Chapter 3
Sequences
Both the main elements of calculus (differentiation and integration) require the
notion of a limit. Sequences will play a central role when we work with limits.
Definition 3.1. A Sequence is a function whose domain is the natural numbers.
We sometimes denote a sequence as {f (n)}n∈N, where f (n) is a function on the
natural numbers, and we do not specify the range.
Example 3.2. Let
1
n
n
.
2
n (−1)n
Then f is a function from the natural numbers into 2 × 2 matrices, and {f (n)}n∈N
is a sequence of 2 × 2 matrices.
f (n) =
Example 3.3. Let
f (n) = xn
on [0, 1]. Then f is a function from the natural numbers to functions on [0, 1], and
{f (n)}n∈N is sequence of functions.
Example 3.4. Let
1
.
n
Then f is a function from the natural numbers to the real numbers, and {f (n)}n∈N
is a sequence of numbers.
f (n) =
Before proceeding to define the limit of a sequence, try to compute the following
limits. If you find the limit, ask yourself how you did it and what does the limit
represent. Imagine explaining the process to a freshman.
n+1
3n + 2
sin n
2. lim
n→∞ n
1. lim
n→∞
3. lim (ln(n + 1) − ln n)
n→∞
4. lim (1 + n)1/n
n→∞
Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.
19
20
3. Sequences
3.1. Convergence
For the rest of this chapter we will only consider sequences of real numbers, typically
denoted {an }n∈N .
Definition 3.5. A sequence of real numbers, {an }n∈N , converges to a real number,
a, if and only if, for each ǫ > 0, there exists a natural number N , such that, for
all n ≥ N , |an − a| < ǫ. When {an }n∈N converges to a we write lim an = a or
n→∞
an → a as n → ∞.
In plain language the definition says an is close to a for large n. We note that
implementing the definition requires a candidate for the limit. Later you will see
that we can establish the existence of a limit without having to actually know the
limit.
1
Example 3.6. Prove lim
= 0.
n→∞ n
Proof. Let ǫ > 0. By the Archimedean Property we can find a natural number N
so that 1ǫ < N , or N1 < ǫ. Thus for all n ≥ N
or
1
1
≤
< ǫ,
n
N
1
− 0 < ǫ,
n
for n ≥ N .
�
Sometimes a limit does not exist. We need methods to deal with limits not
existing. One possibility would be to negate the definition.
Lemma 3.7. A sequence {an }n∈N does not converge to the number a if there exists
a ǫ0 > 0 such that, for all N ∈ N, there exists a n0 ≥ N with |an0 − a| ≥ ǫ0 .
Proof. In logic notation the definition of convergence of a sequence is
or
(∀ǫ > 0)(∃N ∈ N)(∀n ≥ N )(|an − a| < ǫ)
(∀ǫ > 0)(∃N ∈ N)(∀n ∈ N)(n ≥ N → |an − a| < ǫ).
Negating the two, we find
or
(∃ǫ0 > 0)(∀N ∈ N)(∃n0 ≥ N )(|an − a| ≥ ǫ)
(∃ǫ0 > 0)(∀N ∈ N)(∃n0 ∈ N)(n0 ≥ N and |an − a| ≥ ǫ).
�
n
Example 3.8. Prove the sequence {(−1) }n∈N does not have a limit.
Proof. The difficulty in applying the lemma is that we have to show there is no
limit for each real number a. Suppose a = 1. Set ǫ0 = 1. Then, for all N ∈ N,
one of |aN − 1| or |aN +1 − 1| is 2, and for all N ∈ N there exists a n0 ≥ N with
|an0 − 1| ≥ ǫ0 . A similar argument follows for a = −1. Finally suppose a is not
equal to 1 or −1. Let ǫ0 = min{|a − 1|, |a + 1|}n∈N. Then given any N ∈ N we have
|an − a| ≥ ǫ0 for all n ∈ N.
�
3.2. Some Topology
21
Here is another proof which uses contradiction.
Proof. Suppose the limit did exist. Then, for ǫ = 1/2, there exists a N ∈ N such
that |an − a| < 1/2 for n ≥ N . But
1 1
2 = |1 + 1| = |(1 − a) + (a + 1)| ≤ |1 − a| + |a + 1| = |1 − a| + | − 1 − a| < + = 1, n
2 2
a contradiction.
�
Theorem 3.9. The limit of a sequence, if it exists, is unique.
Proof. Suppose there were two limits, say a and b. Given any ǫ > 0 there exists
an N1 and N2 such that |an − a| < ǫ/2 and |an − b| < ǫ/2 for all n ≥ max{N1 , N2 }.
Then
|a − b| = |a − an + an − b| ≤ |an − a| + |an − b| < ǫ,
for all ǫ > 0. By trichotomy, either |a − b| = 0 for |a − b| > 0. If the latter
holds, set ǫ = |a − b|/2. Then on the one hand |a − b| < ǫ, while on the other,
|a − b| > |a − b|/2 = ǫ, a contradiction. Thus |a − b| = 0 or a = b.
