Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Calculus
Infinite Series I
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Outline
1
Sequences of Real Numbers
2
Infinite Series
3
Integral and Comparison Tests
4
Alternating Series
5
Absolute Convergence
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
What is a Sequence? (I)
In mathematics, we use the term sequence to mean an infinite
collection of real numbers, written in a specific order.
Example:
The use of Newton’s method with an initial guess x0 for solving
nonlinear equations like
tan x − x = 0
generates a sequence of successively improved
approximations,
x1 , x2 , x3 , · · · , xn , · · ·
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Basic Concepts and Notations (I)
By sequence, we mean any function whose domain is the set
of integers starting with some integer n0 (often 0 or 1).
Example:
The function
1
a(n) = ,
n
for n = 1, 2, 3, · · · ,
defines the sequence
1 1 1 1
, , , ,···
1 2 3 4
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Basic Concepts and Notations (II)
Consider the sequence
1
a(n) = ,
n
for n = 1, 2, 3, · · · ,
which generates a successively numbers
1 1 1 1
, , , ,···
1 2 3 4
We call
1
the 1st term,
1
1
the 2st term, and so on
2
1
Here, is called the general term.
n
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Basic Concepts and Notations (III)
Notations:
1
Function Notation:
a(n) for n = 1, 2, · · ·
2
Subscript Notation:
an for n = 1, 2, · · ·
3
Set Notation:
{an }∞
n=1 or {a1 , a2 , a3 , · · · }
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Terms of a Sequence
Example (1.1)
Write out the terms of the sequence whose general term is
n+1
given by an =
, for n = 1, 2, 3, . . . .
n
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Graphing a Sequence
To graph a sequence, we plot a number of discrete points,
since a sequence is a function defined only on the integers.
Figure: [7.1] an =
1
.
n2
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Limits and Sequences (I)
Consider the sequence
an =
1
n2
for n = 1, 2, 3, · · ·
It is observed that
1
→0
n2
as
n→0
In this case, we say that the sequence converges to 0 and write
1
=0
n→∞ n2
lim an = lim
n→∞
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Limits and Sequences (II)
In general, we say that the sequence {an }∞
n=1 converges to L,
i.e.,
lim an = L
n→∞
if we can make an a close to L as desired, simply by making n
sufficiently large.
This concept of the limit of a sequence is similar to that used in
the definition of the limit
lim f (x) = L
x→∞
for a function of a real variable x.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Rules for Computing Limits of a Sequence
Most of the usual rules for computing limits of functions of a
real variables also apply to computing the limit of a sequence.
Theorem (1.1)
∞
Suppose that {an }∞
n=n0 and {bn }n=n0 both converge. Then
1
lim (an + bn ) = lim an + lim bn ,
n→∞
2
n→∞
n→∞
lim (an − bn ) = lim an − lim bn ,
n→∞
n→∞
lim (an bn ) = lim an
lim bn and
n→∞
3
n→∞
n→∞
lim an
4
n→∞
an
= n→∞ (assuming lim bn 6= 0).
n→∞ bn
n→∞
lim bn
lim
n→∞
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Finding the Limit of a Sequence
Example (1.2)
Evaluate lim
n→∞
5n + 7
.
3n − 5
Figure: [7.2] an =
5n + 7
.
3n − 5
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Divergent Sequence
Example (1.3)
n2 + 1
.
n→∞ 2n − 3
Evaluate lim
Figure: [7.3] an =
n2 + 1
.
2n − 3
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Divergent Sequence Whose Terms
Do Not Tend to ∞
Example (1.4)
Determine the convergence or
divergence of the sequence
{(−1)n }∞
n=1 .
Figure: [7.4] an = (−1)n .
