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

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2.4 Basic Properties of Limits 43

In the sequence (a

)

∞

n=1

, we ignore terms where b

= 0. There is no

problem doing this because M 6= 0 implies that b

6= 0 for all sufﬁciently large n.

(We use “for all sufﬁciently large n” as shorthand for saying there is some N so

that this holds for all n ≥ N .) We will return to this point in the proof of (4).

PROOF. (1) Notice that

|(a

+ b

) − (L + M)| = |a

− L + b

− M| ≤ |a

− L| + |b

− M|.

Since lim

n→∞

= L, we can ﬁnd N

> 0 so that

− L| <

for all n ≥ N

Similarly, we can ﬁnd N

> 0 so that

− M| <

for all n ≥ N

Thus, if n ≥ max{N

, N

}, then

|(a

+ b

) − (L + M)| ≤ |a

− L| + |b

− M|

= ε

(2) Here, we have

|αa

− αL| = |α||a

− L|.

We employ a simple trick to avoid dividing by 0. Using the deﬁnition of limit, we

can ﬁnd N so that

− L| <

|α| + 1

for all n ≥ N.

Thus, if n ≥ N ,

|αa

− αL| ≤ |α||a

− L| < |α|

|α| + 1

< ε.

So (αa

)

∞

n=1

converges to αL, as required.

(3) First we simplify the difference we are trying to control. Observe that

− LM| = |a

− Lb

+ Lb

− LM|

≤ |a

− Lb

| + |Lb

− LM|

= |a

− L||b

| + |L||b

− M|.

We would like to repeat the method of part (2) for each term on the right-hand side.

For the second term, there is no problem. Choose N

so that

− M| <

2|L| + 1

for all n ≥ N

For the ﬁrst term, there is a problem: |b

| is not a constant. However, since

)

∞

n=1

converges to M , Proposition 2.4.2 implies there is a number B so that

| ≤ B for all n. Thus,

− LM| ≤ |a

− L||b

| + |L||b

− M| ≤ |a

− L|B + |L||b

− M|.

44 The Real Numbers

Now we can choose N

so that

− L| <

2B + 1

for all n ≥ N

Putting everything together, if n ≥ max{N

, N

}, then

− LM| < |a

− L|B + |L||b

− M|

2B + 1

B + |L|

2|L| + 1

= ε.

(4) Although the algebra is more complicated, the same strategy works here.

Assuming b

6= 0, we can write

−

M − Lb

M − LM + LM − Lb

≤

M − LM

LM − Lb

= |a

− L|

+ |M − b

(2.4.4)

We would like to repeat the arguments of the previous case. As in case (3), we must

replace 1/|b

| and |L/(Mb

)| with constant upper bounds.

Since M 6= 0, |M|/2 > 0. Using |M|/2 as our ε, we can ﬁnd some N

so that

− M| <

|M|

for all n ≥ N

The key point is that |b

| ≥ |M |/2. Taking the reciprocals of both sides, we get

≤

|M|

for all n ≥ N

So continuing from (2.4.4), we have

−

≤ |a

− L|

+ |M − b

≤ |a

− L|

+ |M − b

Now we are set to use the ideas of the previous part. Choose N

so that

− L| <

ε|M|

for all n ≥ N

;

and choose N

so that

− M| <

ε|M|

4|L| + 1

for all n ≥ N

2.4 Basic Properties of Limits 45

Now if n ≥ max{N

, N

}, then since b

≥ |M|/2 > 0, we know b

is not

zero and we can use (2.4.4) and previous paragraph to conclude

−

≤

− L

|M|

M − b

2|L|

|M|

ε|M|

|M|

ε|M|

4|L| + 1

2|L|

|M|

= ε.

Exercises for Section 2.4

A. Give a careful proof using the deﬁnition of limit of the fact that when lim

n→∞

= L

and lim

n→∞

= M, then lim

n→∞

+ 3b

= 2L + 3M.

B. Compute the following limits.

(a) lim

n→∞

tan

n sin

(b) lim

n→∞

100+5n

4n−10

n→∞

csc

2tan

−1

logn

C. If lim

n→∞

= L > 0, prove that lim

n→∞

√

L. Be sure to discuss the issue of when

√

makes sense.

HINT: Express |

√

−

√

L| in terms of |a

− L|.