�
3.2. Some Topology
A large part of mathematics consists of recognizing and generalizing patterns. If we
look carefully at the definition of convergence for sequences we see that we really
do not need the field axioms. All we require is the notion of closeness. We can have
that concept without the use of addition or subtraction (or even absolute value).
Definition 3.10. A subset O of R is called open in R if, for each point x ∈ O,
there is a r > 0 such that all points y in R satisfying |x − y| < r also belong to the
set O.
Example 3.11. The set O = {x | 0 < x < 1} is open in R.
Example 3.12. The set O = {x | 0 < x ≤ 1} is not open in R.
We state without proof the following theorems.
Theorem 3.13. Open Set Properties.
(a) The empty set and R are open in R.
(b) The intersection of any two open sets is open in R.
(c) The union of any collection of open sets is open in R.
Definition 3.14. (a) A subset of R is called closed in R if its complement is open
in R.
(b) A neighborhood of x ∈ R is any set containing an open set containing x.
(c) A point x ∈ R is called an boundary point of a set S ⊂ R if every neighborhood
of x contains a point in S and a point not in S.
(d) A point x ∈ R is called an interior point of a set S ⊂ R if there exists a
neighborhood of x contained in S.
(e) A point x ∈ R is called an exterior point of a set S ⊂ R if there is a neighborhood of x which is entirely contained in complement of S.
22
3. Sequences
Example 3.15. The set O = {x | 0 ≤ x ≤ 1} is closed in R.
Example 3.16. Consider the set O = {x | 0 ≤ x ≤ 1}. The boundary points are
x = 0 and x = 1. Moreover x = 1/2 is an interior point.
Again without proof we have
Theorem 3.17. Let O ⊂ R. The following statements are equivalent
(a) O is open in R;
(b) every point in O is an interior point of O.
(c) O is a neighborhood of each of its points.
Theorem 3.18. A set is closed if and only if it contains all of its boundary points.
Theorem 3.19. A subset of R is open if and only if it is the union of a countable
collection of open intervals.
Finally we can recast the definition 3.5. The advantage of doing so it that we
do not need the field axioms to define an open set, and we could, if we had more
time, generalize the notion of convergence to sets with less structure than the ones
we consider here.
A sequence {an }n∈N eventually has a certain property if it has that property
for all n ≥ N for some N ∈ N.
Definition 3.20. A sequence of real numbers {an }n∈N converges to a number a if,
for each neighborhood of a, an is eventually in that neighborhood.
3.3. Algebra of Limits
In this section we establish some familiar properties of sequences.
Theorem 3.21. Suppose {an }n∈N and {bn }n∈N are convergent sequences, with
limits a and b respectively. Then
(a) For all α ∈ R, lim αan = αa.
n→∞
(b) lim (an + bn ) = a + b.
n→∞
(c) lim an bn = ab.
n→∞
(d) If bn = 0 for all n ∈ N and b = 0, then lim
n→∞
an
a
= .
bn
b
Proof. We will prove (c) and (d) and leave the other two as exercises. We are
given the following. Given any ǫ′ > 0 there exists N1 , N2 , natural numbers, such
that |an − a| < ǫ′ and |bn − b| < ǫ′ for all n ≥ max{N1 , N2 }n∈N . We calculate
|an bn − ab| = |(an − a)bn + a(bn − b)| ≤ |bn ||an − a| + |a||bn − b|.
We know |an − a| and |bn − b| are small. We have to worry about |bn | becoming
large. Let us suppose for the moment that |bn | ≤ M for all n ∈ N and some M > 0.
Then
|an bn − ab| ≤ M |an − a| + |a||bn − b|.
23
3.3. Algebra of Limits
Let ǫ > 0 be given. In one case we choose ǫ′ = ǫ/2M , and in the other ǫ′ = ǫ/2a.
Then
ǫ
ǫ
|an bn − ab| ≤ M |an − a| + |a||bn − b| < + = ǫ
2 2
for all n ≥ max{N1 , N2 }. We show convergent sequences are bounded in the next
lemma.
The proof of (d) is similar. Indeed,
an
a
−
bn
b
=
=
≤
ban − abn
bbn
ban − ba + ba − abn
bbn
|an − a| |a||bn − b|
+
.
|bn |
|b||bn |
Again we must worry about 1/|bn | becoming large. Let us suppose for the moment
that 1/|bn| ≤ M for n ≥ N3 , for some N3 ∈ N and for some M . Then, as in Part
(c), let ǫ > 0 be given. We choose ǫ′ = ǫ/(2M ) for the bound on |an − a| and
ǫ′ = |b|ǫ/(2|a|) for the bound on |bn − b|. For n ≥ max{N1 , N2 , N3 }, we have
a
ǫ
ǫ
an
−
< + .
bn
b
2 2
�
We have two bounds to establish.