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Finding lim f (n) by Computing lim f (x)
n→∞
x→∞
Theorem (1.2)
Suppose that lim f (x) = L. Then lim f (n) = L, also.
x→∞
n→∞
A graphical representation of the theorem:
Figure: [7.5] an = f (n), where f (x) → 2, as x → ∞.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
lim f (n) = L ; lim f (x) = L
n→∞
x→∞
The converse of Theorem 1.2 is false, i.e.,
lim f (n) = L
n→∞
need not imply
lim f (x) = L
x→∞
For example,
lim cos(2πn) = 1, but lim cos(2πx) does not exist
n→∞
Figure: [7.6] an = cos(2πn).
x→∞
Figure: [7.7] y = cos(2πx).
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Applying L’Hôpital’s Rule to a Related Function
Example (1.5)
n+1
.
n→∞ en
Evaluate lim
Figure: [7.8] an =
n+1
.
en
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Squeeze Theorem
Theorem (1.3)
∞
Suppose {an }∞
n=n0 and {bn }n=n0 are convergent sequences, both
converging to the limit, L. If there is an integer n1 ≥ n0 such that
for all n ≥ n1 , an ≤ cn ≤ bn , then {cn }∞
n=n0 converges to L, too.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Applying the Squeeze Theorem to a Sequence
Example (1.6)
Determine theconvergence
or
sin n ∞
divergence of
n2 n=1
Figure: [7.9] an =
sin n
.
n2
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Consequence of the Squeeze Theorem
Corollary (1.1)
If lim |an | = 0, then lim an = 0, also.
n→∞
n→∞
This fact is practically useful for sequences with both positive
and negative terms.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Sequence with Terms of Alternating Signs
Example (1.7)
Determine theconvergence
or
(−1)n ∞
divergence of
.
n
n=1
Figure: [7.10] an =
(−1)n
.
n
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Finding Limits of a Sequence involving n!
Definition (1.1)
For any integer n ≥ 1, the factorial, n! is defined as the product
of the first n positive integers,
n! = 1 · 2 · 3 · · · · · n.
We define 0! = 1.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
An Indirect Proof of Convergence
Example (1.8)
Investigate
the convergence of
∞
n!
.
nn n=1
Figure: [7.11] an =
n!
.
nn
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Increasing and Decreasing Sequences
The sequence {an }∞
n=1 is increasing if
a1 ≤ a2 ≤ · · · ≤ an ≤ an+1 ≤ · · · .
The sequence {an }∞
n=1 is decreasing if
a1 ≥ a2 ≥ · · · ≥ an ≥ an+1 ≥ · · · .
If a sequence is either increasing or decreasing, it is called
monotonic.
A useful method to show that a sequence an is monotonic
is by looking at the ratio of the two successive terms an and
an+1 for all n.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
An Increasing Sequence
Example (1.9)
Investigatewhether
∞the
n
sequence
is
n + 1 n=1
increasing, decreasing or
neither.
Figure: [7.12] an =
n
.
n+1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Sequence That Is Increasing for n ≥ 2
Example (1.10)
Investigatewhether
the
n! ∞
sequence
is
en n=1
increasing, decreasing or
neither.
Figure: [7.13] an =
n!
.
en
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Bounded Sequences
We say that a sequence is bounded if there is a number M > 0
(called a bound) for which
|an | ≤ M
for all n.
A bound is not the same as a limit, although the two may
coincide.
A sequence may have a number of bounds.
A sequence may have only one limit or no limit.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Bounded Sequence
Example (1.11)
Show that the sequence
3 − 4n2
n2 + 1
∞
is bounded.
n=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Properties of Sequences and Convergence
Theorem (1.4)
Every bounded, monotonic sequence converges.
Figure: [7.14a] A bounded and
increasing sequence.
Figure: [7.14b] A bounded and
decreasing sequence.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
An Indirect Proof of Convergence
Example (1.12)
Investigate theconvergence
of
2n ∞
the sequence
.
n! n=1
Figure: [7.15] an =
2n
.
n!