D. Let (a

)

∞

n=1

and (b

)

∞

n=1

be two sequences of real numbers such that |a

− b

| <

Suppose that L = lim

n→∞

exists. Show that (b

)

∞

n=1

converges to L also.

E. Find lim

n→∞

log(2 + 3

)

. HINT: log(2 + 3

) = log3

+ log

2+3

F. (a) Let x

√

n − 1. Use the fact that (1 + x

)

= n to show that x

≤ 2/n.

HINT: Use the Binomial Theorem and throw away most terms.

(b) Hence compute lim

n→∞

1/n

G. Show that the set of rational numbers is dense in R, meaning that every real number is

a limit of rational numbers.

H. (a) Show that

b−1

≤ logb ≤ b − 1. HINT: Integrate 1/x from 1 to b.

(b) Apply this to b =

√

a to show that loga ≤ n(

√

a − 1) ≤

√

a log a.

n→∞

√

a − 1

I. Suppose that lim

n→∞

= L. Show that lim

n→∞

+ a

+ ··· + a

= L.

J. Show that the set S = {n + m

√

2 : m, n ∈ Z} is dense in R.

HINT: Find inﬁnitely many of these numbers in [0, 1] and use the Pigeonhole Principle

to ﬁnd two which are close within 10

−k

46 The Real Numbers

2.5. Upper and Lower Bounds

2.5.1. DEFINITION. If S is a nonempty subset of R that is bounded above, the

supremum or least upper bound is the number L such that

s ≤ L for all s ∈ S

and whenever L

is another upper bound for S, then L

≥ L. This value is denoted

supS.

Similarly, if S is a nonempty subset of R that is bounded below, the inﬁmum

or greatest lower bound is the number L such that

s ≥ L for all s ∈ S

and whenever L

is another lower bound for S, then L

≤ L. This value is denoted

infS.

It is not obvious that supremums and inﬁmums will always exist. The next

result will describe precisely when they do exist. However, a ﬁnite set will always

have a supremum and an inﬁmum—just pick the biggest element and the smallest

element of the set respectively. An inﬁnite set may not have biggest and smallest

elements.

2.5.2. EXAMPLES.

(1) If A = {4, −2, 5, 7}, then infA = −2 and supA = 7. Both supremum and

inﬁmum belong to A.

(2) If B = {2, 4, 6, . . .}, then infB = 2 and sup B does not exist. Sometimes we

write supB = +∞ to mean there are arbitrarily large elements of B. The inﬁmum

belongs to B.

(3) If C = {π/n : n ∈ N}, then supC = π and infC = 0. The supremum is the

maximum and belongs to C. The inﬁmum does not.

(4) If D =

(−1)

n + 1

: n ∈ N

, then infD = −1 and supD = 1. Neither 1 nor

−1 belong to D.

In the next result, our deﬁnition of the real numbers as all inﬁnite decimals

plays a crucial role. This result is not true for subsets of rational numbers for this

very reason. For example, the set {s ∈ Q : s

< 2} has no least upper bound in Q,

but

√

2 is the supremum in R.

2.5.3. LEAST UPPER BOUND PRINCIPLE.

Every nonempty subset S of R that is bounded above has a supremum. Similarly,

every nonempty subset S of R that is bounded below has an inﬁmum.

2.5 Upper and Lower Bounds 47

PROOF. We prove the second statement ﬁrst, as it is more convenient. Let M be

some lower bound for S with decimal expansion M = m

. . . . Let s be

some element of S with decimal expansion s = s

. . . . To ensure s

+ 1 > s

for all s ∈ R, if s ∈ Z, we choose the decimal expansion with s

= s. Note that

is a lower bound for S but s

+ 1 is not. There are only ﬁnitely many integers

between m

and s

. Pick the largest of these that is still a lower bound for S, and

call it a

. Since a

+ 1 is not a lower bound, we also choose an element x

in S

such that x

< a

+ 1.

Next pick the greatest integer a

such that y

= a

+ 10

−1

is a lower bound

for S. Since a

= 0 works and a

= 10 does not, a

belongs to {0, 1, . . . , 9}. To

verify our choice, pick an element x

in S such that

< a

+ 10

−1

+ 1).