Lemma 3.22. Suppose {bn }n∈N converges to b. Then a M > 0 exists so that
|bn | ≤ M for all n ∈ N.
Proof. Let ǫ = 1 in Definition 3.5. Then a natural number N exists so that
|bn − b| < 1 for all n ≥ N . The triangle inequality implies, for such n,
|bn | − |b| ≤ |bn | − |b| ≤ |bn − b|.
That is, |bn | ≤ 1 + |b| for n ≥ N . Set M = max{|b1 |, |b2 |, . . . , |bN |, 1 + |b|}, and we
find |bn | ≤ M for n ∈ N.
�
Lemma 3.23. Suppose {bn }n∈N converges to b with bn = 0 for all n ∈ N and b = 0.
1
Then a N ∈ N and a M > 0 exists so that
≤ M for all n ≥ N .
|bn |
Proof. Here we have to worry about |bn | becoming small. Let ǫ = |b|/2 in Definition 3.5. A N ∈ N exists such that |bn − b| < |b|/2 for n ≥ N . The triangle
inequality shows
|b|
|bn | = |bn − b + b| ≥ |b| − |bn − b| >
2
for all n ≥ N . Set M = 2/|b| and the result follows.
�
Theorem 3.24. (Squeeze Theorem) Let {xn }n∈N ,{yn }n∈N , and {zn }n∈N be three
sequences of real numbers such that xn ≤ yn ≤ zn for all n. Suppose further that
{xn }n∈N and {zn }n∈N converge to x. Then {yn }n∈N converges to x.
24
3. Sequences
Proof. A good place to start would be to write down what we know: Given any
ǫ > 0 there exists Ni ∈ N, i = 1, 2, such that |xn − x| < ǫ for all n ≥ N1 and
|zn − x| < ǫ for all n ≥ N2 . The above inequalities imply
−ǫ < xn − x ≤ yn − x ≤ zn − x < ǫ
for n ≥ max{N1 , N2 }. That is, |yn − x| < ǫ for such n.
�
Corollary 3.25. Given any ξ ∈ Qc , there exists sequence of rational numbers
{rn }n∈N which converge to ξ. That is, irrational numbers can by approximated with
rational numbers.
Proof. Clearly ξ < ξ + n1 . By the density of rationals there exists rn such that
�
ξ < rn < ξ + n1 . The squeeze theorem implies {rn }n∈N converges to ξ.
While the supremum of a set may Not be in the set, it can however be approximated by elements in the set.
Corollary 3.26. Let E ⊂ R be non empty and bounded from above. Then a
sequence {xn }n∈N exists in E such that xn → α = sup E as n → ∞.
Proof. The set E has a supremum, α, by the completeness axiom. Since α − n1
is no longer an upper bound there exists wn ∈ E with α − n1 < wn < α (Theorem
�
2.11). The squeeze theorem shows {wn }n∈N converges to α.
3.4. Monotone Convergence Theorem
We will develop two techniques to show a sequence converges without having to find
the limit and apply Definition 3.5. The application of the Monotone Convergence
Theorem is one of the techniques. It is equivalent to the completeness axiom.
Theorem 3.27. (Monotone Convergence Theorem) Suppose {an }n∈N is a monotone increasing, bounded sequence of real numbers. That is, an ≤ an+1 and |an | ≤
M some M and for all n ∈ N. Then {an }n∈N has a limit.
Proof. To apply the definition for convergence we need a candidate for the limit.
Set E = {a1 , a2 , . . .}. The set E is non empty and bounded. By the completeness
axiom E has a supremum. Set a = sup E. Let ǫ > 0. Then a − ǫ is no longer an
upper bound for E. Hence a aN exist such that a − ǫ < aN ≤ a. Since the sequence
is increasing a − ǫ < an ≤ a for all n ≥ N . That is 0 ≤ a − an < ǫ or |a − an | < ǫ
for all n ≥ N .
�
Example 3.28. Consider the sequence
an =
1
1
1
+
+ ···+
.
n+1 n+2
2n
The sequence is bounded since
|an | ≤
1
1
n
1
+
+ ··· +
=
≤ 1.
n+1 n+1
n+1
n+1
25
3.4. Monotone Convergence Theorem
It is also monotone since
an+1 − an
1
1
1
1
1
+
...+
+
+
n+2 n+3
2n 2n + 1 2n + 2
1
1
1
+
+ ···+
−
n+1 n+2
2n
1
1
2
=
−
=
2n + 1 2(n + 1)
(n + 1)(2n + 1)
≥ 0
=
By the monotone convergence theorem, the sequence {an }n∈N has a limit (in fact,
we will see later that lim an = ln 2).
n→∞
Example 3.29. Suppose 0 < a < 1. Prove lim an = 0.
n→∞
n
Proof. Set an = a . The sequence is a monotone decreasing sequence in this case.