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Infinite Series: An Example
Consider the repeating decimal expansion of
1
3
1
= 0.3333333̄
3
Alternatively, we think of this expansion as
1
3
= 0.3 + 0.03 + 0.003 + 0.0003 + · · ·
= 3(0.1) + 3(0.1)2 + 3(0.1)3 + · · · + +3(0.1)k + · · ·
For convenience, we write the above expression using
summation notation as
∞
1 X
=
3(0.1)k
3
k=1
(an infinite sum?)
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Meaning of a Infinite Sum
Consider the expression
∞
1 X
=
3(0.1)k
3
k=1
We can add two things at a time but we can not add infinite
many things together.
The infinite sum expression means that as you add
together more and more terms, the terms gets closer and
closer to 1/3
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Partial Sums (I)
Consider a sequence {ak }∞
k=1 . We define the individual sums by
S1 = a1
S2 = a1 + a2 = S1 + a2
S3 = a1 + a2 +a3 = S2 + a3
| {z }
S2
S4 = a1 + a2 + a3 +a4 = S3 + a4
{z
}
|
S3
..
.
Sn = a1 + a2 + · · · + an−1 +an = Sn−1 + a4
{z
}
|
Sn−1
and so on. We refer to Sn as the nth partial sum.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Partial Sums (II)
Consider the sequence
S1 =
S2 =
S2 =
S4 =
1
2k
∞
. The partial sums are:
k=1
1
2
1
1
3
1
+ 2 = =1− 2
2 2
4
2
3
1
7
1
+ 3 = =1− 3
4 2
8
2
1
15
1
7
+ 4 =
=1− 4
8 2
16
2
and so on. We notice that
n
X
1
1
lim
= lim Sn = lim 1 − n = 1
n→∞
n→∞
n→∞
2k
2
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Partial Sums (III)
The expression
n
X
1
1
lim
= lim Sn = lim 1 − n = 1
n→∞
n→∞
n→∞
2k
2
k=1
implies thatas we add together more and more terms of the
1 ∞
sequence
, the partial sums are drawing closer and
2k k=1
closer to 1. In this instance, we write
∞
X
1
=1
2k
k=1
∞
X
k=1
1
is called a series (or infinite series)
2k
∞
X
1
is not a sum in the usual sense, but rather, the limit
2k
k=1
of the sequence of partial sums.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Definition of Infinite Series
Definition
For any sequence {an }∞
k=1 , we can write down the series
a1 + a2 + · · · + ak + · · · =
∞
X
ak
k=1
If the sequence of partial sums Sn =
n
X
ak
k=1
∞
X
number S, then we say that the series
converges to some
ak converges to S.
k=1
We write
∞
X
k=1
ak = lim ak = lim Sn = S
n→∞
n→∞
We say that the series diverges if lim Sn doesn’t exist.
n→∞
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Convergent Series
Example (2.1)
Determine if the series
∞
X
1
converges or diverges.
2k
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Divergent Series
Example (2.2)
Investigate the convergence or divergence of the series
∞
X
k=1
k2 .
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A (Telescope) Series with a Simple Expression for the
Partial Sums
Example (2.3)
Investigate the convergence or
divergence of the series
∞
X
1
.
k(k + 1)
k=1
Figure: [7.16] Sn =
n
X
k=1
1
.
k(k + 1)
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Geometric Series
Theorem (2.1)
For a 6= 0, the geometric series
(
a
∞
X
if |r| < 1
converges to
k
1−r
ar
diverges if |r| ≥ 1
k=0
Here, r is referred to as the ratio.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Convergent Geometric Series
Example (2.4)
Investigate the convergence or
divergence of the series
k
∞
X
1
.
5
3
k=2
n+1 k
X
1
.
Figure: [7.17] Sn =
5
3
k=2
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Divergent Geometric Series
Example (2.5)
Investigate the convergence or
divergence of the series
k
∞
X
7
6 −
.