Continue in this way recursively. At the kth stage, we ﬁnd the largest integer

in {0, 1, . . . , 9} such that y

= a

. . . a

i=0

−i

is a lower bound

for S. Since y

+ 10

−k

is not a lower bound, we also pick an element x

in S

such that x

< y

+ 10

−k

to verify our choice. The integer a

k+1

is now chosen by

induction. Figure 2.2 shows how a

and x

would be chosen.

.(a

+ 1)

FIGURE 2.2. The second stage (k = 2) in the proof.

Let L = a

··· = lim

k→∞

. We claim that L is the inﬁmum of S. First

note that if s ∈ S, then y

≤ s for all k ≥ 0. Therefore

L = lim

k→∞

≤ lim

k→∞

s = s.

So L is a lower bound.

Now if L

= b

··· > L, there is some ﬁrst integer k such that b

> a

and b

= a

for 0 ≤ i < k. But then

= a

. . . a

k−1

≥ y

+ 10

−k

> x

Thus L

is not a lower bound for S. Hence L is the greatest lower bound.

To deal with upper bounds, notice that L is an upper bound for S precisely

when −L is a lower bound for −S = {−s : s ∈ S}. Thus supS = −inf(−S)

exists. ¥

A sequence (a

) is (strictly) monotone increasing if a

≤ a

n+1

< a

n+1

)

for all n ≥ 1. Similarly, we deﬁne (strictly) monotone decreasing sequences.

2.5.4. MONOTONE CONVERGENCE THEOREM.

A monotone increasing sequence that is bounded above converges. A monotone

decreasing sequence that is bounded below converges.

48 The Real Numbers

PROOF. Suppose (a

)

∞

n=1

is an increasing sequence that is bounded above. Then

by the Least Upper Bound Principle, there is a number

L = sup{a

: n ∈ N}.

We will show lim

n→∞

= L.

Let ε > 0 be given. Since L − ε is not an upper bound for A, there is some

integer N so that a

> L −ε. Then because the sequence is monotone increasing,

L − ε < a

≤ a

≤ L for all n ≥ N.

So |a

− L| < ε for all n ≥ N as required. Therefore, lim

n→∞

= L.

The case of a decreasing sequence is similar. It is less work to deduce it as

a consequence of the result for increasing sequences. That is, if a

is decreasing

and bounded below by B, then the sequence −a

is increasing and bounded above

by −B. Thus the sequence (−a

)

∞

n=1

has a limit L = lim

n→∞

− a

. Therefore

−L = lim

n→∞

. ¥

2.5.5. EXAMPLE. Consider the sequence given recursively by

= 1 and a

n+1

2 +

√

for all n ≥ 1.

Evaluating the ﬁrst few terms, we obtain

1, 1.7320508076, 1.8210090645, 1.8301496356, 1.8310735189,

1.831166746, 1.8311761518, 1.8311771007, 1.8311771965, . . . .

It appears that this sequence increases to some limit.

First we show by induction that

1 ≤ a

< a

n+1

< 2 for all n ≥ 1.

Since 1 = a

√

3 = a

< 2, this is valid for n = 1. Suppose that it holds for

some n. Then

n+2

2 +

√

n+1

2 +

√

= a

n+1

≥ 1,

and

n+2

2 +

√

n+1

2 +

√

2 < 2.

This veriﬁes our claim at the next step, and hence by induction, it is valid for each

n ≥ 1.

Therefore, (a

) is a monotone increasing sequence. So by the Monotone Con-

vergence Theorem (Theorem 2.5.4), it follows that there is a limit L = lim

n→∞

It is not clear that there is a nice expression for L. However, once we know the

sequence converges, it is not hard to ﬁnd a formula for L. Notice that

L = lim

n→∞

n+1

= lim

n→∞

2 +

√

2 +

lim

n→∞

2 +

√

2.5 Upper and Lower Bounds 49

Here we use the fact that the limit of square roots is the square root of the limit

(see Exercise 2.4.C). Squaring both sides gives L

−2 =

√

L, and further squaring

yields

0 = L

− 4L

− L + 4 = (L − 1)(L

+ L

− 3L − 4).

Since L > 1, it must be a root of the cubic p(x) = x

+ x

−3x −4 in the interval

(1, 2). There is only one such root, as graphing the curve shows (see Figure 2.3).

