
2
b. The sum of terms of a GP, a + ar + ar + · · · + ar
n−1
, can be written as
n
X
k=1
ark−1 .
6
Theorem 1 If r 6= 1, then
n
X
ark−1 = a + ar + ar2 + · · · + arn−1 =
k=1
a(rn − 1)
a(1 − rn )
=
.
r−1
1−r
Proof
Remark If r = 1, then
n
X
k=1
ark−1 = a
{z· · · + a} = na.
|+a+
n copies
Example 3 Compute each of the following sums:
a. 1 + 13 + 19 +
1
27
+
1
81
+
1
.
243
b. 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024.
Definition Let {an } be a sequence. An infinite series
∞
X
an is defined to be
n=1
∞
X
an = lim Sk ,
n=1
where Sk denotes the partial sum Sk =
k
X
k→∞
an = a1 + a2 + a3 + · · · + ak−1 + ak . In other words,
n=1
∞
X
n=1
an = lim Sk = lim
k→∞
k→∞
k
X
an .
n=1
The Geometric Series Test
Let an be a GP with initial term a and common ration r. Then
( a
∞
∞
X
X
,
if − 1 < r < 1
an =
arn−1 = 1 − r
diverges, otherwise
n=1
n=1
Proof
7
Example 4 Compute each of the following infinite geometric series, if it converges.
a. 1 − 13 + 19 −
1
27
+
b. 1 + 21 + 14 + 18 +
1
81
1
16
−
+
1
243
1
32
+ ···
+ ···
c. 1 − 2 + 4 − 8 + 16 − 32 + · · ·
Example 5 In the following picture, find the infinite sum
∞
X
Pn Pn+1 . Here ∠Pn Pn+1 Pn+2 is the
n=1
right angle for all n.
y
1
P2
P6
P3
O
P4
P5
√
P1
3
x
8
Section 9.3 Convergence of Series
Recall the following definition:
Definition An infinite series
∞
X
an is defined to be
n=1
∞
X
an = lim
k→∞
n=1
The k th partial sum Sk is defined to be Sk =
k
X
k
X
an .
n=1
an . Therefore,
n=1
∞
X
an = lim Sk .
n=1
k→∞
In this section and the next, we are mostly interested in determining whether a given infinite series
is convergent or divergent. We will develop several tests. Before we get started, note that changing
finitely many terms in a series does not change whether or not it converges, although it may change
the value of its sum if it does converge.
Example 1 Recall that
1+
1 1 1
1
1
1
+ + +
+
+
+ · · · = 2.
2 4 8 16 32 64
If you change it to
1
1
1
1
+2+
+
+
+ ··· ,
4
16 32 64
then still the new infinite sum converges, although the value of the sum is now different.
−1 + 3 +
Theorem 1 If
∞
X
n=1
an and
∞
X
bn both converge and if c is a constant, then
n=1
∞
∞
∞
X
X
X
a.
(an ± bn ) converges to
an ±
bn .
n=1
b.
∞
X
n=1
Proof
n=1
can converges to c
∞
X
n=1
an .
n=1
9
Example 2 Let c be a nonzero constant. If
∞
X
an diverges, then does
n=1
∞
X
can diverge as well?
n=1
The nth Term Test
a. If lim |an | =
6 0 or lim |an | does not exist, then
n→∞
n→∞
∞
X
an diverges.
n=1
b. If lim |an | = 0, then the test fails.
n→∞
Proof
Example 3 Does the series
∞
X
n+2
n=1
n
∞
∞
X
X
1
(−1)n
converge? How about
? Or
?
n
n2
n=1
n=1
The Integral Test
Suppose an = f (n), where f (x) is decreasing and positive for x ≥ 1.
Z ∞
∞
X
a. If
f (x) dx converges, then
an converges, too.
1
Z
b. If
∞
f (x) dx diverges, then
1
Proof
n=1
∞
X
n=1
an diverges, too.
10
∞
X
1
Example 4 Does the series
converge?
en
n=1
Example 5 Does the series
∞
X
n=2
1
converge?
n ln n
∞
X
1
converge?
Example 6 For what values of p does the series
np
n=1
We can summarize Example 6 as follows:
The p-series Test
The p-series
∞
X
1
converges if p > 1 and diverges if p ≤ 1.
p
n
n=1
11
Section 9.4 Tests for Convergence
In this section, we will develop more tests for convergence of a series.
Comparison Test
Suppose 0 ≤ an ≤ bn for all n.
a. If
∞
X
bn converges, then
n=1
b. If
∞
X
∞
X
an converges, too.
n=1
an diverges, then
n=1
∞
X
bn diverges, too.
n=1
Proof
Example 1 Decide whether the following series converge:
a.
∞
X
2n − 1
n=1
b.
n3
+3
∞
X
6n2 + 1
n=1
2n3 − 1
c.
∞
X
n=1
n2
1
+n
12
Example 2 Determine whether
∞
X
2n + 10
n=1
5n3 − 3
converges. Can we apply comparison test?