2
k=0
k
n−1 X
7
Figure: [7.18] Sn =
6 −
.
2
k=0
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
KTH-Term Test for Divergence (I)
To determine whether a series is convergent of divergent
usually involves a lot of hard work. Here, we introduce a
method called KTH-term test for divergence.
Theorem (2.2)
If
∞
X
k=1
ak converges, then lim ak = 0.
k→∞
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
KTH-Term Test for Divergence (II)
The following result follows directly from Theorem 2.2.
Theorem (kth-term test for divergence)
If lim ak 6= 0, then the series
k→∞
∞
X
ak diverges.
k=1
The kth-term test is very simple to use since all we have to
do is check whether
ak → 0 or not as k → ∞.
Note that
ak → 0 doesn’t imply
∞
X
k=1
and we need additional testing.
ak converges,
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Series Whose Terms Do Not Tend to Zero
Example (2.6)
Investigate the convergence or
divergence of the series
∞
X
k
.
k+1
k=1
Figure: [7.19] Sn =
n
X
k=1
k
.
k+1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Harmonic Series (I)
An Example where the kth-Term Test Tells Nothing about the Convergence
Example (2.7)
Investigate the convergence or
divergence of the harmonic
∞
X
1
series:
.
k
k=1
Figure: [7.20] Sn =
n
X
1
k=1
k
.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Harmonic Series (II)
The Integral Test
Note that the nth partial sum
Sn =
n
X
1
k=1
k
=
1 1 1
1
+ + + ··· +
1 2 3
n
corresponds to the sum of the area of
the n rectangles superimposed on the
graph y = 1/x.
Figure: [7.21] y =
1
.
x
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Harmonic Series (III)
The Integral Test
Notice that since each of the indicated
rectangles lies partly above the curve,
we have
Sn = Sum of areas of n rectangles
≥ Area under the curve
Z n+1
dx
=
x
1
n+1
= ln |x|
= ln(n + 1)
1
Figure: [7.21] y =
1
.
x
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Harmonic Series (IV)
The Integral Test
Since
Sn ≥ ln(n + 1)
and
lim ln(n + 1) = ∞
n→∞
we conclude that
∞
X
1
k=1
k
= lim Sn = ∞
n→∞
Figure: [7.21] y =
1
.
x
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Some Useful Theorems for Determining the
Convergence of a Sequence
Theorem (2.3)
1
If
∞
X
ak converges to A and
k=1
series
∞
X
bk converges to B, then the
k=1
(ak ± bk ) converges to A ± B and
k=1
2
∞
X
∞
X
(cak )
k=1
converges to cA, for any constant, c.
∞
∞
∞
X
X
X
If
ak converges and
bk diverges, then
(ak ± bk )
k=1
diverges.
k=1
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
An Overview
In general we are unable to determine the sum of a convergent
series. In fact, for most series we can not determine whether
they converges and diverge or not by simply looking at the
sequence of partial sums. Most of the time, we will need to test
a series for convergence in some indirect way. If we find that
the series is convergent, we can then approximate its sum by
numerically computing some partial sums. And this section is
devoted to the development of additional testing methods for
convergence of series. The two methods introduced in this
section are:
1
The Integral Test
2
Comparison Tests
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Integral Test (I)
For a given series
∞
X
ak , suppose that
k=1
there is a function f for which
f (k) = ak , for k = 1, 2, · · · ,
where f is continuous, decreasing and
f (x) ≥ 0 for all x ≥ 1. Consider the n-th
partial sum
Sn =
n
X
ak = a1 + a2 + a3 + · · · + an .
{z
}
|
k=1
sum of areas of
n − 1 rectangles
Figure: [7.22] (n − 1)
rectangles, lying beneath
the curve.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Integral Test (II)
Notice that
Z
n
0≤
f (x)dx
Sn − a1
≤
.
| {z }
1
| {z }
Sum of areas of
Area under the
n − 1 rectangles
curve
Z ∞
Suppose that
f (x)dx converges.
