6Real numbers
IA Numbers and Sets
6.2 Sequences
Here we will look at sequences, with series in the next chapter. Only a brief
introduction to these topics will be provided here, as these topics will be studied
later in Analysis I.
Definition
(Sequence)
.
A sequence is a function
N → R
. If
a
is a sequence,
instead of
a
(1)
, a
(2)
, ···
, we usually write
a
1
, a
2
, ···
. To emphasize it is a
sequence, we write the sequence as (a
n
).
We want to capture the notion of a sequence tending to a limit. For example,
we want to say that 1
,
1
2
,
1
3
, ···
tends to 0, while 1
,
2
,
3
, ···
does not converge to
a limit.
The idea is that if
a
n
→ l
, then we can get as close to
l
as we like, as long
as we are sufficiently far down the sequence. More precisely, given any “error
threshold”
ε
, we can find a (possibly large) number
N
such that whenever
n ≥ N
,
we have a
n
− l < ε.
Definition
(Limit of sequence)
.
The sequence (
a
n
) tends to
l ∈ R
as
n
tends
to infinity if and only if
(∀ε > 0)(∃N ∈ N)(∀n ≥ N) a
n
− l < ε.
If
a
n
tends to
l
as
n
tends to infinity, we write
a
n
→ l
as
n → ∞
;
lim
n→∞
a
n
=
l
;
or a
n
converges to l.
Intuitively, if
a
n
→ l
, we mean given any
ε
, for sufficiently large
n
,
a
n
is
always within l ± ε.
The definition a
n
6→ l is the negation of the above statement:
(∃ε > 0)(∀N ∈ N)(∃n ≥ N) a
n
− l ≥ ε.
Definition
(Convergence of sequence)
.
The sequence (
a
n
) converges if there
exists an l such that a
n
→ l. The sequence diverges if it doesn’t converge.
Every proof of
a
n
→ l
looks like: Given
ε >
0, (argument to show
N
exists,
maybe depending on ε), such that ∀n ≥ N, a
n
− l < ε.
Example. Show that a
n
= 1 −
1
n
→ 1.
Given
ε >
0, choose
N >
1
ε
, which exists by the Axiom of Archimedes. If
n ≥ N , then a
n
− 1 =
1
n
≤ ε. So a
n
→ 1.
Example. Let
a
n
=
(
1
n
n is prime
1
2n
n is not prime
.
We will show that
a
n
→
0. Given
ε >
0. Choose
N >
1
ε
. Then
∀n ≥ N
,
a
n
− 0 ≤
1
n
< ε.
Example. Prove that
a
n
=
(
1 n is prime
0 n is not prime
diverges.
Let
ε
=
1
3
. Suppose
l ∈ R
. If
l <
1
2
, then
a
n
− l > ε
when
n
is prime. If
l ≥
1
2
, then
a
n
− l > ε
when
n
is not prime. Since the primes and nonprimes
are unbounded, (∀N )∃n > N such that a
n
− l > ε. So a
n
diverges.
An important property of R is the following:
Theorem. Every bounded monotonic sequence converges.
In case of confusion, the terms are defined as follows: (
a
n
) is increasing if
m ≤ n
implies
a
m
≤ a
n
. Decreasing is defined similarly. Then it is monotonic if
it is increasing or decreasing. (
a
n
) is bounded if there is some
B ∈ R
such that
a
n
 ≤ B for all n.
Proof.
wlog assume (
a
n
) is increasing. The set
{a
n
:
n ≥
1
}
is bounded and
nonempty. So it has a supremum
l
(least upper bound axiom). Show that
l
is
the limit:
Given any
ε >
0,
l − ε
is not an upper bound of
a
n
. So
∃N
such that
a
N
≥ l − ε
. Since
a
n
is increasing, we know that
l ≥ a
m
≥ a
N
> l − ε
for all
m ≥ N . So ∃N such that ∀n ≥ N , a
n
− l < ε. So a
n
→ l.
We can show that this theorem is equivalent to the least upper bound axiom.
Definition
(Subsequence)
.
A subsequence of (
a
n
) is (
a
g(n)
) where
g
:
N → N
is
strictly increasing. e.g. a
2
, a
3
, a
5
, a
7
··· is a subsequence of (a
n
).
Theorem. Every sequence has a monotonic subsequence.
Proof.
Call a point
a
k
a “peak” if (
∀m ≥ k
)
a
m
≤ a
k
. If there are infinitely
many peaks, then they form a decreasing subsequence. If there are only finitely
many peaks,
∃N
such that no
a
n
with
n > N
is a peak. Pick
a
N
1
with
N
1
> N
.
Then pick
a
N
2
with
N
2
> N
1
and
a
N
2
> a
N
1
. This is possible because
a
N
1
is
not a peak. The pick
a
N
3
with
N
3
> N
2
and
a
N
3
> a
N
2
, ad infinitum. Then we
have a monotonic subsequence.
We will now prove the following basic properties of convergence:
Theorem.
(i) If a
n
→ a and a
n
→ b, then a = b (i.e. limits are unique)
(ii) If a
n
→ a and b
n
= a
n
for all but finitely many n, then b
n
→ a.
(iii) If a
n
= a for all n, then a
n
→ a.
(iv) If a
n
→ a and b
n
→ b, then a
n
+ b
n
→ a + b
(v) If a
n
→ a and b
n
→ b, then a
n
b
n
→ ab
(vi) If a
n
→ a 6= 0, and ∀n(a
n
6= 0). Then 1/a
n
→ 1/a.
(vii)
If
a
n
→ a
and
b
n
→ a
, and
∀n
(
a
n
≤ c
n
≤ b
n
), then
c
n
→ a
. (Sandwich
theorem)
Many students are confused as to why we should prove these “obvious”
properties of convergence. It seems “obvious” that if
a
n
converges to
a
and
b
n
converges to
b
, then the sum converges to
a
+
b
. However, it is not obvious (at least
for the firsttime learners) that if (
∀ε >
0)(
∃N
)(
∀n ≥ N
)
a
n
−a < ε
and (
∀ε >
0)(
∃N
)(
∀n ≥ N
)
b
n
−b < ε
, then (
∀ε >
0)(
∃N
)(
∀n ≥ N
)