The monotone convergence theorem still applies (see exercise 3.20). We need to
show the sequence is bounded from below. This is easy since an = an = |a|n ≥ 0.
We show the sequence is monotone decreasing. That is, an+1 ≤ an . We do this
by induction. Since a < 1, a2 < a. This is the base step. Suppose an < an−1 .
That is, an < an−1 . Multiplying by a, we find an+1 < an . That is, an+1 ≤ an .
By induction an+1 ≤ an for all n ∈ N, and the sequence is decreasing. By the
monotone convergence theorem, {an }n∈N has a limit.
To find the limit, we apply the algebra of limits (we had to know the limit
existed to do this!). Set lim an = L. Then
n→∞
L = lim an+1 = lim aan = a lim an = aL.
n→∞
n→∞
n→∞
That is, L(1 − a) = 0. This implies L = 0.
�
Example 3.30. Suppose the sequence {xn }n∈N is defined recursively by xn+1 =
2 − 1/xn with x1 = 2. Prove that {xn }n∈N has a limit and find the limit.
Proof. Notice 1 ≤ xn ≤ 2 for all n ∈ N. Indeed, the inequality is true for n = 1.
Suppose 1 ≤ xn−1 ≤ 2. Then xn = 2 − 1/xn−1 , and under the assumption on xn−1 ,
1 ≤ xn ≤ 2. By induction, the inequality is true for all n.
The sequence is also decreasing. Indeed, for all n
(xn − 1)2
1
− xn = −
<0.
xn
xn
By the monotone convergence theorem the sequence has a limit. Moreover, by the
algebra of limits, the limit satisfies x = 2 − 1/x, or x = 1.
�
√
Example 3.31. Find a rational sequence converging to 2.
xn+1 − xn =
2−
Proof. We know by Corollary 3.25 such a sequence exists. Newton’s method for
finding zeros of a differentiable function is
f (xn )
xn+1 = xn − ′
.
f (xn )
26
3. Sequences
We do not need to study Newton’s method - we just use it to find a candidate for
the sequence. Applying the method to the function f (x) = x2 − 2, we find
xn+1 =
(3.1)
1
2
xn +
2
xn
.
We arbitrary choose x0 = 2. Then x1 = 3/2, x2 = 17/12, x3 = 577/408 .... You
√
can check that the sequence of rational numbers seems to converge (rapidly) to 2.
To prove the convergence we apply the monotone√convergence theorem. We
first show the sequence {xn }n∈N is bounded. Clearly, 2 ≤ x0 = 23 ≤ 2. Suppose
√
2 ≤ xn ≤ 2. Then, using (3.1),
√ min
2≤x≤2
The function g(x) =
bounded by
1
2
1
2
x+
x+
2
x
1
2
x+
.
x
2≤x≤2 2
√
√
has a minimum of 2 at x = 2. The right side is
2
x
≤ xn+1 ≤ √max
√
1
2
1
2+ 2
x
1
x+
≤ √max
+ √max
≤1+ √ ≤
≤2.
√max
x
2
2
2≤x≤2 2
2≤x≤2 2
2≤x≤2 x
√
By induction 2 ≤ xn ≤ 2 for all n ∈ N.
Next we prove {xn }n∈N is monotone decreasing. Clearly x1 − x0 = −1/2 ≤ 0.
Suppose xn − xn−1 ≤ 0. Again using (3.1),
xn+1 − xn = (xn − xn−1 )
Since xn ≥
√
2 for all n ∈ N,
1
1
−
2 xn xn−1
.
1
1
1
1
≥ −√ √ ≥0.
−
2 xn xn−1
2
2 2
Hence, xn+1 − xn ≤ 0, and, by induction, xn − xn−1 ≤ 0 for all n ∈ N.
The monotone convergence applies, and {xn }n∈N has a limit. Suppose the limit
is L. Then, using the algebra of limits,
L = lim xn+1 = lim
n→∞
n→∞
Solving for L, we find L =
√
2.
1
2
xn +
2
xn
=
1
2
L+
2
L
.
�
We have one final application of the monotone convergence theorem. As you
will see, we need to know when an infinite intersection of intervals is non empty.
Example 3.32. Set In = (0, 1/n) = {x | 0 < x < 1/n}. The set {In }n∈N is a
sequence of intervals. We will be interested in the infinite intersection
E :=
∞
n=1
In this case note E = ∅.
= {x | x ∈ In , ∀n ∈ N}.
27
3.5. Subsequences and The Bolzano-Weierstrass Theorem
Example 3.33. Set In = (n, ∞) = {x | n < x}. Again the set {In }n∈N is a
sequence of intervals. Here
∞
E :=
n=1
= {x | x ∈ In , ∀n ∈ N} = ∅.
Definition 3.34. A sequence of intervals {In }n∈N is called nested if I1 ⊆ I2 ⊆ I3 ⊆
. . ..
Here are sufficient conditions for the intersection to be non empty.
