2Convergence of sequences
IA Analysis I
2.2 Sums, products and quotients
Here we prove the things that we think are obviously true, e.g. sums and products
of convergent sequences are convergent.
Lemma (Sums of sequences). If a
n
→ a and b
n
→ b, then a
n
+ b
n
→ a + b.
Proof.
Let
ε >
0. We want to find a clever
N
such that for all
n ≥ N
,
|a
n
+
b
n
−
(
a
+
b
)
| < ε
. Intuitively, we know that
a
n
is very close to
a
and
b
n
is
very close to b. So their sum must be very close to a + b.
Formally, since
a
n
→ a
and
b
n
→ b
, we can find
N
1
, N
2
such that
∀n ≥ N
1
,
|a
n
− a| < ε/2 and ∀n ≥ N
2
, |b
n
− b| < ε/2.
Now let N = max{N
1
, N
2
}. Then by the triangle inequality, when n ≥ N ,
|(a
n
+ b
n
) − (a + b)| ≤ |a
n
− a| + |b
n
− b| < ε.
We want to prove that the product of convergent sequences is convergent.
However, we will not do it in one go. Instead, we separate it into many smaller
parts.
Lemma (Scalar multiplication of sequences). Let
a
n
→ a
and
λ ∈ R
. Then
λa
n
→ λa.
Proof. If λ = 0, then the result is trivial.
Otherwise, let
ε >
0. Then
∃N
such that
∀n ≥ N
,
|a
n
− a| < ε/|λ|
. So
|λa
n
− λa| < ε.
Lemma. Let (a
n
) be bounded and b
n
→ 0. Then a
n
b
n
→ 0.
Proof.
Let
C 6
= 0 be such that (
∀n
)
|a
n
| ≤ C
. Let
ε >
0. Then
∃N
such that
(∀n ≥ N) |b
n
| < ε/C. Then |a
n
b
n
| < ε.
Lemma. Every convergent sequence is bounded.
Proof.
Let
a
n
→ l
. Then there is an
N
such that
∀n ≥ N
,
|a
n
− l| ