Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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3.4 Absolute and Conditional Convergence 83

place of the c

s. Thus

∞

n=1

converges if

∞

n=1

does. However, 0 ≤ x

≤ |a

So by the Comparison Test (3.2.3), this series converges. The series for (c

) is

handled in the same way.

Now suppose that the sequence converges conditionally. Hence

∞

n=1

| =

∞

n=1

∞

n=1

| = +∞.

Therefore, at least one of the series

∞

n=1

and

∞

n=1

| diverges.

However, if one of these two series is convergent, we can obtain a contradiction.

For convenience, suppose that

∞

n=1

| converges to L. Since

∞

n=1

diverges, for

each R > 0, there is an integer N so that

n=1

> R + L.

Choose M sufﬁciently large so that the ﬁrst M terms of (a

)

∞

n=1

contains the ﬁrst

N terms of (b

). Then for all m ≥ M

k=1

≥

n=1

−

∞

k=1

| ≥ (R + L) − L = R.

Therefore, the series

∞

n=1

diverges, contrary to fact. This contradiction shows

that

∞

n=1

| must diverge. The case of

∞

n=1

converging is similar. ¥

3.4.8. REARRANGEMENT THEOREM.

∞

n=1

is a conditionally convergent series, then for every real number L, there

is a rearrangement that converges to L.

PROOF. Let us write the positive terms as b

, b

, . . . and the negative terms as

, c

, . . . . By Lemma 3.4.7,

∞

k=1

= +∞ and

∞

k=1

= −∞.

Also by Theorem 3.1.4,

lim

n→∞

= 0 = lim

n→∞

= lim

n→∞

84 Series

Now ﬁx a real number L. Choose the least positive integer m

such that

i=1

> L.

Then choose the least positive integer n

such that

i=1

j=1

< L.

We continue in this way, adding just enough positive terms to make the total greater

than L and then switching to negative terms until the total is less than L. In this

way, we deﬁne increasing sequences m

and n

to be the least positive integers

greater than m

k−1

and n

k−1

, respectively, such that

i=1

k−1

j=1

> L >

i=1

j=1

= v

This new series is

+···+b

+ c

+···+c

+ b

+···+b

+ c

+···+c

+ ··· .

We shall show that this series has limit L. From the construction of the new series,

we have the inequalities

− b

≤ L < u

and v

< L ≤ v

− c

Therefore,

L + c

≤ v

< L < u

≤ L + b

Since lim

n→∞

L + b

= lim

n→∞

L + c

= L, the Squeeze Theorem (Theorem 2.3.6)

shows that

lim

i→∞

= L = lim

j→∞

Finally, let s

be the partial sums of the new series. It is evident that

i−1

≤ s

≤ u

for m

i−1

+ n

i−i

≤ k ≤ m

+ n

i−1

and

≤ s

≤ u

for m

+ n

i−1

≤ k ≤ m

+ n

A second application of the Squeeze Theorem shows that our series converges to L

as desired. ¥

We complete this section with yet another convergence test. The proof uti-

lizes a rearrangement technique called summation by parts, which is analogous

to integration by parts.

3.4 Absolute and Conditional Convergence 85

3.4.9. SUMMATION BY PARTS LEMMA.

Suppose (x

) and (y

) are sequences of real numbers and deﬁne X

k=1

and

k=1

. Then

n=1

n+1

= X

m+1

PROOF. The argument is essentially an exercise in reindexing summations. Let

= 0 and notice that the left-hand side (LHS) equals

LHS =

n=1

− X

n−1

n=1

n+1

− Y

)

n=1

−

n=1

n−1

n=1

n+1

−

n=1

= −X

+ X

m+1

= X

m+1

Thus, providedthat lim

m→∞

m+1

exists, the two series

and

either both converge or both diverge.

3.4.10. DIRICHLET’S TEST.

Suppose that (a

)

∞

n=1

is a sequence of real numbers with bounded partial sums:

k=1

≤ M < ∞ for all n ≥ 1.

If (b

)

∞

n=1

is a sequence of positive numbers decreasing monotonically to 0, then

the series

∞

n=1

converges.

PROOF. We use the Summation by Parts Lemma to rewrite a

. Let y

= b

− b

n−1

for n > 1, and let x

= a

for all n. Deﬁne X

and Y

as in the

lemma. Observe that Y

= b

and so a

= x

Notice that |X

| =

k=1

≤ M for all n. Since |X

n+1

| ≤ Mb

n+1

, the

Squeeze Theorem shows that X

n+1

converges to zero. Furthermore,

k=1

k+1

| ≤

k=1

M|y

k+1

| = M (b

− b

n+1

) ≤ M b

86 Series

Thus

∞

k=1

k+1

converges absolutely. Using the Summation by Parts Lemma,

convergence follows from

∞

n=1

= lim

m→∞

n=1

= lim

m→∞

−

n=1

n+1

= −

∞

k=1

k+1

3.4.11. EXAMPLE. Consider the series

∞

n=1

sinnθ

for 0 ≤ θ ≤ 2π. At the

points θ = 0, π, and 2π, the series is 0. For θ in (0, 2π)\{π}, we will show that

this series converges conditionally.

Let a

= sinnθ and b

. To evaluate the partial sums of the a

s, use the

following trigonometric identities:

cos(k ±

)θ = coskθ cos

∓ sinkθ sin

whence

2sinkθ sin

= cos(k −

)θ − cos(k +

)θ.

Therefore, we obtain a telescoping sum

k=1

sinkθ =

2sin

k=1

cos(k −

)θ − cos(k +

)θ

csc

cos

− cos(n +

)θ

Consequently,

k=1

≤ csc

θ < ∞.

So Dirichlet’s Test applies and the series converges.

To see that convergence is conditional, it sufﬁces to show that most of the

terms behave like the harmonic series. Out of every two consecutive terms sinkθ,

at most one term has absolute value less than |sinθ|/2 (look at (cosa, sina) for

a = kθ, (k + 1)θ and use |sinθ|/2 ≤ min{|sin

|, |cos

|}). Thus, we obtain

|sin(2k − 1)θ|

2k − 1

|sin2kθ|

≥

|sinθ|

By comparison with a multiple of the harmonic series,

∞

k=1

|sinkθ|

diverges.

Exercises for Section 3.4

A. Find the series in Exercise 3.2.S that converge conditionally but not absolutely.

B. Decide which of the following series converge absolutely, conditionally, or not at all.

(a)

∞

n=1

(−1)

n log(n + 1)

(b)

∞

n=1

(−1)

(2 + (−1)

(c)

∞

n=1

(−1)

sin(

)

3.4 Absolute and Conditional Convergence 87

C. Compute the sum of the series

∞

n=1

(2n − 1)

given that

∞

n=1

HINT:

(2n−1)

2n(2n−1)

−

D. Show that the series

∞

n=1

cos(

2nπ

)

converges absolutely. Find the sum, given that

∞

n=1

= π

/6. (See Example 13.6.5.)

E. Show that a conditionally convergent series has a rearrangement converging to +∞.

F. Show that the Alternating Series Test is a special case of Dirichlet’s Test.

G. Use summation by parts to prove Abel’s Test: Suppose that

∞

n=1

converges and (b

)

is a monotonic convergent sequence. Show that

∞

n=1

converges.

H. Determine the values of θ for which the series

∞

n=1

inθ

log(n + 1)

converges.

I. Let a

(−1)

for (k − 1)

< n ≤ k

and k ≥ 1. Decide if the series

∞

n=1

converges.

CHAPTER 4

Topology of R

The space R

is the right setting for many problems in real analysis. For example,

in many situations, functions of interest depend on several variables. This puts us

into the realm of multivariable calculus, which is naturally set in R

. We will study

normed vector spaces further in Chapter 7, building on the properties and concepts

we study here. The space R

is the most important normed vector space, after the

real numbers themselves.

4.1. n-Dimensional Space

The space R

is the set of n-vectors x = (x

, x

, . . . , x

) with arbitrary real

coefﬁcients x

for 1 ≤ i ≤ n. Generally, vectors in R

will be referred to as points.

This space has a lot of structure, most of which should be familiar from advanced

calculus or linear algebra courses. In particular, we should mention that the zero

vector is (0, 0, . . . , 0), which we denote as 0.