Theorem 3.35. (Nested-Cell Theorem) Let {In }n∈N be a sequence of nested, closed,
bounded intervals. Then
∞
n=1
= {x | x ∈ In , ∀n ∈ N} is non empty.
Proof. By assumption In = [an , bn ] with a1 ≤ a2 ≤ . . . and b1 ≥ b2 ≥ . . .. For any
natural numbers n and k
ak ≤ an ≤ b n
for
k<n
ak ≤ bk ≤ bn for k ≥ n.
The two inequalities imply ak ≤ bn for all k ∈ N and any n. Thus the monotone
increasing sequence {an }n∈N is bounded by bn for any n. By the monotone convergence theorem lim an = a := sup{an }. The same two inequalities above now imply
n→∞
that a is a lower bound for {bn }n∈N . Since {bn }n∈N is monotone decreasing and
bounded below, the monotone convergence theorem implies lim bn = b := inf{bn }.
n→∞
Moreover a ≤ b. Finally, let x ∈ [a, b]. Then ai ≤ a ≤ x ≤ b ≤ bi for all i ∈ N.
�
That is, x ∈ ∞
n=1 In .
3.5. Subsequences and The Bolzano-Weierstrass Theorem
A subsequence of a sequence {an }n∈N is a subset of {an }n∈N which is ordered. More
specifically,
Definition 3.36. Let {an }n∈N be a sequence and {nk }k∈N be any sequence of
natural numbers such that n1 < n2 < n3 < . . .. The sequence {ank }k∈N is called a
subsequence of {an }n∈N .
n
does not converge. However, if
Example 3.37. The sequence with an = 1+(−1)
2
nk = 2k, {a2k }k∈N = {0, 0, 0, . . .}. Similarly, nk = 2k − 1 produces the subsequence
{a2k−1 }k∈N = {1, 1, 1, . . .}.
Theorem 3.38. A sequence converges to a if and only if each of its subsequences
converges to a. In fact, if every subsequence converges, they all converge to the
same limit.
Proof. If all subsequences converge to a, the sequence converges to a since the
sequence is a subsequence of itself.
Conversely, suppose the sequence converges to a and {ank }k∈N is a subsequence.
Let ǫ > 0. There exists a natural number N such that |an − a| < ǫ for all n ≥ N .
By the construction of the {nk }k∈N we have k ≤ nk for all k ∈ N. Thus, for k ≥ N ,
28
3. Sequences
nk ≥ N . This implies |ank − a| < ǫ for all k ≥ N , and the subsequence {ank }k∈N
converges to a.
�
A restatement of part of the theorem gives
Corollary 3.39. If {an }n∈N converges, all of its subsequences converge to the same
limit.
This gives us another technique for showing a sequence does not converge. We
need only find two convergent subsequences with distinct limits.
Example 3.40. Show the sequence an = (−1)n does not converge using three
different methods.
Proof. We have already shown non convergence using two difference methods in
Example 3.8. Note that {a2k }k∈N and {a2k−1 }k∈N converge to 1 and −1 respectively. Thus {an }n∈N does not converge.
�
The importance of the following theorem cannot be overstated.
Theorem 3.41. (Bolzano-Weierstrass) Every bounded sequence of real numbers
has a convergent subsequence.
Proof. Let us denote the sequence {xn }n∈N . We apply the Nested-Cell Theorem,
Theorem 3.35. We will construct inductively a sequence of intervals such that
I1 ⊃ I2 ⊃ I3 ⊃ . . ..
Since the sequence is bounded, a ≤ xn ≤ b for all n ∈ N and some a and b. Set
I1 = [a, b]. Choose a xn1 ∈ I1 . We define |I1 | = b − a. Consider
I1 = I ′ ∪ I ′′ = a,
a+b
a+b
∪
,b .
2
2
One of I ′ or I ′′ contains infinitely many of the xn . Call it I2 . Then |I2 | = (b − a)/2.
Choose xn2 ∈ I2 with n2 > n1 . Note that I2 ⊃ I1 . This is the base step.
Suppose I1 , . . . , Ik−1 have been chosen so that I1 ⊃ I2 ⊃ I3 ⊃ . . . Ik−1 with
b−a
|Ik−1 | = k−2 ,
2
and xn1 ∈ I1 , xn2 ∈ I2 , . . ., xnk−1 ∈ Ik−1 , with n1 < n2 < . . . < nk−1 . By
construction Ik−1 contains infinitely many of the xn . Bisect it as before,
I1 = I ′ ∪ I ′′
with one of the I ′ or I ′′ containing infinitely many of the xn . Call the interval Ik .
Choose xnk ∈ Ik ⊆ Ik−1 with nk > nk−1 . Note that
b−a
|Ik | = k−1 .
2
By induction this defines Ik and xnk for all k ∈ N.
