Chapter 6

SEQUENCES

6.1 Sequences

A sequence is a function whose domain is the set of natural numbers N. If s is a

sequence, we usually denote its value at n by sn instead of s (n) . We may refer to thesequence s as {sn} or by listing the elements {s1, s2, ....} . This notation is somewhatambiguous though, since a sequence is an ordered set, and should not be confused withthe range of s = {s

n: n ∈ N} . Some authors prefer the use of parentheses instead

of braces, to emphasize this particular feature of sequences. Our notation will use

braces for both.

We call sn the nth term of the sequence and we often describe a sequence by

giving a formula for the nth term. Thus {1/n} is an abbreviation for the sequence{1,

1

2,

1

3, ....

}. Sometimes we may wish to change the domain of a sequence from N to

N∪{0} , or to {n ∈ N : n ≥ m} . In this case we write {sn}∞

n=0or {sn}

∞

n=mrespectively.

If no mention is made to the contrary, we assume that the domain is just N.

Sequences are sometimes also expressed in what we call recursive form. This

means that a generic term sn is not expressed explicitly as a function of n, but

instead as a function of some subset of previous terms {sn−1, sn−2..., sn−k} . Thus{sn : sn =

sn−1

sn−1+1, s1 = 1

}is again an abbreviation for the sequence

{1, 1

2, 13...}— Note

that whenever the recursive definition for sn depends on k previous terms, for the

sequence to be well defined it is necessary to provide the first k terms —.

6.2 Convergence

6.2.1 Convergent sequences

Intuitively, we say that a given sequence {sn} converges to a certain value s wheneverthe sn become arbitrarily close to s for n arbitrarily large. It is clear from the abovethat we need a metric space to define what “close” means. In this section we willrestrict to the metric space (R, |·|) .The formal definition of convergent sequence inan arbitrary metric space is the following:

Definition 224 A sequence {sn} in a metric space (X,d) is said to converge if

there exists s ∈ X such that for every ε > 0 there exists a real number Nε such that

for all n ∈ N and n ≥ Nε, d (sn, s) < ε.

76 Sequences

In this case we say that {sn} converges to s, or that s is the limit of the

sequence {sn} , and write s

n→ s, lim s

n= s, or lim

n→∞sn= s. If a sequence does

not converge, it is said to diverge.

Remark 225 Notice that for a sequence to be convergent the definition requires its

limit to belong to X, the set in which the sequence is defined. Take for example the

sequence {1/n}. Even though it converges to 0 in X = R, it fails to do so in X = R++.

In the same fashion, the sequence{xn : xn+1 =

x2n+2

2xn, x1 = 1

}converges to

√2 in R

,but it fails to converge in Q.

Example 226 The sequence {sn} with sn = 1 for all n ∈ N converges to 1 in R.

The proof is trivial since d(sn, 1) = 0 < ε for all sn and for all ε > 0.

Example 227 Let us show that lim 1/n = 0 in (R, |·|) . According to the definition,

we have to show that for all ε > 0, ∃Mε∈ R, s.t. ∀n ∈ N, n � M

ε, |1/n − 0| =

1/n < ε. So take an arbitrary ε > 0, ε ∈ R. From the Archimedean property of R

there exits Mε ∈ N ⊆ R such that Mε > 1/ε. Then it follows that for all n ≥ Mε we

have 1/n � 1/Mε < ε, which concludes the proof.

Example 228 To prove that the sequence {1 + (−1)n} is divergent, let us suppose

that {sn} converges to some real number s. Letting ε = 1 in the definition of conver-

gence, we find that there exists a number Mε such that |1 + (−1)n − s| < 1. If n > Mε

and n is odd, then |1 + (−1)n − s| = |s| < 1, so that −1 < s < 1. On the other hand, if

n > Mεand n is even, then we have |1 + (−1)n − s| = |2− s| < 1, so that 1 < s < 3.

Since s cannot satisfy both inequalities, we have reached a contradiction. Thus the

sequence {sn} must be divergent.

