sequences

Chapter 1

Sequences of real numbers

Outline:

• The limit of a sequence of real numbers, bounded/Cauchy/convergent sequences

• Cauchy’s theorem for sequences

• Passing to the limit in inequalities

• Sufficient conditions for convergence: convergence of monotonic sequences, squeeze theorem

• Operations with sequences which have a limit

• Césaro-Stolz theorem and the Root Criterion for sequences

1.1 Sequences of real numbers

Definition 1.1.1 A sequence of real numbers is a function a : N→ R.

For the convenience of the notation, it is customary to identify the function (sequence) f with itsvalues a0 = a (0) , a1 = a (1) , a2 = a (2) , a3 = a (3) , .... We will adopt this convention, and we willhenceforth write sequences as (an)n∈N, (an) , or simply an.

Definition 1.1.2 Given a sequence (an)n∈N, we say that the sequence is:a) bounded above, if there exists M ∈ R such that an ≤M , for any n ∈ N;b) bounded below, if there exists m ∈ R such that m ≤ an, for any n ∈ N;c) bounded, if it is bounded above and below;d) increasing, if an ≤ an+1 for any n ∈ N (if the previous inequality is strict, we say that the

sequence is strictly increasing);e) decreasing, if an ≥ an+1 for any n ∈ N (if the previous inequality is strict, we say that the

sequence is strictly decreasing);f) monotone, if it is either increasing or decreasing;g) Cauchy sequence, if for any ε > 0 there exists N (ε) ∈ N such that |an − am| < ε for any

m,n ≥ N (ε);h) convergent to l ∈ R, if for any ε > 0 there exists N (ε) ∈ N such that |an − l| < ε for any

n ≥ N (ε). The number l is called the limit of the sequence (an)n∈N, and we write limn→∞ an = l oran −→

n→∞l.

Remark 1.1.3 If for any ε > 0 there exists N (ε) ∈ N such that an ≥ ε for any n ≥ N (ε), we saythat the sequence an has the limit +∞. Similarly, if for any ε > 0 there exists N (ε) ∈ N such thatan ≤ −ε for any n ≥ N (ε), we say that the sequence an has the limit −∞. We write limn→∞ an = +∞,respectively limn→∞ an = −∞, but we do not say that the sequence an is convergent (this requires thelimit to be finite! We may say that the sequence an diverges to +∞, respectively −∞).

2

CHAPTER 1. SEQUENCES OF REAL NUMBERS 3

We have the following:

Proposition 1.1.4 If the limit of a sequence exists, it is unique.

Proof. Assume there exists l1, l2 ∈ R such that the sequence (an)n∈N converges to l1 and l2. Givenε > 0, there exists N1 (ε) and N2 (ε) such that

|an − l| < ε, n ≥ N1 (ε) ,

and|an − l| < ε, n ≥ N2 (ε) .

It follows that we have

|l1 − l2| = |l1 − an − (l2 − an)|≤ |an − l1|+ |an − l2|< ε+ ε

= 2ε,

for any n ≥ max {N1 (ε) , N2 (ε)} .Thus |l1 − l2| < 2ε for any ε > 0, and therefore |l1 − l2| = 0, that is l1 = l2.A subsequence of a given sequence (an)n∈N is a sequence (akn)n∈N, where kn are natural numbers

with 0 ≤ k1 < k2 < ...

Proposition 1.1.5 Let (an)n∈N be a sequence of real numbers. We have the following:a) If (an)n∈N converges to l, then any subsequence (akn)n∈N converges to l.b) If (an)n∈N is a convergent sequence, then (an)n∈N is also a Cauchy sequence.c) If (an)n∈N is a Cauchy sequence which has a convergent subsequence, then (an)n∈N is also a

convergent sequence.

Lemma 1.1.6 (Cesaro’s Lemma) Any bounded sequence contains a convergent subsequence.

From the previous proposition and lemma, we obtain the following:

Theorem 1.1.7 (Cauchy’s theorem for sequences) A sequence is convergent if and only if it isCauchy.