Then
1
Z
0 ≤ Sn − a1 ≤
n
Z
f (x)dx ≤
1
∞
f (x)dx
1
Figure: [7.22] (n − 1)
rectangles, partially above
the curve.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Integral Test (III)
Equivalently
Z
a1 ≤ Sn ≤ a1 +
∞
f (x)dx
1
This implies that
{Sn }∞
n=1 is bounded
Moreover, {Sn }∞
n=1 is a increasing
sequence. We conclude that {Sn }∞
n=1 is
convergent, and so does
∞
X
k=1
ak
Figure: [7.22] (n − 1)
rectangles, partially above
the curve.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Integral Test (IV)
A similar approach leads to
Z n
.
0≤
f (x)dx
≤
Sn−1
|{z}
1
| {z }
Sum of areas of
Area under the
n
− 1 rectangles
curve
Z
∞
If
f (x)dx diverges, then
1
Z
lim
n→∞ 1
n
f (x)dx = ∞ ⇒ lim Sn−1 = ∞
n→∞
and consequently
∞
X
n=1
ak = ∞
Figure: [7.23] (n − 1)
rectangles, partially above
the curve.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Integral Test
Theorem (3.1)
If f (k) = ak for each k = 1, 2, . . . and f is continuous, decreasing
Z ∞
∞
X
and f (x) ≥ 0, for x ≥ 1, then
ak either both
f (x)dx and
1
converge or both diverge.
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Integral Test
Example (3.1)
Investigate the convergence or
divergence of the series
∞
X
1
.
k2 + 1
k=0
Figure: [7.24] Sn =
n−1
X
k=0
k2
1
.
+1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The p-Series (I)
Example (3.2)
Determine for which values of p the series
converges.
∞
X
1
(a p-series)
kp
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The p-Series (II)
Theorem
∞
X
1
converges if p > 1 and diverges if p ≤ 1.
The p-series
kp
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Remainder
Consider a series S =
partial sum Sn =
n
X
∞
X
ak and its n-th
k=1
ak . we define the
k=1
remainder Rn as
Rn = S − Sn =
=
∞
X
ak −
k=1
∞
X
n
X
ak
k=1
ak
k=n+1
| {z }
sum of the indicated
rectangles (see figure)
Notice that Rn is the error in
approximating S by Sn .
Figure: [7.25] Estimate of
the remainder.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Error Estimate for the Integral Test
Theorem (3.2)
Suppose that f (k) = ak for all
k = 1, 2, . . ., where f is continuous,
decreasing and f (x) ≥Z0 for all x ≥ 1.
∞
Further, suppose that
f (x)dx
1
converges. Then, the remainder Rn
satisfies
Z ∞
∞
X
0 ≤ Rn =
ak ≤
f (x)dx.
k=n+1
n
Figure: [7.25] Estimate of
the remainder.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Estimating the Error in a Partial Sum
Example (3.3)
Estimate the error in using the partial sum S100 to approximate
∞
X
1
.
the sum of the series
k3
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Finding the Number of Terms Needed for a Given
Accuracy
Example (3.4)
Determine the number of terms needed to obtain an
∞
X
1
correct to within
approximation to the sum of the series
k3
10−5 .
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Comparison Test
Theorem (3.3)
Suppose that 0 ≤ ak ≤ bk , for all k.
∞
∞
X
X
1
If
bk converges, then
ak converges, too.
k=1
2
If
∞
X
k=1
k=1
ak diverges, then
∞
X
k=1
bk diverges, too.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Comparison Test for a Convergent Series
Example (3.5)
Investigate the convergence or
∞
X
1
divergence of
.
k3 + 5k
k=1
Figure: [7.26] Sn =
n
X
k=1
1
.
k3 + 5k
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Comparison Test for a Divergent Series
Example (3.6)
Investigate the convergence or
∞
X
5k + 1
divergence of
.