So far we have seen examples where algebraic manipulation was enough tofind an M

εand establish whether the definition was satisfied or not. However, for

more complicated cases such a simple technique may not work.

Example 229 Take for example the case of sn =n2+2n

n3−5

. We want to show that

lim sn = 0. From the definition, given any ε > 0, we want to make

∣∣∣n2+2n

n3−5

∣∣∣ < ε. By

restricting n > 3,we can remove the absolute value since n2+2n

n3−5> 0 . Thus we want to

know how big n has to be in order to make n2+2n

n3−5

< ε. Solving for this inequality would

be very messy, so we try to find some estimate of how large the left hand side can

be. To do this we seek an upper bound for the numerator and a lower bound for the

denominator. On the one hand we have n2 + 2n < an2 for some a > 1 if n > 2

(a−1).

On the other we have n3−5 >

n3

bfor some b > 1 if n > 3

√5

b

b−1. Taking an arbitrary

pair a, b > 1 and restricting n > max

{3, 2

(a−1),

3

√5

b

b−1

}we have that

∣∣∣∣

n2 + 2n

n3 − 5

∣∣∣∣=

n2 + 2n

n3 − 5

≤an

2

n3

b

=

ab

n.

Convergence 77

The right hand side can be easily made less than any ε. Just pick n ≥ab

ε. So

now we are ready to organize this into a formal proof. Given ε > 0, take Mε=

max

{3, 2

(a−1),

3

√5

b

b−1,ab

ε

}. Then for all n ∈ N and n > Mε we have

∣∣∣∣

n2 + 2n

n3 − 5

− 0

∣∣∣∣=

n2 + 2n

n3 − 5

≤an

2

n3

b

=ab

n< ε

Hence limn2+2n

n3−5

= 0.

The use of upper bounds can be generalized by the following theorem

Exercise 230 Suppose that lim sn= 0. If {t

n} is a bounded sequence, prove that

lim (sntn) = 0. Use this result to prove that lim sin(n)

n= 0.

Theorem 231 Let {sn} and {a

n} be sequences of real numbers and let s ∈ R . If for

some k > 0 and some m ∈ N we have

|sn− s| ≤ k |a

n| , for all n > m

and if lim an = 0, then it follows that lim sn = s.

Proof. Given any ε > 0, since lim an = 0 there exists N1 ∈ R such that n ∈ Nand n > N1 implies that |an| < ε/k. Now let N = max {m,N1} . Then for n > N we

have n > m and n > N1, so that

|sn− s| ≤ k |a

n| < k

ε

k= ε

Thus lim sn = s.

Another useful theorem to show convergence of sequences by use of bounding

sequences is the following.

Theorem 232 (“Sandwich” theorem) Suppose xn� y

n� z

nfor n > M, where

M is fixed, limn→∞

xn= l, and lim

n→∞zn= l, then lim

n→∞yn= l.

Exercise 233 Prove the Sandwich theorem.

There is also a necessary condition for convergence that is very useful both

in proofs of convergence and in proving that a sequence does not converge. It states

that for n large enough, any two terms in a sequence must be arbitrarily close to each

other.

78 Sequences

Theorem 234 If a sequence {sn} converges, then for all ε > 0 there is N ∈ R such

that for all n1, n2 ∈ N , n1, n2 > N , |sn1− s

n2| < ε.

Proof. Since {sn} converges, for any ε > 0 there is N ∈ R such that for all

n ∈ N , n > N we have |sn− s| < ε

2. In particular, if we take n1 > N and n2 > N we

have by the triangle inequality

|sn1− s

n2| ≤ |s

n1− s|+ |s− s

n2| <

ε

2+

ε

2= ε.

Exercise 235 Prove that the reciprocal may not be true. That is, prove that you can

have a sequence whose terms are arbitrarily close for n sufficiently large, but still the

sequence does not converge.

6.2.2 Properties of Convergent Sequences

In this section we derive two important properties of convergent sequences. But first

we need a definition.