Remark 1.1.8 We can use the previous theorem to show for example that the sequence (an)n∈N withan = 1+

12 + ...+ 1

n is not convergent. By contradiction, if (an)n∈N were convergent, then it were alsoCauchy, and therefore |an − am| can be made arbitrarily small for all n and m sufficiently large. But

|a2n − an| =

¯̄̄̄1

n+ 1+

1

n+ 2+ ...+

1

2n

¯̄̄̄=

1

n+ 1+

1

n+ 2+ ...+

1

2n

>1

2n+1

2n+ ...+

1

2n

=1

2,

for any n ≥ 1, which shows that (an)n∈N is not Cauchy, and therefore not convergent.

The following proposition gives sufficient conditions for the convergence of a sequence:

Proposition 1.1.9 If the sequence (an)n∈N is increasing and bounded above (or decreasing and boundedbelow), then it is convergent.

Mihai N. Pascu — Mathematical Analysis lecture notes


Proof. Suppose that the sequence (an)n∈N is increasing, and there existsM ∈ R such that an ≤Mfor any n ∈ N.

Let l = supn∈N an ∈ R (note that since an ≤M , we have l ≤M < +∞).By the definition of the supremum, given ε > 0, there exists N (ε) ∈ N such that

l − ε < aN(ε).

Since the sequence an is increasing, for any n ≥ N (ε) we have

l − ε < aN(ε) ≤ an ≤ l < l + ε,

or equivalent|an − l| < ε, n ≥ N (ε) ,

which shows that the sequence (an)n∈N is convergent to l.Similar proof for the case when the sequence (an)n∈N is decreasing and bounded below.

Remark 1.1.10 If the sequence is not monotone, or if it is not bounded, then it may not be convergent.To see this, consider for example the sequences (an)n∈N with an = (−1)

n (bounded, but not monotone)or an = n (increasing, but not bounded above).

From the previous remark we see that not any sequence (an)n∈N is convergent. However, wecan define two important “limits” for any sequence, using the previous proposition (by noticing that¡supk≥n ak

¢n∈N is a decreasing sequence, respectively (infk≥n ak)n∈N is an increasing sequence), as

follows.

Definition 1.1.11 We define the superior limit / lim sup of a sequence (an)n∈N by

lim sup an = limn→∞

supk≥n

ak ∈ R ∪ {+∞} ,

and the inferior limit / lim inf by

lim sup an = limn→∞

infk≥n

ak ∈ R ∪ {−∞} .

Remark 1.1.12 It can be shown that lim inf an ≤ lim sup an for any sequence (an)n∈N, and that thesequence (an)n∈N is convergent if and only if we have equality, that is

lim inf an = lim sup an

³= lim

n→∞an

´.

It can also be shown that lim sup an is the largest limit of a convergent subsequence of the sequence(an)n∈N, and lim inf an is the smallest limit of a convergent subsequence of a the sequence (an)n∈N.

Example 1.1.13 The sequence (an)n∈N given by an = (−1)n is not convergent, so limn→∞ an doesnot exist.

However, it is easy to see that supk≥n ak = 1 and infk≥n ak = −1 for all n ∈ N, and therefore

lim inf an = −1 < +1 = lim sup an.

Remark 1.1.14 (Passing to the limit in inequalities) If (an)n∈N and (bn)n∈N are sequences suchthat an ≤ bn for any n ∈ N, it is easy to see that we have

lim inf an ≤ lim inf bn

andlim sup an ≤ lim sup bn.

If in particular limn→∞ an and limn→∞ bn exists, then we also have

limn→∞

an ≤ limn→∞

bn.



Remark 1.1.15 It is important to note that even if an < bn for all n ∈ N (the inequalities are strict),the resulting inequalities may not be strict, as it can be see by considering an = 1 < 1 + 1

n = bn, forwhich we have limn→∞ an = 1 = limn→∞ bn.

As another sufficient condition for convergence, we have the following:

Theorem 1.1.16 (Squeeze theorem) If an ≤ bn ≤ cn for all n ∈ N and the sequences (an)n∈N and(cn)n∈N are convergent to the same limit l, then the sequence (bn)n∈N is also convergent and it hasthe limit l.