2k − 1
k=1
Figure: [7.27] Sn =
n
X
5k + 1
k=1
2k − 1
.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Comparison That Does Not Work
Example (3.7)
Investigate the convergence or
divergence of the series
∞
X
1
.
3
k − 5k
k=3
Figure: [7.28] Sn =
n
X
k=3
1
.
k3 − 5k
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Limit Comparison Test
Theorem (3.4)
Suppose that ak , bk > 0 and that for some (finite) value, L,
∞
∞
X
X
ak
ak and
bk both converge or
lim
= L > 0. Then,either
k→∞ bk
k=1
k=1
they both diverge.
The Limit Comparison Test can be used to resolve
convergence questions for a great number of series. The
first step in using this (like the Comparison Test) is to find
another series (whose convergence or divergence is
known) that "looks like" the series in question.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Limit Comparison Test
Example (3.8)
Investigate the convergence or divergence of the series
∞
X
1
.
3
k − 5k
k=3
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Limit Comparison Test
Example (3.9)
Investigate the convergence or
divergence of the series
∞
X
k2 − 2k + 7
.
k5 + 5k4 − 3k3 + 2k − 1
k=1
Figure: [7.29]
n
X
k2 − 2k + 7
Sn =
.
k5 + 5k4 − 3k3 + 2k − 1
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Alternating Series: An Overview
Consider a sequence {ak }∞
k=1 and ak > 0 for all k.
An alternating series is a series of the form
∞
X
(−1)k+1 ak = a1 − a2 + a3 − a4 + a5 − a6 + · · · ,
k=1
i.e., a series whose terms alternate back and forth from positive
to negative.
The reasons for studying the convergence or divergence of an
alternative series are:
1
Alternating series appear frequently in applications.
2
Alternating series are surprisingly simply to deal with and
studying them will yield significant insight into how series
works.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Alternating Harmonic Series (I)
Example (4.1)
Investigate the convergence or
divergence of the alternating
harmonic series
∞
X
(−1)k+1
k=1
k
=1−
1 1 1
+ − + ··· .
2 3 4
The partial sums of the series
suggest that the series might
converges to about 0.7.
Figure: [7.30]
Sn =
n
X
(−1)k+1
.
k
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Alternating Harmonic Series (II)
The first eight partial sums are:
S5 =
7
1
47
+ = ,
15 5
60
1
1
= ,
2
2
S6 =
37
47 1
− = ,
60 6
60
S3 =
1 1
5
+ = ,
2 3
6
S7 =
37 1
319
+ =
,
60 7
420
S4 =
5 1
7
319 1
533
− = , S8 =
− =
,
6 4
12
420 8
840
S1 = 1,
S2 = 1 −
Figure: [7.31] Partial sums of
∞
X
(−1)k+1
.
k
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
The Alternating Harmonic Series (III)
We observe that
1
the partial sums are bouncing back and forth and
2
the partial sums seem to be zeroing-in on some value.
Figure: [7.31] Partial sums of
∞
X
(−1)k+1
.
k
k=1
To definitely resolve the question of convergence of the series
we need the following theorem.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Alternating Series Test (I)
Theorem (4.1)
Suppose that lim ak = 0 and 0 < ak+1 ≤ ak for all k ≥ 1. Then,
k→∞
the alternating series
∞
X
k=1
(−1)k+1 ak converges.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Alternating Series Test (II)
S1 = 0 + a1
S2 = S1 − a2 (a2 < a1 ⇒ 0 < S2 < S1 )
S3 = S2 + a3 (a3 < a2 ⇒ S2 < S3 < S1 )
Continuing in this fashion, we get
S2 < S4 < S6 < · · · < S5 < S3 < S1 .
Figure: [7.32] Convergence of the partial sums of an alternating
series.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Alternating Series Test (II)
1
2
All of the odd-index partial sums (S2n+1 for n = 1, 2, ...) are
larger than all of the even-index partial sums (S2n for
n = 1, 2, ...).