Definition 236 A sequence {sn} is bounded if its range {sn : n ∈ N}is a bounded

set, that is, if there exists M ≥ 0 such that |sn| ≤M for all n ∈ N.

Theorem 237 (Boundedness) Every convergent sequence is bounded.

Proof. Let {sn} be a convergent sequence and let s = lim sn. From the defi-

nition of convergence with ε = 1, we obtain N ∈ R such that |sn − s| < 1 whenever

n > N. Thus for n > N the triangle inequality implies that |sn| < |s| + 1. Let

M = max{|s1| , . . . , |sN | , |s| + 1}, then we have |sn| ≤ M for all n ∈ N, so {s

n} is

bounded.

It is easy to see that the converse of this theorem is not true. We saw in aprevious example the sequence {1 + (−1)n}, which is bounded but diverges.

Theorem 238 (Uniqueness) If a sequence {sn} converges, then its limit is unique.

Proof. Let {sn} be a sequence and suppose that it converges both to x and y,

with x �= y. Let |x− y| = d > 0 be the distance between x and y. Convergence of

{sn} to x implies that for any ε > 0, there exists N1 ∈ R such that

|sn − x| < ε, for every n > N1.

In particular this holds for ε = d

3. Similarly, convergence of {s

n} to y implies that for

any ε > 0, there exists N2 ∈ R such that

|sn− y| < ε, for every n > N2,

and in particular this holds for ε = d

3. Therefore if n > max {N1, N2} we have

0 < d = |x− y| = |x− sn + sn − y| ≤ |x− sn|+ |sn − y| < 2d

3

But this implies d < 2

3d, which is a contradiction given that d > 0. So it must be

d = 0, and thus x = y.

Convergence 79

6.2.3 Convergence and topological properties of sets

In the previous chapter, we saw that a set X was closed if and only if it contained all

its accumulation points. Here we will present the same result based on the theory of

sequences. But in order to do that, we have first to prove two other theorems. The

first one states that if xn→ x, then x is an accumulation point for the set described

by the range of the sequence, and hence every neighborhood of x contains infinitely

many points of {xn} . The second theorem establishes that it is equivalent to say that

a point belongs to the closure of a given set E and to say that such a point is the

limit of a sequence completely contained in E.

Theorem 239 {xn} ⊆ X converges to x ∈ X if and only if every neighborhood of x

contains {xn} for all but finitely many n.

Proof. If {xn} converges to x then for each ε > 0 there exists N ∈ R such

that for all n ∈ N, n > N d (xn, x) < ε. This means that any neighborhood of x of

radius ε has at least one (hence infinitely many) elements of {xn}. Hence there mustbe at most only finitely many elements of {xn} outside any neighborhood of x.

Suppose now that every neighborhood of x contains {xn} for all but finitelymany n. Hence for each neighborhood N(x, ε), ∃ N ∈ R such that for all n ∈ N,

n > N, xn∈ N(x, ε). This suffices to conclude that {x

n} converges to x.

Theorem 240 Let E ⊆ X. Then x ∈ E if and only if there exists a sequence {xn}in E such that x = lim

n→∞xn.

Proof. Suppose there exists a sequence {xn} ⊆ E that is convergent to x ∈ Xbut x /∈ E. Therefore x ∈ EC.Since EC is open, there exists a neighborhood N (x, ε) ⊆EC such that N (x, ε)∩{xn : n ∈ N} = ∅. Therefore, for such ε there is no N ∈ R such

that for all n > N |xn − x| < ε, contradicting the assumption that x = limn→∞ xn.

Then x ∈ E.

Suppose now that x ∈ E. Then x ∈ E or x ∈ bd E. This implies that for every

ε > 0, N (x, ε) ∩ E �= ∅. We can therefore construct a sequence {xn} in E, being x

n

any point such that xn ∈ N(x, 1

n

)∩E. By the Archimedean property of the reals, for

all ε > 0 there exists N ∈ N such that 1/N < ε, and by construction of the sequence

there exists xN ∈ {xn} such that |xN − x| < 1/N < ε. Therefore x = limn→∞ xn.