Proof. Passing to the limit (inferior limit) in the given inequality, we obtain

lim inf an ≤ lim inf bn ≤ lim inf cn,

and since the sequences an and cn are convergent to l, we obtain

l ≤ lim inf bn ≤ l,

that is lim inf bn = l.A similar proof shows that lim sup an = l, and therefore lim inf bn = lim sup bn = l, which shows

that the sequences bn is convergent and limn→∞ bn = l.The following proposition contains some properties of convergent sequences:

Proposition 1.1.17 If (an)n∈N and (bn)n∈N are convergent sequences, then:

1. The sequence (an ± bn)n∈N is also convergent, and we have

limn→∞

(an ± bn) = limn→∞

an ± limn→∞

bn

2. The sequence (an · bn)n∈N is also convergent, and we have

limn→∞

(an · bn) = limn→∞

an · limn→∞

bn

3. If bn 6= 0 and limn→∞ bn 6= 0, then the sequence³anbn

´n∈N

is also convergent, and we have

limn→∞

(an ± bn) = limn→∞

an ± limn→∞

bn

4. The sequence (|an|)n∈N converges to |a|.

Proof. 1. If limn→∞ an = a and limn→∞ bn = b, then for any ε > 0 we have

|(an ± bn)− (a± b)| ≤ |an − a|+ |bn − b|< ε/2 + ε/2

= ε,

for any n ≥ N (ε), since an and bn are convergent sequences to a, respectively b, proving the claim2. First note that since (an)n∈N and (bn)n∈N are convergent sequences, they are bounded, so there

exists M > 0 such that|an| , |bn| < M, n ∈ N,

and passing to the limit we see that we also have |a| , |b| ≤M .Given ε > 0 there exists N (ε) such that

|an − a| , |bn − b| < ε

2M, n ∈ N.



For any n ≥ N (ε) we have:

|anbn − ab| = |an (bn − b) + (an − a) b|≤ |an| |bn − b|+ |b| |an − a|≤ M

ε

2M+M

ε

2M

=ε

2+

ε

2= ε,

which shows that the sequence (an · bn)n∈N converges to a · b.3. Similar to the previous proof, for 0 < ε < |b|

2 , we have by writing¯̄̄̄anbn− a

b

¯̄̄̄=

|anb− abn||bnb|

=|an (b− bn) + bn (an − a)|

|b2 + b (bn − b)|

≤̇ |an| |bn − b|+ |bn| |an − a||b|2 − |b| |bn − b|

≤ Mε+Mε

b2 − |b| ε

≤ 2Mε

b2 − |b| |b|2=

4M

|b|2ε,

which can be made arbitrarily small (here we need b 6= 0), showing that limn→∞anbn= a

b .The following theorem is often useful in exercises for computing limits of the type limn→∞

anbn:

Theorem 1.1.18 (Cesaro-Stolz theorem) Let bn be a sequence of positive numbers which is in-creasing to +∞. Then

limn→∞

anbn= lim

n→∞an+1 − anbn+1 − bn

,

provided the last limit exists.

Proof. We will consider the case when l = limn→∞an+1−anbn+1−bn is finite (when l = ±∞ the proof is

similar).From the definition of the limit it follows that for every ε > 0 there is Nε such that for any n ≥ Nε,

we have :l − ε <

an+1 − anbn+1 − bn

< l + ε.

Because bn is strictly increasing we can multiply the last equation with bn+1 − bn > 0 to get:

(l − ε)(bn+1 − bn) < an+1 − an < (l + ε)(bn+1 − bn).

Let k ≥ Nε be a natural number. Summing the above relations for n = Nε, Nε + 1, . . . , k we get :

(l − ε)kX

n=Nε

(bn+1 − bn) <kX

n=Nε

(an+1 − an) < (l + ε)kX

n=Nε

(bn+1 − bn),

or equivalent(l − ε)(bk+1 − bNε) < ak+1 − aNε < (l + ε)(bk+1 − bNε).

Dividing by bk+1 we obtain:

(l − ε)(1− bNε

bk+1) <

ak+1bk+1

− aNε

bk+1< (l + ε)(1− bNε

bk+1),



or equivalent

(l − ε)(1− bNε

bk+1) +

aNε

bk+1<

ak+1bk+1

< (l + ε)(1− bNε

bk+1) +

aNε

bk+1.