The {Ik }k∈N satisfy the hypothesis of the Nested-Cell Theorem. Thus their
intersection is non empty and there is an x ∈
∞
In . We need to show the subse-
n=1
quence {xnk } converges to x as k → ∞. Let ǫ > 0. By Example 3.29 there is a N
3.6. Cauchy Sequences
29
such that
b−a
<ǫ
2k
for all k ≥ N . Since xnk ∈ Ik and x ∈ Ik
b−x
|xnk − x| ≤ |Ik | = k−1 < ǫ
2
(see HW 2.1) for all k ≥ N .
�
There is a nice geometric interpretation of the Bolzano-Weierstrass Theorem.
Definition 3.42. Let E ⊂ R. A point a ∈ R is called a cluster point or accumulation point of E if E ∩ (a − r, a + r) contains infinitely many points for all r > 0.
That is, a is a cluster point of E if every neighborhood of a contains a point in E
distinct from a.
Example 3.43. If E = (0, 1), the cluster points are [0, 1]. Note that a cluster
point need not be in E.
Example 3.44. If E = Q, the cluster points are all of R, since Q is dense in R.
Example 3.45. The set E = {1, 2, 3} has no cluster points.
Example 3.46. Suppose {xn }n∈N is a convergent sequence of real numbers converging to x ∈ R. Then x is a cluster point of the set {x1 , x2 , . . .}.
We may restate the Bolzano-Weierstrass Theorem as follows.
Theorem 3.47. (Bolzano-Weierstrass for Sets) Every bounded infinite subset of
R has a cluster point.
Proof. We will call the infinite set E. We first construct a sequence of distinct
points in E. Since E is non empty, a x1 ∈ E exists. Consider the set E/{x1 }.
It too is non empty. Thus a x2 exists in E/{x1 }. Suppose the distinct points
x1 , x2 , . . . , xk−1 have been similarly chosen. Then the set E/{x1 , x2 , . . . , xk−1 } is
non empty, and there exists a xk ∈ E/{x1 , x2 , . . . , xk−1 }. By induction we have
constructed a sequence {xn }n∈N of distinct points in E.
The sequence {xn }n∈N is bounded since E is bounded. By the BolzanoWeierstrass theorem, a subsequence of {xn }n∈N , {xnk }k∈N exists and converges
to some x ∈ R. Given any neighborhood of x, the sequence {xn }n∈N is eventually
in that neighborhood. Thus every neighborhood of x contains a point in E, and x
is a cluster point of E.
�
3.6. Cauchy Sequences
There is yet another way to show a sequence has a limit without finding the limit
(the monotone convergence theorem was the other).
Definition 3.48. A sequence {xn }n∈N is called Cauchy (pronounced Co She) if
and only if for all ǫ > 0 there exists a N ∈ N such that |xn − xm | < ǫ for all
n, m ≥ N .
30
3. Sequences
Theorem 3.49. (Cauchy) A sequence {xn }n∈N converges if and only if it is Cauchy.
Proof. Suppose {xn }n∈N converges. Let ǫ > 0. Then there exists N ∈ N such that
|xn − x| < ǫ/2 for all n ≥ N . Then
ǫ
ǫ
|xm − xn | = |xm − x + x − xn | ≤ |xm − x| + |xn − x| < + = ǫ
2 2
for all n, m ≥ N . That is, {xn }n∈N is Cauchy.
Suppose {xn }n∈N is Cauchy. We need a candidate for the limit. We apply the
Bolzano-Weierstrass theorem to find the candidate. To do so we need to show the
sequence is bounded. We copy the proof of Lemma 3.22 for this. Let ǫ = 1 in
the definition of a Cauchy sequence. Then a natural number N exists such that
|xm − xn | < 1 for all m, n ≥ N . That is,
|xm | = |xm − xN + xN | ≤ |xm − xN | + |xN | < 1 + |xN |
for all m ≥ N . Thus |xm | ≤ max{|x1 |, |x2 |, . . . , |xN −1 |, 1 + |xN |} for all m ∈ N.
The Bolzano-Weierstrass theorem applies and there exists a subsequence
{xnk }k∈N converging to some x ∈ R. We need to show the whole sequence converges
to x. Let ǫ > 0. There exists N1 and N2 such that
ǫ
∀m, n ≥ N1 ,
|xm − xn | <
2
and
ǫ
|xnk − x| <
∀k ≥ N2 .
2
Set N = max{N1 , N2 }, and fix K ≥ N . Then nK ≥ N , and
|xn − x|
=
≤
<
for all n ≥ N .
|xn − xnK + xnK − x|
|xn − xnK | + |xnK − x|
ǫ
ǫ
+ = ǫ,
2 2
�
Example 3.50. Consider the sequence given recursively by
xn−2 + xn−1
xn =
2
with x1 = 1 and x2 = 2. One can check by writing out the first few terms that the
sequence is not monotone. Indeed, the sequence oscillates. We could try to obtain
a candidate for the limit by assuming it exists and using the algebra of limits. If
lim xn = L, taking the limit of xn above, we would find L = (L + L)/2, which
n→∞
gives us no useful information. Our only hope is to show the sequence is Cauchy
(if we believe it converges).