(
a
n
+
b
n
)
−
(
a
+
b
)
 < ε
.
In some sense, what we are trying to prove is that our attempt at defining
convergence actually satisfies the “obvious” properties we think convergence
should satisfy.
In proving this, we will make frequent use of the triangle inequality:
x
+
y ≤
x + y.
Proof.
(i)
Suppose instead
a < b
. Then choose
ε
=
b−a
2
. By the definition of the
limit,
∃N
1
such that
∀n ≥ N
1
,
a
n
− a < ε
. There also
∃N
2
st.
∀n ≥ N
2
,
a
n
− b < ε.
Let
N
=
max{N
1
, N
2
}
. If
n ≥ max{N
1
, N
2
}
, then
a − b ≤ a − a
n

+
a
n
− b < 2ε = b − a. Contradiction. So a = b.
(ii)
Given
ε >
0, there
∃N
1
st.
∀n ≥ N
1
, we have
a
n
− a < ε
. Since
b
n
=
a
n
for all but finitely many n, there exists N
2
such that ∀n ≥ N
2
, a
n
= b
n
.
Let
N
=
max{N
1
, N
2
}
. Then
∀n ≥ N
, we have
b
n
−a
=
a
n
−a < ε
. So
b
n
→ a.
(iii) ∀ε, take N = 1. Then a
n
− a = 0 < ε for all n ≥ 1.
(iv)
Given
ε >
0,
∃N
1
such that
∀n ≥ N
1
, we have
a
n
− a < ε/
2. Similarly,
∃N
2
such that ∀n ≥ N
2
, we have b
n
− b < ε/2.
Let
N
=
max{N
1
, N
2
}
. Then
∀n ≥ N
,

(
a
n
+
b
n
)
−
(
a
+
b
)
 ≤ a
n
− a
+
b
n
− b < ε.
(v) Given ε > 0, Then there exists N
1
, N
2
, N
3
such that
∀n ≥ N
1
: a
n
− a <
ε
2(b + 1)
∀n ≥ N
2
: b
n
− b <
ε
2a
∀n ≥ N
3
: b
n
− b < 1 ⇒ b
n
 < b + 1
Then let N = max{N
1
, N
2
, N
3
}. Then ∀n ≥ N,
a
n
b
n
− ab = b
n
(a
n
− a) + a(b
n
− b)
≤ b
n
a
n
− a + ab
n
− b
< (b+ 1)a
n
− a + ab
n
− b
<
ε
2
+
ε
2
= ε
(vi) Given ε > 0, then ∃N
1
, N
2
such that a
n
− a <
a
2
2
ε and a
n
− a <
a
2
.
Let N = max{N
1
, N
2
}. The ∀n ≥ N,
1
a
n
−
1
a
=
a
n
− a
a
n
a
<
2
a
2
a
n
− a
< ε
(vii)
By (iii) to (v), we know that
b
n
− a
n
→
0. Let
ε >
0. Then
∃N
such that
∀n ≥ N
, we have
b
n
− a
n
 < ε
. So
c
n
− a
n
 < ε
. So
c
n
− a
n
→
0. So
c
n
= (c
n
− a
n
) + a
n
→ a.
Example. Let x
n
=
n
2
(n+1)(2n+1)
n
4
+1
. Then we have
x
n
=
(1 + 1/n)(2 + 1/n)
1 + 1/n
4
→
1 · 2
1
= 2
by the theorem (many times).
Example.
Let
y
n
=
100
n
n!
. Since
y
n+1
y
n
=
100
n+1
<
1
2
for large
n >
200, we know
that 0 ≤ y
n
< y
200
·
2
200
2
n
. Since y
200
·
2
200
2
n
→ 0, we know that y
n
→ 0 as well.