We have therefore seen that any limit point of E can be expressed as a limit

of a convergent sequence in E. So, this theorem tells us that any nonempty closed

set contains convergent sequences. Moreover, we know that if in a given set E every

convergent sequence has a limit in the set, then the set is closed. Therefore we can

define the closedness of a set in a metric space using convergent sequences.

Theorem 241 A set E ⊆ X is closed if and only if every convergent sequence in X

completely contained in E has its limit in E.

Proof. Let E be a closed set and let {xn} be a sequence in E converging to

x ∈ X. By the previous theorem we know that x ∈ ¯E. Since E is closed, then E =¯E

and therefore x ∈ E.

80 Sequences

Suppose now that every convergent sequence in X completely contained in E

has its limit in E, but E is not closed. Then there exists x ∈ E ′ ⊆ ¯E and x /∈ E. As

in the previous theorem there exists a sequence {xn} ⊆ E such that x

n→ x ∈ ¯E−E.

But this contradicts that every convergent sequence {xn} ⊆ E must have its limit in

E. Therefore E is closed.

6.3 Limit and comparison theorems

In the previous section, we saw that the definition of convergence may sometimes be

messy to use even for sequences given by relatively simple formulas. In this section

we will derive some basic results that will greatly simplify our work.

The first theorem is a very important result showing that algebraic operations

are compatible with taking limits.

Theorem 242 Consider two sequences {xn} and {y

n} such that x

n→ x and y

n→ y.

Then the following properties hold:1) x

n+ y

n→ x+ y;

2) xnyn → xy;3) If xn �= 0,∀n and x �= 0, then yn/xn → y/x.Proof. 1) Using the triangular inequality, we know that |xn + yn − x− y| �

|xn− x| + |y

n− y|. Now given ε > 0, since x

n→ x there exists N1 ∈ R such

that n > N1implies |xn− x| < ε

2. Similarly there exists such N2 ∈ R such that

n > N2 implies |yn − y| < ε

2. Thus, if we let N = max{N1, N2}, then n > N implies

that

|(xn + yn)− (x+ y)| � |xn − x|+ |yn − y| < ε.

Therefore we conclude that lim (xn + yn) = x+ y.

2) Observe that xnyn− xy = x

n(y

n− y) + y(x

n− x). Since {x

n} is convergent

then it is bounded. Let M1 = sup |xn| and M = max {M1, |y|}. Hence

|xnyn− xy| �M1 |yn − y|+ |y| |xn − x| ≤M ( |yn − y|+ |xn − x| ) .

Now given ε > 0 there exist Nx and Ny such that |xn − x| < ε

2Mand |yn − y| < ε

2M

for all n > Nx and n > Ny respectively. Let N = max{Nx, Ny}. Then n > N implies

that

|xnyn − xy| � M ( |yn− y|+ |x

n− x| )

< Mε

2M+M

ε

2M= ε.

Thus lim xnyn= xy.

3) Since yn/x

n= y

n(1/x

n), it suffices from (2) to show that lim (1/x

n) = 1/x.

That is, given ε > 0, we must make∣∣∣∣

1

xn−

1

x

∣∣∣∣=

∣∣∣∣

x− xn

xnx

∣∣∣∣< ε

Limit and comparison theorems 81

for n sufficiently large. To get a lower bound on the denominator, we note that since

x �= 0, there exists N1 such that n > N1 implies |xn− x| < |x|

2. Thus for n > N1 we

have

|x− xn| <

|x|

2⇒ |x| − |x

n| <

|x|

2⇒

|x|

2< |x

n|

and

∣∣∣∣

x− xn

xnx

∣∣∣∣=

|xn − x|

|xn| |x|<2 |xn − x|

|x|2

But since xn → x, for every ε > 0 there exists also N2 such that for n > N2 we have

|xn − x| < |x|2ε2. Let N = max {N1, N2} . Then n > N implies

∣∣∣∣

x− xn

xnx

∣∣∣∣<

2

|x|2|x

n− x| < ε

Hence lim 1/xn= 1/x.