Note that
1
xn−1 + xn
= |xn − xn−1 |.
|xn − xn+1 | = xn −
2
2
Homework Exercise 3.29 shows that the sequence {xn }n∈N is Cauchy. Moreover,
Cauchy’s theorem shows that the sequence has a limit. We need only find the limit.
31
3.6. Cauchy Sequences
Recall that, if the sequence converges, all subsequences converge to the same
limit (Corollary 3.39). We will find the limit of the subsequence {x2n+1 }n∈N . We
claim that
1
x2n − x2n−1 = n−1 .
4
Indeed, the equality is true for n = 1. Suppose it is true for n. Then
1
x2(n+1) − x2(n+1)−1 = x2n+2 − x2n+1 = (x2n + x2n+1 ) − x2n+1
2
1
1
1
=
x2n − (x2n−1 + x2n )
(x2n − x2n+1 ) =
2
2
2
1
1
=
(x2n − x2n−1 ) = n ,
4
4
and the result follows by induction.
We have discovered that x2n = x2n−1 +
1
4n−1 .
Thus
x1
= 1
1
1
(1 + 2) = 1 +
x3 =
2
2
1 11
1
1
1
x3 + 1 + x3 = 1 + +
(x4 + x3 ) =
x5 =
2
2
4
2 24
1
1 11 1 1
1
1
x7 =
x5 + 2 + x5 = 1 + +
(x6 + x5 ) =
+
2
2
4
2 2 4 2 42
1
1
1
1+ + 2 .
= 1+
2
4 4
Proceeding inductively, we find
1
1
1
1
1 + + 2 + · · · + n−1
x2n+1 = 1 +
2
4 4
4
=
1+
Taking the limit, we find
1
2
1 − 41n
1 − 41
.
lim xn = lim x2n+1 =
n→∞
n→∞
5
.
3
�
32
3. Sequences
3.7. Review
• Techniques to show limit converges
– Apply the definition of a limit
– Algebra of limits
– Show sequence monotone and bounded
– Show sequence Cauchy
• Techniques to show no limit
– Negate the definition of a limit
– Show sequence not bounded
– Show sequence not Cauchy
– Find a contradiction
– More than one convergent subsequence with different limits
• Be able to state (and negate) the definitions of
– Supremum and infimum
– completeness axiom
– lim xn = x.
n→∞
– a sequence and a Cauchy (pronounced Coe She) Sequence
– lim f (x) = L.
x→a
– Continuity at a point and a continuous function.
• Be able to state and prove
– Archimedean principle (any version you want)
– Density of rationals (just the case 0 < x, i.e. prove the existence of r ∈ Q
s.t. 0 < r < x.)
– Monotone convergence Theorem
• Be able to state and apply
– Bolzano-Weierstrass Theorem
– Cauchy’s theorem
• The
–
–
–
–
–
–
–
–
–
–
minimal level of proficiency means you can solve, quickly, problems like
sup[a, b) = b.
|a| < ǫ for all ǫ > 0 implies a = 0.
If {an }n∈N and {bn }n∈N converge to a and b respectively and an ≤ bn for
n ≥ 1. Prove that a ≤ b.
1/n → 0 as n → ∞.
xn → x as n → ∞. Then ∃M > 0, such that |xn | ≤ M for all n ≥ 1.
If 0 < xn < 1 and xn converges to x, where must x live?
Show that n does not converge.
If {an }n∈N and {bn }n∈N converge to a and b respectively, then an bn converges to ab as n → ∞.
Prove that
n3 − 12n + 1
sin(n3 )
lim 4
n→∞ n + 15n2 + 5
has a limit. Give the details.
Look over assigned homework problems!
33
3.7. Review
Summary of theorems so far. Try to understand the idea behind the proof of
each theorem.
The Completeness Axiom
❄
MTC
❄
Nested Cell
❅
❅
❘
❅
Archimedean Principle
❆
❯❆
Denisty of Q
❄
Bolzano-Weierstrass
❍❍
❥
❍
Cauchy Thm
34
3. Sequences
3.8. Homework
Exercise 3.1. Give example of two sequences that do not converge but whose
(a) sum converges
(b) product converges
(c) quotient converges
Exercise 3.2. Give an example of a convergent sequence {xn } ⊂ (0, 1], but whose
limit is not in (0, 1].
Exercise 3.3. Use the Definition 3.5 to establish the following limits.
1
(a) lim 2 = 0
n→∞ n
2n
=2
(b) lim
n→∞ n + 3
2
2n + 4
=
(c) lim
n→∞ 5n + 1
5
Exercise 3.4. If a convergent sequence is rearranged, prove that the new sequence
has the same limit.