Limit Comparison Test
Suppose an > 0 and bn > 0 for all n.
∞
∞
X
X
an
bn either both converge or both diverge.
an and
= c with c > 0, then
n→∞ bn
n=1
n=1
a. If lim
an
= 0.
n→∞ bn
b. Suppose lim
i. If
ii. If
∞
X
n=1
∞
X
bn converges, then
an diverges, then
∞
X
an converges, too.
n=1
∞
X
bn diverges, too.
n=1
n=1
an
= ∞.
n→∞ bn
c. Suppose lim
i. If
ii. If
∞
X
n=1
∞
X
an converges, then
bn diverges, then
n=1
Proof
Example 3 Complete Example 2.
∞
X
n=1
∞
X
bn converges, too.
an diverges, too.
n=1
13
Example 4 Decide whether the following series converge:
a.
∞
X
n=1
n2 − 5
n3 + n + 4
c.
∞
X
ln n
n=1
∞
X
n2
∞
X
1
d.
ln n
n=2
1
b.
sin
n
n=1
We introduce several more tests. Note that the next three tests can be applied to series with both
positive and negative terms.
The Ratio Test
Suppose that the limit of
a. If L < 1, then
∞
X
|an+1 |
|an+1 |
exists, say lim
= L.
n→∞
|an |
|an |
an converges.
n=1
b. If L > 1 (including L = ∞), then
∞
X
n=1
c. If L = 1, then the test fails.
Proof
an diverges.
14
Example 5 Decide whether the following series converge:
a.
∞
X
n2
n=1
b.
2n
∞
X
(−1)n nn
n=1
n!
The Absolute Convergence Test
If
∞
X
|an | converges, then so does
∞
X
an .
n=1
n=1
Proof
∞ ∞
∞
X
X
(−1)n X
1
(−1)n
=
Example 6 Since
converges,
converges, too.
n2 n2
n2
n=1
n=1
n=1
15
Example 7 Decide whether the following series converge:
∞
X
(−1)n
√
a.
n
n
n=1
b.
∞
X
sin n
n=1
n2
The Alternating Series Test
If an
∞
X
is a positive decreasing sequence with lim an = 0, then
(−1)n an converges.
n→∞
Proof
Example 8 Decide whether
∞
X
(−1)n
n=1
n
converges.
∞
X
1 · 3 · 5 · · · · · (2n − 1)
Example 9 Show that
(−1)n
converges.
2 · 4 · 6 · · · · · 2n
n=1
n=1
16
Remark For any infinite series
∞
X
an , exactly one of the following is true,
n=1
a.
∞
X
|an | converges (so
n=1
b.
∞
X
c.
an also converges by the Absolute Convergence Test),
n=1
|an | diverges but
n=1
∞
X
∞
X
∞
X
an still converges, or
n=1
an diverges.
n=1
Definition
a. If
∞
X
|an | converges, then
n=1
b. If
∞
X
∞
X
an is said to converge absolutely.
n=1
an converges but
n=1
Example 10
∞
X
|an | diverges, then
n=1
∞
X
(−1)n
n=1
n
∞
X
an is said to converge conditionally.
n=1
∞
X
(−1)n
√ converges absolutely.
converges conditionally, while
n n
n=1
Example 11 Determine whether
∞
X
(−1)n
n=1
ln n
converge absolutely, conditionally, or diverges.
Remark There are still many convergence tests not discussed here, such as Root test, Kummer’s
test, Gauss’s test, Raabe’s test, and so on.
17
Summary
The Geometric Series Test
Let an be a GP with initial term a and common ration r. Then
( a
∞
∞
X
X
,
if − 1 < r < 1
an =
arn−1 = 1 − r
diverges, otherwise
n=1
n=1
The nth Term Test
a. If lim |an | =
6 0 or lim |an | does not exist, then
n→∞
n→∞
∞
X
an diverges.
n=1
b. If lim |an | = 0, then the test fails.
n→∞
The Integral Test
Suppose an = f (n), where f (x) is decreasing and positive for x ≥ 1.
Z ∞
∞
X
an converges, too.
a. If
f (x) dx converges, then
1
Z
b. If
n=1
∞
f (x) dx diverges, then
1
∞
X
an diverges, too.
n=1
The p-series Test
∞
X
1
The p-series
converges if p > 1 and diverges if p ≤ 1.
np
n=1
18
Comparison Test
Suppose 0 ≤ an ≤ bn for all n.
a. If
∞
X
bn converges, then
n=1
b. If
∞
X
∞
X
an converges, too.
n=1
an diverges, then
n=1
∞
X
bn diverges, too.
n=1
Limit Comparison Test
Suppose an > 0 and bn > 0 for all n.