First, it is a vector space. Recall (see Section 1.4) that vectors may be added

and subtracted and multiplied by (real) scalars. Indeed, for any x and y in R

and

scalars t ∈ R,

x + y = (x

, x

, . . . , x

) + (y

, y

, . . . , y

)

= (x

+ y

, x

+ y

, . . . , x

+ y

)

and

tx = t(x

, x

, . . . , x

) = (tx

, tx

, . . . , tx

Vector spaces are reviewed in Section 1.4, but we are assuming that you know the

basics of linear algebra. Instead, we concentrate on the properties of R

that build

on the ideas of distance and convergence.

There is the notion of length of a vector, given by

kxk = k(x

, x

, . . . , x

)k =

i=1

1/2

4.1 n-Dimensional Space 89

This is called the Euclidean norm on R

, and kxk is the norm of x. This conforms

to our usual notion of distance in the plane and in space. Moreover, it is the natural

consequence of the Euclidean distance in the plane using the Pythagorean formula

and induction on the number of variables. (See Exercise 4.1.A.) The distance

between two points x and y is then determined by

kx − yk =

i=1

− y

1/2

An important property of distance is the triangle inequality: The distance from

point A to point B and then on to a point C is at least as great as the direct distance

from A to C. This is interpreted geometrically as saying that the sum of the lengths

of two sides of a triangle is greater than the length of the third side (Figure 4.1).

(Equality can occur if the triangle has no area.)

x + y

FIGURE 4.1. The triangle inequality.

To verify this algebraically, we need an inequality involving the dot product,

which is useful in its own right. Recall that the dot product or inner product of

two vectors x and y is given by

hx, yi = h(x

, . . . , x

), (y

, . . . , y

)i =

i=1

There is a close connection between the inner product and the Euclidean norm

because of the evident identity

hx, xi = kxk

The inner product is linear in both variables:

hrx + sy, zi = rhx, zi + shy, zi for all x, y, z ∈ R

and r, s ∈ R

and

hx, sy + tzi = shx, yi + thx, zi for all x, y, z ∈ R

and s, t ∈ R

4.1.1. SCHWARZ INEQUALITY.

For all x and y in R

|hx, yi| ≤ kxkkyk.

Equality holds if and only if x and y are collinear.

90 Topology of R

PROOF. Let x = (x

, . . . , x

) and y = (y

, . . . , y

). Then

2kxk

kyk

− 2|hx, yi|

= 2

i=1

j=1

− 2

i=1

j=1

+ x

−

i=1

j=1

i=1

j=1

− 2x

+ x

i=1

j=1

− x

)

≥ 0.

This establishes the inequality because a sum of squares is positive.

Equality holds precisely when x

− x

= 0 for all i and j. If both x and y

equal 0, there is nothing to prove. So we may suppose that at least one coefﬁcient

is nonzero. There is no harm in assuming that x

6= 0, as the proof is the same in

all other cases. Then

for all 1 ≤ j ≤ n.

Hence y =

x. ¥

4.1.2. TRIANGLE INEQUALITY.

The triangle inequality holds for the Euclidean norm on R

kx + yk ≤ kxk + kyk for all x, y ∈ R

Moreover, equality holds if and only if either x = 0 or y = cx with c ≥ 0.

PROOF. Use the relationship between the inner product and norm to compute

kx + yk

= hx + y, x + yi

= hx, xi + hx, yi + hy, xi + hy, yi

≤ hx, xi + |hx, yi| + |hy, xi| + hy, yi

≤ kxk

+ kxkkyk + kxkkyk + kyk

kxk + kyk

If equality holds, then we must have hx, yi = kxkkyk. In particular, the

Schwarz inequality holds. So either x = 0 or y = cx. Substituting y = cx

into hx, yi = kxkkyk gives c = kyk/kxk ≥ 0. ¥

Colinearity does not imply equality for the triangle inequality in all cases be-

cause hx, yi could be negative. For example, x and −x are collinear for any

nonzero vector x, but

0 = kx + (−x)k < kxk + k − xk = 2kxk.

4.1 n-Dimensional Space 91

When we write elements of R

in vector notation, we are implicitly using the

standard basis {e

: 1 ≤ i ≤ n}, as deﬁned in Section 1.4. (Recall that e

is the

vector with a single 1 in the ith position and zeros in the other coordinates.) A set

, . . . , v

} in R

is orthonormal if hv

, v

i = δ

for 1 ≤ i, j ≤ m, where

= 0 when i 6= j and δ

= 1. If, in addition, {v

, . . . , v

} spans R

, it is called

an orthonormal basis. In particular, {e

, . . . , e

} is an orthonormal basis for R

4.1.3. LEMMA. Let {v

, . . . , v

} be an orthonormal set in R

. Then

i=1

1/2

An orthonormal set in R

is linearly independent. So an orthonormal basis for R

is a basis and has exactly n elements.