Indeed,

(x) = 3x

+ 2x − 3 = 3(x

− 1) + 2x

is positive on [1, 2], and so p is strictly increasing. As p(1) = −5 and p(2) = 2,

it follows that p has exactly one root in between. (See the Intermediate Value

Theorem, Theorem 5.6.1.)

FIGURE 2.3. Graph of x

+ x

− 3x − 4.

For the amusement of the reader, we give an explicit algebraic formula:

L =

79+

√

2241

79−

√

2241

− 1

Notice that we proved that the sequence converged ﬁrst and then evaluated the

limit afterwards. This is important, for consider the sequence given by a

= 2 and

n+1

= (a

+ 1)/2. This is a monotone increasing sequence. Suppose we let L

denote the limit and compute

L = lim

n→∞

n+1

= lim

n→∞

+ 1)/2 = (L

+ 1)/2.

Thus (L −1)

= 0, which means that L = 1. This is an absurd conclusion because

this sequence is monotone increasing and greater than 2. The fault lay in assuming

that the limit L actually exists, because instead it diverges to +∞.

Exercises for Section 2.5

A. We say that lim

n→∞

= +∞ if for every real number R, there is an integer N so

that a

> R for all n ≥ N. Show that a divergent monotone increasing sequence

converges to inﬁnity in this sense.

50 The Real Numbers

B. Let a

= 0 and a

n+1

√

5 + 2a

for n ≥ 1. Show that this sequence converges and

ﬁnd the limit.

C. Is the set S = {x ∈ R : 0 < sin(

) <

} bounded above (below)? Find supS and

infS.

D. (a) Evaluate lim

n→∞

√

+ 5

(b) Show that this sequence is monotone decreasing.

E. Let a, b be positive real numbers. Set x

= a and x

n+1

−1

+ b

for n ≥ 0.

(a) Prove that x

is monotone decreasing.

(b) Prove that the limit exists and ﬁnd it.

F. Suppose (a

) is a sequence of positive real numbers such that a

n+1

−2a

+ a

n−1

> 0

for all n ≥ 1. Prove that the sequence either converges or tends to +∞.

G. Euler’s constant, denoted γ, is deﬁned as γ = lim

n→∞

1 +

+ ··· +

− logn.

Let a

k=1

− logn for n ≥ 1. Show that (a

)

∞

n=1

is decreasing and bounded

below by zero, and so this limit exists.

HINT: Prove the inequality

n + 1

≤ log(n + 1) − log n ≤

(It is unknown whether γ is rational or not. It is known that if γ is rational, then the

denominator has more than 244,000 decimal digits [as of 2001]. So it is suspected to

be irrational.)

H. Let x

1 +

2 +

3 + ··· +

√

(a) Show that x

< x

n+1

(b) Show that x

n+1

≤ 1 +

√

HINT: Square x

n+1

and factor a 2 out of the square root.

is bounded above by 2. Deduce that lim

n→∞

exists.

I. (a) Let (a

)

∞

n=1

be a bounded sequence. Deﬁne a sequence b

= sup{a

: k ≥ n} for

n ≥ 1. Prove that (b

) converges. This limit is called the limit superior of (a

almost always abreviated to limsup a

(b) Without redoing the proof, do the same for the limit inferior of (a

), which is

deﬁned as liminf a

:= lim

n→∞

inf

k≥n

J. Show that a sequence (a

)

∞

n=1

converges if and only if limsupa

= liminfa

K. Suppose that (a

)

∞

n=1

is a sequence of positive real numbers. Show that

limsup

liminfa

L. Suppose that (a

)

∞

n=1

and (b

)

∞

n=1

are sequences of positive real numbers and that

limsup

< ∞. Prove that there is a constant M so that a

≤ Mb

for all n ≥ 1.

2.6 Subsequences 51

M. Suppose that the real numbers were deﬁned using Dedekind cuts as at the end of Sec-

tion 2.2. Show that if S is a family of Dedekind cuts that is bounded above, then the

union of all of these cuts is also a cut and represents the least upper bound.

2.6. Subsequences

Even if a sequence does not converge, by picking out certain terms of the se-

quence, it may be possible to ﬁnd a new convergent sequence.

2.6.1. DEFINITION. A subsequence of a sequence (a

)

∞

n=1

is a new sequence

)

∞

k=1

= (a

, a

, . . .), where n

< n

< . . ..