A particular case of (1) is “a+ xn → a+ x,∀a ∈ R when xn → x”, and of (3)is “ax

n→ ax,∀a ∈ R when x

n→ x”.

Another useful fact is that the order relation “ ≤ ” is preserved when takinglimits. The proof is left as an exercise.

Theorem 243 Suppose that {yn} and {zn} are convergent sequences with lim yn = s

and lim zn= z. If y

n≤ z

nfor all n > M, for certain M ∈ N , then y ≤ z.

Exercise 244 Prove the previous theorem.

6.3.1 Infinite Limits

The sequence sn = n is clearly not convergent, since it is not bounded. But its

behavior is not the least erratic: the terms get larger and larger. Although there is

no real number that the terms approach, we would like to say that sn“goes to ∞”.

We make this precise in the following definition.

Definition 245 A sequence {xn} of R diverges to +∞ if ∀M > 0, ∃N > 0 such

that ∀n > N, xn > M. We write xn → +∞ or limn→∞ xn = +∞. A sequence

{xn} diverges to −∞ if ∀M > 0, ∃N > 0 such that ∀n > N, xn < −M. We write

xn →−∞ or limn→∞ xn = −∞.

The technique of developing proofs for infinite limits is similar to that for finite

limits, and is closely related to theorem243. We leave the proof as an exercise.

82 Sequences

Theorem 246 Suppose that {sn} and {tn} are sequences such that sn ≤ tn for alln > M, for a certain M ∈ N.

(a) If lim sn= +∞, then lim t

n= +∞.

(b) If lim tn= −∞, then lim s

n= −∞.

Exercise 247 Prove the previous theorem.

Exercise 248 Prove the following1) lim

(√n+ 1 −

√n

)= 0

2) lim n4+13

2n5+3= 0

3) lim n!

nn= 0

6.4 Monotone sequences and Cauchy sequences

In the preceding two sections, we have seen a number of results that enable us to

show that a sequence converges. Unfortunately, most of these techniques depend on

our knowing what the limit of the sequence is before we begin. Often in applications

it is desirable to be able to show that a given sequence is convergent without knowing

precisely the value of the limit. In this section we obtain two important theorems

that do just that.

6.4.1 Monotone Sequences

Definition 249 A sequence {xn} of real number is increasing if xn+1 � xn (strictlyincreasing if xn+1 > xn) ∀n ∈ N; decreasing if xn+1 � xn (strictly decreasing if

xn+1 < x

n), ∀n ∈ N and monotone if it is increasing or decreasing (strictly mono-

tone if it is strictly increasing or strictly decreasing).

Example 250 The sequence of real numbers {1/n} is decreasing (thus monotone)and bounded. The constant sequence x1 = x2 = ... = 1 is increasing, decreasing, andbounded. Both of them converge.

Example 251 The sequence {(−1)n} is bounded, but it is not monotone. The se-

quence {2n} is monotone but it is not bounded. None of them converge.

As you may have imagined, we can generalize these examples in a theorem

Theorem 252 (Monotone Convergence Theorem) A monotone sequence of real

numbers {xn} converges if and only if it is bounded.

Proof. We already know that a convergent sequence is bounded. Hence we just

need to show that a monotone and bounded sequence of real numbers is convergent.

Suppose {xn} is a bounded increasing sequence. Let S denote the nonempty bounded

set {xn : n ∈ N}. By the completeness axiom S has a least upper bound, and we let

x = supS. We claim that limxn = x. Given any ε > 0, x− ε is not an upper bound

Monotone sequences and Cauchy sequences 83

for S. Thus there exists an integer N such that xN > x− ε. Furthermore, since {xn}is increasing and x is an upper bound for S we have

x− ε < xN ≤ xn ≤ x < x+ ε

or equivalently

|xn − x| < ε

for all n ≥ N. Hence limxn = n. In the case when the sequence is decreasing, let

x = inf S and proceed in a similar manner.