As the partial sums oscillate back and forth, they should be
converging to some limit S, and
S2 < S4 < S6 < · · · < S < · · · < S5 < S3 < S1 .
Figure: [7.32] Convergence of the partial sums of an alternating
series.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Alternating Series Test
Example (4.2)
Reconsider the convergence of the alternating harmonic series
∞
X
(−1)k+1
.
k
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Alternating Series Test
Example (4.3)
Investigate the convergence or
divergence of the alternating
∞
X
(−1)k (k + 3)
series
k(k + 1)
k=1
Figure: [7.33]
Sn =
n
X
(−1)k (k + 3)
.
k(k + 1)
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Divergent Alternating Series
Example (4.4)
Determine whether the alternating series
∞
X
(−1)k k
k=1
or diverges.
k+2
converges
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Estimating the Sum of an Alternating Series (I)
Once we know that a series converges, we can always
approximate the sum of the series by computing some
partial sums.
However, in finding an approximate sum of a convergence
sum, how close is close enough?
In what follow, we give an estimate for the error in
approximating a alternative series by the n-th partial sum.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Estimating the Sum of an Alternating Series (II)
Consider a convergent alternating series
S=
∞
X
(−1)k ak
k=1
We have shown that
S2 < S4 < S6 < · · · < S < · · · < S5 < S3 < S1 .
For even n
Sn ≤ S ≤ Sn+1 ⇒ 0 ≤ S − Sn ≤ Sn+1 − Sn = an+1
Since an > 0, we have
−an+1 ≤ 0 ≤ S − Sn ≤ an+1 ⇒ |S − Sn | ≤ an+1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Estimating the Sum of an Alternating Series (III)
From
S2 < S4 < S6 < · · · < S < · · · < S5 < S3 < S1 .
and for n odd, we have
Sn+1 ≤ S ≤ Sn
Subtracting Sn , we get
−an+1 ≤ Sn+1 − Sn ≤ S − Sn ≤ 0 ≤ an+1
or
|S − Sn | ≤ an+1 , for n odd
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Estimating the Sum of an Alternating Series (IV)
Theorem (4.2)
Suppose that lim ak = 0 and 0 < ak+1 ≤ ak for all k ≥ 1. Then,
k→∞
the alternating series
∞
X
(−1)k+1 ak converges to some number
k=1
S and the error in approximating S by the nth partial sum Sn
satisfies
|S − Sn | ≤ an+1 .
The theorem states that the absolute value of the error in
approximating S by Sn does not exceed an+1 , the absolute
value of the first neglected term.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Estimating the Sum of an Alternating Series (V)
Example (4.5)
Approximate the sum of the alternating series
∞
X
(−1)k+1
k=1
the 40th partial sum and estimate the error in this
approximation.
k4
by
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Finding the Number of Terms Needed for a Given
Accuracy
Example (4.6)
For the convergent alternating series
∞
X
(−1)k+1
k=1
k4
, how many
terms are needed to guarantee that Sn is within 1 × 10−10 of the
actual sum S?
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
An Overview (I)
So far, except the Alternating Series Test, the other tests for
convergence of series apply only to series all of whose terms
are positive.
What do we do if we are faced with a series that has both
positive and negative terms, but that is not an alternating
series? For instance, the series
∞
X
sin k
k=1
k3
= sin 1 +
1
1
1
sin 2 +
sin 3 +
sin 4 + · · · ,
8
27
64
has both positive and negative terms, but the terms do not
alternate signs.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Absolutely and Conditionally Convergent
To analyze such a series, we introduce the following definition.
Definition
Consider a general series
∞
X
ak that has both positive and
k=1
negative terms.
∞
X
1
If
|ak | converges then we say that the original series is
k=1
2
absolutely convergent (or converges absolutely).