For example, (a

)

∞

n=1

and (a

)

∞

n=1

are subsequences. It is easy to verify that

if (a

)

∞

n=1

converges to a limit L, then (a

)

∞

k=1

also converges to the same limit.

The sequence a

= n does not have a limit, nor does any subsequence because

any subsequence tends to +∞. However, we will show that as long as a sequence

remains bounded, it has subsequences that converge.

2.6.2. NESTED INTERVALS LEMMA.

Suppose that

= [a

, b

] = {x ∈ R : a

≤ x ≤ b

}

are nonempty closed intervals such that I

n+1

is contained in I

for each n ≥ 1.

Then the intersection

n≥1

is nonempty.

2.6.3. REMARK. This proof of this result is so easy that a couple of warning

remarks in advance are appropriate. This result depends in a crucial way on the

completeness of the reals implicit in the existence of a real number for every inﬁnite

decimal expansion. This was used in establishing the Least Upper Bound Principle,

which we use here in the guise of the convergence of monotone sequences.

The result is false for intervals of rational numbers. In Example 2.6.8, we con-

struct a sequence of closed intervals with rational endpoints that intersect exactly

in the point

√

8, which is not rational. So the corresponding intervals of rational

numbers intersects in the empty set.

Likewise, this result is false for open intervals. Indeed,

n≥1

(0,

) = ∅.

PROOF. Notice that since I

n+1

is contained in I

, it follows that

≤ a

n+1

≤ b

n+1

≤ b

Thus (a

) is a monotone increasing sequence bounded above by any b

; and like-

wise (b

) is a monotone decreasing sequence bounded below by any a

. Hence

by Corollary 2.5.4, a = lim

n→∞

exists; as does b = lim

n→∞

. By Exercise 2.3.D,

a ≤ b. Thus

≤ a ≤ b ≤ b

Consequently, the point a belongs to each I

for each k ≥ 1. ¥

52 The Real Numbers

This is just the tool needed to establish the key result of this section.

2.6.4. BOLZANO–WEIERSTRASS THEOREM.

Every bounded sequence of real numbers has a convergent subsequence.

PROOF. Let (a

) be a sequence bounded by B. Thus the interval [−B, B] contains

the whole (inﬁnite) sequence. Now if I is an interval containing inﬁnitely many

points of the sequence (a

), and I = J

∪ J

is the union of two smaller intervals,

then at least one of them contains inﬁnitely many points of the sequence too.

So let I

= [−B, B]. Split it into two closed intervals of length B, namely

[−B, 0] and [0, B]. One of these halves contains inﬁnitely many points of (a

);

call it I

. Similarly, divide I

into two closed intervals of length B/2. Again pick

one, called I

, that contains inﬁnitely many points of our sequence. Recursively,

we construct a decreasing sequence I

of closed intervals of length 2

2−k

B so that

each contains inﬁnitely many points of our sequence. Figure 2.4 shows the choice

of I

and I

, where the terms of the sequence are indicated by vertical lines.

−B B

FIGURE 2.4. Choice of intervals I

and I

By the Nested Interval Lemma, we know that

k≥1

contains a point L. Now

choose an increasing sequence n

such that a

belongs to I

. This is possible since

it contains inﬁnitely many points in the sequence, and only ﬁnitely many are less

than n

k−1

. It will be shown that

lim

k→∞

= L.

Indeed, both a

and L belong to I

, and hence

− L| ≤ |I

| = 2

−k

(4B).

The right-hand side tends to 0, and thus lim

k→∞

= L. ¥

2.6.5. EXAMPLE. Consider the sequence (a

) = (sign(sinn))

∞

n=1

, where the

sign function takes values ±1 depending on the sign of x except for sign0 = 0.

Without knowing anything about the properties of the sin function, we can observe

that the sequence (a

) takes at most three different values. At least one of these

values is taken inﬁnitely often. Thus it is possible to deduce the existence of a

subsequence that is constant and therefore converges.

Using our knowledge of sin allows us to get somewhat more speciﬁc. Now

sinx = 0 exactly when x is an integer multiple of π. Since π is irrational, kπ is

never an integer for k > 0. Therefore, a

takes only the values ±1. Note that