Exercise 253 Consider the sequence{xn: x

n+1 =x2n+2

2xn, x1 = 1

}. Show that for n ≥

2 ,{xn} is decreasing and bounded below. In particular, show that√2 < xn+1 <

xn∀n ≥ 2 (try induction).Thus we know that {x

n} is convergent in R. Show also

that {xn} converges to

√2(Hint: use theorem 234).

6.4.2 Cauchy sequences

When a sequence is convergent, we saw in theorem 234 that the terms get close to eachother as n gets large. It turns out that this property ( called the Cauchy property)in some cases is actually sufficient to imply convergence.

Definition 254 A sequence {xn} in a metric space (X, d) is said to be a Cauchy

sequence if for all ε > 0, there exists a number Mεs.t. for all n,m ∈ N, n,m �M

ε

⇒ d(xn, xm) < ε.

Example 255 The sequence {1/n} ⊆ R++ is Cauchy. For each ε > 0 take Mε=

2

ε,

implying that for all n,m ∈ N and n,m > Mε, |1/n − 1/m| < |1/n|+ |1/m| < 21

Mε

<ε. Then the sequence is Cauchy. However, it is not convergent since 0 does not belong

to R++.

Exercise 256 Prove that the sequence {(∑

n

i=11/n)} is not Cauchy.

As we saw in theorem 234, convergence is a sufficient condition for a sequence

to be a Cauchy sequence.

Theorem 257 Every convergent sequence is a Cauchy sequence.

Proof. See theorem 234.

Exercise 258 Show that every Cauchy sequence is bounded.

Its contrapositive “If a sequence is not Cauchy, then it is not convergent” is

very a useful tool to prove that a sequence does not converge. Its converse however, is

not true as we have shown in the previous example. However, there are some metric

spaces where Cauchy sequences do converge. These spaces are called complete.

84 Sequences

Definition 259 A metric space (X, d) is called complete if every Cauchy sequence

in X converges in X.

Theorem 260 The metric space (R, d) is complete.

Proof. Suppose that {sn} ⊆ R is a Cauchy sequence and let S = {sn : n ∈ N}be the range of the sequence. We consider two cases, depending on whether S is finite

or infinite.

If S is finite, then the minimum distance ε between distinct points of S is

positive. Since {sn} is Cauchy, there exists a number N such that m,n > N implies

that |sn − sm| < ε. Let n0 be the smallest integer greater than N. Given any m > N ,

sm and sn0 are both in S, so if the distance between them is less than ε, it must be

zero (since ε is the minimum distance between distinct points in S). Thus sm= s

n0

for all n > N. It follows that lim sn= s

n0.

Now suppose that S is infinite. From the previous exercise we know that Sis bounded. Thus from the Bolzano Wierestrass theorem, there exists a point s in Rthat is an accumulation point of S. We claim that {sn} converges to s. Given any

ε > 0, there exists N ∈ R , such that |sn− s

m| < ε whenever n,m > N. Since s

is an accumulation point of S, the neighborhood N(s, ε

2

)=

(s−

ε

2, s+

ε

2

)contains

infinitely many points of S. Thus in particular there exists an integer m > N such

that sm ∈ N(s, ε

2

). Hence, for any n > N we have

|sn− s| = |s

n− s

m+ s

m− s|

≤ |sn− s

m|+ |s

m− s| <

ε

2+

ε

2= ε

and lim sn = s. Therefore every Cauchy sequence in R is a convergent sequence, which

implies that (R, d) is a complete space.

Corollary 261 Every Cauchy sequence in R is convergent.

Note that in the proofs of both theorem, the completeness axiom of R is thekey factor. So, you can naturally guess that these theorems do not hold for a sequenceof Q, as it is the case.

6.5 Subsequences

Definition 262 Let {sn}∞n=1

be a sequence and let {nk}∞

k=1be any sequence of natu-

ral numbers such that n1 < n2 < n3 < ... . The sequence {xnk}∞k=1

is called a subse-

quence of {xn}∞n=1

.