∞
X
If
ak is convergent but not absolute convergent, then
k=1
the original series is conditionally convergent.
Notice that we can use methods
P introduced previously to test
the convergence of the series ∞
k=1 |ak |.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Testing for Absolute Convergence
Example (5.1)
Determine if
∞
X
(−1)k+1
k=1
2k
Figure: [7.34] Sn =
is absolutely convergent.
n
X
(−1)k+1
.
2k
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Conditionally Convergent Series
Example (5.2)
Determine if the alternating harmonic series
∞
X
(−1)k+1
k=1
is absolutely convergent.
k
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Convergence and Absolutely Convergence (I)
Theorem (5.1)
If
∞
X
k=1
|ak | converges, then
∞
X
ak converges.
k=1
The result says that is a series converges absolutely, then
it must also converge. So when we test series, we first test
for absolute convergence, and if the series converges
absolutely, then we need not test any further to establish
convergence.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Convergence and Absolutely Convergence (II)
Proof of the Theorem 5.1
For any k, we have
−|ak | ≤ ak ≤ ak
Adding |ak | to all the terms, we get
0 ≤ ak + |ak | = bk ≤ 2|ak |.
Since
∞
X
ak is absolutely convergent, the series
k=1
is also convergent. By the Comparison Test,
∞
X
∞
X
2|ak | = 2
k=1
∞
X
|ak |
k=1
bk is convergent.
k=1
Observe that we may write
∞
X
k=1
ak =
∞
X
k=1
(ak + |ak | − |ak |) =
∞
X
k=1
bk −
∞
X
|ak |
k=1
Since the two series on the right-hand side are convergent, it follows
∞
X
that
ak must also be convergent.
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Testing for Absolute Convergence
Example (5.3)
Determine whether
∞
X
sin k
k=1
k3
Figure: [7.35]Sn =
is convergent or divergent.
n
X
sin k
.
k3
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Ratio Test (I)
We now introduce a very powerful tool for testing a series for
absolute convergence.
Theorem (5.2)
Given
∞
X
k=1
ak+1 = L.
ak , with ak =
6 0 for all k, suppose that lim k→∞ ak Then,
1
if L < 1, the series converges absolutely,
2
if L > 1 (or L = ∞), the series diverges and
3
if L = 1, there is no conclusion.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Ratio Test (II)
Recall that in the statement of the Ratio Test, we said that if
ak+1 =1
lim
k→∞ ak then the Ratio Test yields no conclusion. By this we mean
that in such cases, the series may or may not converge
and further testing is need.
The Ratio Test is particularly useful when the general term
of a series contains an exponential terms or a factorial.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Ratio Test
Example (5.4)
Test
∞
X
(−1)k k
k=1
2k
Figure: [7.36] Sn =
for convergence.
n
X
k
.
k
2
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Ratio Test
Example (5.5)
Test
∞
X
(−1)k k!
k=0
ek
Figure: [7.37] Sn =
for convergence.
n−1
X
(−1)k k!
.
ek
k=0
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Divergent Series for Which the Ratio Test Fails
Example (5.6)
Use the Ratio Test for the harmonic series
∞
X
1
k=1
k
.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
A Convergent Series for Which the Ratio Test Fails
Example (5.7)
Use the Ratio Test to test the series
∞
X
1
.
k2
k=0
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Root Test
We now present one final test for convergence of series.
Theorem (5.3)
Given
∞
X
k=1
ak , suppose that lim
k→∞
p
k
|ak | = L. Then,
1
if L < 1, the series converges absolutely,
2
if L > 1 (or L = ∞), the series diverges and
3
if L = 1, there is no conclusion.
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence
Using the Root Test
Example (5.8)
Use the Root Test to determine the convergence or divergence
of the series
∞ X
2k + 4 k
.
5k − 1
k=1
Sequences of Real Numbers Infinite Series Integral and Comparison Tests Alternating Series Absolute Convergence