PROOF. Use the inner product to compute

i=1

j=1

, a

i=1

j=1

i=1

In particular, if

i=1

= 0, we ﬁnd that

i=1

= 0 and thus a

= 0 for

1 ≤ i ≤ m. This shows that {v

, . . . , v

} is linearly independent. Finally, a basis

for R

is a linearly independent set of vectors that spans R

. An orthonormal basis

spans by deﬁnition and is independent, as shown. A basic result of linear algebra

shows that every basis for R

has exactly n elements. ¥

Exercises for Section 4.1

A. Establish the Pythagorean formula: If x and y are orthogonal vectors, prove that

kx + yk =

kxk

+ kyk

1/2

B. (a) Suppose that x =

i=1

is a vector in R

with nonzero coefﬁcients only in the

ﬁrst j positions. Apply the Pythagorean formula to the orthogonal vectors x and

y = x

j+1

(b) Show by induction that the Pythagorean formula yields the norm of a vector in all

dimensions.

C. Show that kx + yk

+ kx −yk

= 2kxk

+ 2kyk

for all vectors x and y in R

. This

is called the parallelogram law. What does it mean geometrically?

D. Prove that if x and y are vectors in R

, then

kxk − kyk

≤ kx − yk.

E. Prove by induction that kx

+ ··· + x

k ≤ kx

k + ··· + kx

k for vectors x

in R

F. Suppose that x and y are unit vectors in R

. Show that if

x+y

= 1, then x = y.

92 Topology of R

G. Let x and y be two nonzero vectors in R

such that the angle between them is θ. Prove

that hx, yi = kxkkykcos θ.

HINT: If x makes the angle α to the positive x-axis, then x = (kxkcos α, kxksinα).

H. For nonzero vectors x and y in R

, deﬁne θ by kxkkykcos θ = hx, yi, and call this

the angle between them.

(a) Prove the cosine law: If x and y are vectors and θ is the angle between them, then

kx + yk

= kxk

+ 2kxkkykcos θ + kyk

(b) Prove that hx, yi can be deﬁned only in terms of norms of related vectors.

I. Suppose that U is a linear transformation from R

into R

that is isometric, meaning

that kUxk = kxk for all x ∈ R

(a) Prove that hUx, Uyi = hx, yi for all x, y ∈ R

(b) If {v

, . . . , v

} is an orthonormal set in R

, show that {Uv

, . . . , Uv

} is also

orthonormal.

J. (a) Let U be an isometric linear transformation of R

onto itself. Show that the n

columns of the matrix of U form an orthonormal basis for R

(b) Conversely, if {v

, . . . , v

} is an orthonormal basis for R

, show that the linear

transformation Ux =

i=1

is isometric.

K. Let M be a subspace of R

with an orthonormal basis {v

, . . . , v

}. Deﬁne a linear

transformation on R

by P x =

i=1

hx, v

(a) Show that P x belongs to M, and P y = y for all y ∈ M . Hence show P

= P .

(b) Show that hP x, x − P xi = 0.

= kP xk

+ kx − P xk

(d) If y belongs to M, show that kx − yk

= ky − P xk

+ kx − P xk

(e) Hence show that P x is the closest point in M to x.

4.2. Convergence and Completeness in R

The notion of norm for points in R

immediately allows us to discuss conver-

gence of sequences in this context. The deﬁnition of limit of a sequence of points

in R

is virtually identical to the deﬁnition of convergence in R. The only

change is to replace absolute value, which is the measure of distance in the reals,

with the Euclidean norm in n-space.

4.2.1. DEFINITION. A sequence of points (x

) in R

converges to a point a if

for every ε > 0, there is an integer N = N(ε) so that

− ak < ε for all k ≥ N.

In this case, we write lim

k→∞

= a.

The parallel between the two deﬁnitions of convergence allows us to reformu-

late the deﬁnition of limit of a sequence of points in n-space to the consideration

of a sequence of real numbers, namely the Euclidean norms of the points.