Exercise 3.5. Let 0 < xn < 4 for all n ∈ N. Show that the sequence {xn }n∈N has
a convergent subsequence. In what interval will the limit of this subsequence lie?
Exercise 3.6. Suppose an → a > 0 as n → ∞ and that an > 0 for all n ∈ N.
Show that a m > 0 exists so that an ≥ m for all n ∈ N.
Exercise 3.7. (Cesàro Summable) If the sequence {xn }n∈N converges to x, then
the sequence {σn } also converges to x. Here
1
(x1 + x2 + · · · + xn ).
n
Exercise 3.8. Show that the sequence {sin n}n∈N does not converge.
σn =
Exercise 3.9. Let an =
1
n
1
n odd
. Does the limit exist? Proof required.
n even
Exercise 3.10. Suppose {xn }n∈N converges to x and satisfies 2 < xn < 3 for all
n ∈ N. Show that the limit satisfies 2 ≤ x ≤ 3.
Exercise 3.11. Let {an }n∈N be a sequence of real numbers converging to a ∈ R.
(a) Prove lim |an | = |a|.
n→∞
(b) If lim |an | = 0, then lim an = 0.
n→∞
n→∞
(c) Show by example that lim |an | may exist, while {an }n∈N may not converge.
n→∞
Exercise 3.12. Let xn ≥ 0 for all n ∈ N with xn → x as n → ∞. Prove
as n → ∞.
√
√
xn → x
Exercise 3.13. Suppose {xn }n∈N is a convergent sequence of integers. Show that
the sequence is eventually constant.
35
3.8. Homework
Exercise 3.14. Let {an }n∈N be a sequence of positive real numbers with lim
n→∞
L. If L < 1, show that lim an = 0.
an+1
=
an
n→∞
Exercise 3.15. Suppose {xn }n∈N is a bounded sequence (not necessarily convergent), and {yn }n∈N is a sequence converging to zero. Prove that {xn yn }n∈N
converges to zero. Show by example that the result is not true if either hypothesis
on {xn }n∈N or {yn }n∈N is dropped.
Exercise 3.16. Establish, by using any definition or theorem, the limits of the
following sequences.
(a)
(b)
sin n
n
π+
(c)
n∈N
2
1
n
(d)
2n + 4
5n + 1 n∈N
√
√
n− n+1
n∈N
n∈N
Exercise 3.17. Prove Theorem 3.18.
Exercise 3.18. Show that every bounded, non empty closed subset of R has a
maximum. (Hint: show the supremum of a set is a boundary point of the set).
Exercise 3.19. What are the cluster points of the irrational numbers?
Exercise 3.20. Prove the Monotone Convergence Theorem for a decreasing sequence which is bounded below.
Exercise 3.21. Suppose 0 < a. Prove that the sequence {a1/n }n∈N converges to
one.
Exercise 3.22. Show the following recursively defined sequences are convergent
and find the limit
(a) Let x1 = 1 and xn = 41 (2xn−1 + 3).
(b) Let x1 = 3 and xn = 2 − 1/xn−1 .
√
√
(c) Let x1 = 2 and xn = 2 + xn−1 .
Exercise 3.23. Give an example of a sequence whose set of convergent subsequences converge to {1, 1/2, 1/3, 1/4 . . .} ∪ {0}.
Exercise 3.24. Give an example of a sequence with no convergent subsequences.
Exercise 3.25. Give an example of two sequences {xn }n∈N and {yn }n∈N that
satisfy {xn } ⊂ {yn } as sets, but with {xn }n∈N not a subsequence of {yn }n∈N .
Exercise 3.26. The next two statements are equivalent - they are contrapositives.
Prove one of them.
(a) Suppose {xn }n∈N is a bounded sequence. If all its convergent subsequences
have the same limit, then the sequence is convergent.
(b) Show that a nonconvergent bounded sequence has two convergent subsequences each with distinct limits.
36
3. Sequences
Exercise 3.27. Suppose {xn }n∈N is unbounded. Show there exists a subsequence
{xnk } such that 1/xnk → 0 as k → ∞.
Exercise 3.28. Give an example of a Cauchy sequence with {xn }n∈N ⊂ (0, 1], but
whose limit is not in (0, 1].
Exercise 3.29. Let {xn }n∈N be a sequence and 0 < r < 1. Suppose that
|xn+1 − xn | ≤ r|xn − xn−1 |
for n ≥ 2. Show that the sequence is Cauchy.
Exercise 3.30. Use Exercise 3.29 to show the sequence xn = 1/(2 + xn−1 ) for
n ≥ 2 and x1 > 0 is Cauchy. Find the limit.
Exercise 3.31. (a) Negate the definition of a Cauchy sequence.
(b) Show that the sequence
n
xn =
k=1
1
k
is not Cauchy.
(c) Show that |xn+1 − xn | → 0 as n → ∞ (compare this to the definition of a
Cauchy sequence).