When viewed as a function, a subsequence of a sequence s : N → R is the

composition of s and a strictly increasing function n : N → N. That is, s ◦ n :

N → R is a subsequence of the sequence s : N → R. Since this functional notation

is cumbersome, we prefer to use subscripts and write snkinstead of s (n (k)) . It is

important to note that the index of the subsequence is k, not nk and not n.

Subsequences 85

If we delete a finite number of terms in a sequence and renumber the remainingones in the same order, we obtain a subsequence. In fact, we may delete infinitelymany of them as long as there are infinitely many terms left. It is possible to do sobecause a countable set has countable subsets (theorem 157).

Example 263 Given the sequence {x1, x2, x3, . . . , xn, . . . } the following represents a

subsequence {x1, x3, x5, . . . , x2n+1, . . .} and it is obtained by taking into account only

the odd terms of the original sequence.

Exercise 264 Let {nk}∞

k=1be a sequence of natural numbers such that nk < nk+1 for

all k ∈ N. Use induction to show that nk ≥ k for all k ∈ N.

Theorem 265 A sequence {sn} converges to a real number s if, and only if, every

subsequence of {sn} also converges to s.

Proof. Let {sn} converge to s and take any subsequence {s

nk}. We know that

for each ε > 0 there exists a number N such that n > N implies d(sn, s) < ε. Thus

when k > N, we apply the result of the previous exercise to obtain that nk ≥ k > N

and therefore d(snk,x) < ε.

Suppose now that every subsequence {snk} converges to s. Hence {sn} con-

verges as well because it is also a subsequence of itself.

Example 266 One application of this theorem is in finding the limit of a convergent

sequence. Suppose that 0 < x < 1 and consider the sequence defined by sn = x1/n.

Since 0 < x1/n < 1 for all n, {sn} is bounded. Since

x1/n+1 − x

1/n= x

1/n+1(1− x

n+1

n

)> 0, for all n

then {sn} is an increasing sequence. Thus by the monotone convergence theorem

{sn} converges to some number, say s. Now, for each n, s2n = x1/(2n)

=

√sn. Since

{s2n} is a subsequence of {sn} then it also converges to s. Moreover, we have that

lim√

sn=√lim s

n, and this implies

s = lim sn = lim s2n = lim√sn =

√lim sn =

√s.

It follows that s2 = s, so that s = 0 or s = 1. Since s1 = x > 0 and sn is increasing,

so s �= 0. Hence limx1/n

= 1

Exercise 267 Prove that if {sn} is a sequence of nonnegative terms and s

n→ s,

then lim√

sn=√lim s

n.

Example 268 Another application of the previous theorem is in showing that a se-

quence is divergent. For example, if the sequence sn= (−1)n were convergent to some

number s, then every subsequence would also converge to s. But {s2n} converges to 1

and {s2n+1} converges to -1. We conclude that {sn} is not convergent.

86 Sequences

So far we know that every subsequence of a convergent sequence converges

to the same limit and that in a divergent sequence there exists a subsequence that

diverges. Now the question is whether we can find a convergent subsequence in a

divergent sequence. A version of the Bolzano-Wierestrass theorem for sequences says

that if the sequence is bounded this is so.

Theorem 269 Every bounded sequence of real numbers has a convergent subsequence.

Proof. The proof is essentially the same as the Bolzano-Weierstrass theorem

we have seen in the previous chapter and is left as an exercise.

While an unbounded sequence may not have any convergent subsequence, it

will contain a subsequence that has an infinite limit. We leave the result as an

exercise.

Exercise 270 Show that every unbounded sequence contains a monotone subsequence

that has either +∞ or -∞ as a limit.

6.5.1 Limit Superior and Limit Inferior

Definition 271 Let {sn} be a bounded sequence. A subsequential limit s is any

real number that is the limit of some subsequence of {sn}. If S is the set of all

subsequential limits of {sn}, then we define the limit superior to be

lim supn→∞

sn= supS.

Similarly, we define the limit inferior of {sn} to be

lim infn→∞

sn

= inf S.