Since bk increases to +∞, bNεbk+1

and aNεbk+1

tend to 0 as k → ∞, and therefore there exists K suchthat

(l − ε)(1− ε)− ε <ak+1bk+1

< (l + ε)(1− ε) + ε,

for all k ≥ K, which shows that

limk→∞

ak+1bk+1

= l = limn→∞

an+1 − anbn+1 − bn

,

concluding the proof.As an application, we have the following theorem, useful for computing limits of the type limn→∞ n

√an:

Theorem 1.1.19 (Root criterion for sequences) Let an be a sequence of positive numbers. Then

limn→∞

n√an = lim

n→∞an+1an

,

provided the last limit exists.

Proof. Considerln n√an = ln

³a1/nn

´=1

nln an =

ln ann

,

and apply Césaro-Stolz theorem above with an and bn replaced by ln an, respectively n.We have

limn→∞

ln ann

= limn→∞

ln an+1 − ln an(n+ 1)− n

= limn→∞

(ln an+1 − ln an) = limn→∞

lnan+1an

= ln limn→∞

an+1an

.

Since limn→∞an+1an

exists, from the Césaro-Stolz theorem we obtain

ln limn→∞

n√an = lim

n→∞ln ann

= ln limn→∞

an+1an

,

or equivalentlimn→∞

n√an = lim

n→∞an+1an

,

concluding the proof.

1.2 Exercises

1. Show that if (an)n≥1 converges, then (|an|)n≥1 also converges. Is the converse true? (eitherprove it or give a counterexample).

2. Calculate limn→∞³√

n2 + n− n´

3. If a1 =√2 and an+1 =

p2 +√an, n ≥ 1, show that the sequence (an)n≥1 converges and find

its limit.

4. Compute the limits of the following sequences:

(a) an =√n+ 1−√n

(b) an =√n+1−√n

n

5. Let a1 >√α and an+1 =

12

³an +

αan

´, n ≥ 1. Show that (an)n≥1 is a decreasing, bounded

sequence and its limit is limn→∞ an =√α.



6. Find the limit of the following sequences:

an =3n+ 1

2n2 + 5nbn =

3n2 + 4n

2n− 1 cn =4n2 + 2n− 15n2 + 10n


an =4

3n+ 1bn =

2n+ 3

5n− 1 cn =5n− 23n+ 1


an =√n2 − 3n+ 2−

√n2 + 2n+ 1 bn =

√2n− 1−

√n+ 2

cn =√4n+ 3−

√4n− 2 dn = n−

√n2 − 3n+ 2


an =

µ2 +

3

n

¶2bn =

µ5 +

4

n

¶n


an =n+ 2√n3 + 4

bn =n2 − 3√2n3 + 1

cn =n2 − 2n− 1√4n4 − 3n3

dn = n3 −pn2 + 1


an =

µ1 +

4

n− 2

¶5n2+2nbn =

µn− 2n+ 3

¶ n2+2n3+4n

cn =

µn2 − 3n+ 2n2 + 4n+ 4

¶n2+2nn


an =3n + 4 · 5n − 6 · 7n2n + 3 · 4n + 5 · 7n bn =

2n + 3 · 4n − 5 · 6n4n − 2 · 5n + 7n

cn =¡1 + 3 + 32 + . . .+ 3n

¢ 3n + 4n5n

dn =

µ1 +

1

3+1

32. . .+

1

3n

¶(−1)n + 4n2 · 4n


an =pn3 + 2n

¡√n+ 2−

√n− 1

¢bn =

pn2 + 1

³pn2 + 2n−

pn2 − 3n

´14. Find the limit of the following sequences:

an =n

s(n+ 1)!

(2n+ 1)! · 3n bn =n

s((n+ 1)!)2

(2n+ 1)! · 3n


an =1 + 1

2 + . . .+ 1n

nbn =

1 + 1√2+ . . .+ 1√

n√n

cn =1ln 2 +

1ln 3 + . . .+ 1

lnn

n2

16. Find the limit of the sequence limn→∞

1k + 2k + . . .+ nk

nk+1where k ∈ N is a natural number.


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