∞
∞
X
X
an
bn either both converge or both diverge.
an and
= c with c > 0, then
a. If lim
n→∞ bn
n=1
n=1
an
= 0.
n→∞ bn
b. Suppose lim
i. If
ii. If
∞
X
n=1
∞
X
bn converges, then
an diverges, then
∞
X
an converges, too.
n=1
∞
X
bn diverges, too.
n=1
n=1
an
= ∞.
n→∞ bn
c. Suppose lim
i. If
ii. If
∞
X
n=1
∞
X
n=1
an converges, then
bn diverges, then
∞
X
n=1
∞
X
bn converges, too.
an diverges, too.
n=1
19
The Ratio Test
Suppose that the limit of
a. If L < 1, then
∞
X
|an+1 |
|an+1 |
exists, say lim
= L.
n→∞ |an |
|an |
an converges.
n=1
b. If L > 1 (including L = ∞), then
∞
X
an diverges.
n=1
c. If L = 1, then the test fails.
The Absolute Convergence Test
If
∞
X
|an | converges, then so does
∞
X
an .
n=1
n=1
The Alternating Series Test
If an
∞
X
(−1)n an converges.
is a positive decreasing sequence with lim an = 0, then
n→∞
n=1
20
Section 9.5 Power Series and Interval of Convergence
Definition A power series about x = a is a series of the form
∞
X
Cn (x − a)n = C0 + C1 (x − a) + C2 (x − a)2 + · · · + Cn (x − a)n + · · · ,
n=0
where Cn are constants. In this case, a is called the center of the power series.
Example 1 2 + (x − 1) +
and Cn = n1 for n ≥ 1.
Example 2 1 + x +
for all n.
x2
2
+
(x−1)2
2
+
(x−1)3
3
x3
6
x4
24
+
+
x5
120
+
(x−1)4
4
+ · · · is a power series about x = 1. Here C0 = 2
+ · · · is a power series with the center 0. Here Cn =
1
n!
Remark A power series is an infinite series which can be regarded as a polynomial of infinite degree. Since a power series is an infinite series with a variable x in it, its convergence depends on the
value of x. Our first interest is in determining the values of x for which a given power series converges.
Example 3 Determine whether the power series
∞
X
xn
n=0
2n
converges or diverges for
a. x = −1
b. x = 3
Example 4 For which values of x does the power series
∞
X
xn
n=0
2n
converge?
21
Example 5 For which values of x does the power series
∞
X
n!(x − 3)n converge?
n=0
Example 6 For which values of x does the power series
∞
X
(x + 2)n
n=0
n!
converge?
We observed that a power series about (x − a) converges for at least one x, say x = a. In fact, the
following is known:
Theorem 1 Let a power series
∞
X
Cn (x − a)n be given. Then exactly one of the following is true
n=0
for the power series:
a. There is a positive number R, called the radius of convergence, such that the series converges
for |x − a| < R and diverges for |x − a| > R.
b. The series converges only for x = a. In this case we say that the radius of convergence R equals
0.
c. The series converges for all values of x. In this case we say that the radius of convergence R
equals ∞.
Example 7 Determine the radius of convergence for the power series in Example 4,5,6 above.
22
There is an easy method for computing the radius of convergence:
Theorem 2 Let a power series
∞
X
Cn (x − a)n be given. Let an = Cn (x − a)n .
n=0
|an+1 |
= ∞, then R = 0.
n→∞ |an |
a. If lim
|an+1 |
= 0, then R = ∞.
n→∞ |an |
b. If lim
|an+1 |
1
= K|x − a|, then R = .
n→∞ |an |
K
c. If lim
Proof
Example 8 Show that the following power series converges for all x:
1+x+
x2 x3 x4
xn
+
+
+ ··· +
+ ···
2!
3!
4!
n!
Example 9 Determine the radius of convergence of the series
(x − 1) −
(x − 1)2 (x − 1)3 (x − 1)4
(x − 1)n
+
−
+ · · · + (−1)n+1
+ ···
2
3
4
n
23
Suppose that a power series
∞
X
Cn (x − a)n has a finite radius of convergence R. That is to say, we
n=0
know that the power series converges for |x − a| < R and diverges for |x − a| > R. In other words,
the power series converges for a − R < x < a + R and diverges for x < a − R or x > a + R. What
happens at the boundary points, say x = a − R and x = a + R? There is no simple theorem that
answers this question. To investigate the behavior of the power series at these boundary points, one
needs to plug the boundary points into the power series.
Example 10 We know that the radius of convergence of
(x − 1) −
(x − 1)n
(x − 1)2 (x − 1)3 (x − 1)4
+
−
+ · · · + (−1)n+1
+ ···
2
3
4
n
equals 1. This tells us that the power series converges for 0 < x < 2 and diverges for x < 0 or x > 2.
What happens if x = 0 or x = 2?
Definition The interval of convergence of a power series is the collection of all x values for which
the given power series converges.
Example 11 Let a power series
∞
X
Cn (x − a)n be given. If the radius of convergence R = ∞,
n=0
then the interval of convergence is (−∞, ∞). If R = 0, then the interval of convergence is a single
point {a}. If the radius of convergence R is positive number, then the interval of convergence is one
of the following intervals:
(a − R, a + R),
[a − R, a + R),
(a − R, a + R],
or [a − R, a + R].
Example 12 The interval of convergence of the power series in Example 10 is (0, 2].