Note that in the definition we require {sn} to be bounded. Thus theorem 269

implies that {sn} contains a convergent subsequence, so that the set S of subsequen-

tial limits will be nonempty. It will also be bounded, since {sn} is bounded. The

completeness axiom then implies that sup S and inf S both exist as real numbers.

Example 272 Let sn = (−1)n + (1/n). Then

lim supn→∞

sn = 1, lim infn→∞

sn = −1.

It should be clear that we always have lim infn→∞

sn≤ lim sup

n→∞sn. Now,

if {sn} converges to some number s, then all its subsequences converge to s, so we

have lim infn→∞ sn = limsupn→∞

sn. The converse is also true. If it happens that

lim infn→∞ sn < lim supn→∞

sn, then we say that {sn} oscillates.

Subsequences 87

Theorem 273 For a real valued sequence {sn}, limn→∞ sn = s if and only if lim supn→∞

sn =

lim infn→∞

sn= s.

Proof. If snconverges to s so do all its subsequences. Hence S is singleton

and supS = inf S. Suppose now that supS = inf S = s. This means that S is singleton

so that every converging subsequence of {sn} converges to s. Hence {sn} converges to

s.

Theorem 274 Let xn, yn be bounded sequences and xn � yn for n > M, where M is

fixed. Then lim supn→∞

xn� lim sup

n→∞yn, and lim inf

n→∞xn� lim inf

n→∞yn.

Proof. Let X = lim supn→∞

xnand Y = limsup

n→∞ynand suppose that X >

Y. Hence for some {xnk} and {ynk} subsequences of {xn} and {yn} we have that for

every ε > 0 there exist an Nε,x and Nε,y such that for all nk > N = max{Nε,x, Nε,y},|xnk −X| < ε and |ynk − Y | < ε. Take ε < (X − Y )/2 and let n = max{N,M}.Hence, for all nk > n, we have that

xnk − ynk> X − Y − 2ε > 0

which gives a contradiction, meaning that lim supn→∞

xn� lim sup

n→∞yn.

The part relative to lim inf can be proved in a similar way.

Exercise 275 Let sn = n sin2(nπ

2

). Find the set S of subsequential limits, the limit

superior and the limit inferior of {sn} .

Unbounded sequences

There are some occasions when we wish to generalize the notion of the limit supe-

rior and the limit inferior to apply to unbounded sequences. There are two cases

to consider for the limit superior, with analogous definitions applying to the limit

inferior.

1. Suppose that {sn} is unbounded above. Then exercise 270 implies that there ex-

ists a subsequence having +∞ as its limit. This prompt us to define lim supn→∞

sn =

+∞.

2. Suppose that {sn} is bounded above but not below. If some subsequence con-

verges to a finite number, we define lim supn→∞

snto be the supremum of the set

of subsequential limits. Essentially, this coincides with the original definition. If

no subsequence converges to a finite number, then we must have lim sn = −∞,

so we define lim supn→∞

sn = −∞.

Thus for any sequence {sn} , lim sup

n→∞snalways exists as either a real num-

ber, +∞ or −∞. When k ∈ R and α = lim supn→∞

sn, then writing α > k means

that α is a real number greater than k or that α = +∞. Similarly, α < k means that

α is a real number less than k or that α = −∞.

88 Sequences

6.5.2 Some special sequences

The following are very common sequences. You need to be able to prove the results

stated.

Exercise 276 Prove the following statements

1. If p > 0 then limn→∞

( 1n)p = 0;

2. If p > 0 then limn→∞n

√p = 1;

3. limn→∞

n

√n = 1;

4. If p > 0 and α ∈ R, then limn→∞

nα

(1+p)n= 0;

5. If |x| < 1, then limn→∞ xn= 0.

6.6 References

S.R. Lay, Analysis with an Introduction to Proof .Chapter 4. Third Edition. Pren-

tice Hall.

A. Matozzi, Lecture Notes Econ 897 University of Pennsylvania Summer

2001.

Rudin, Principles of Mathematical Analysis . Chapter 3.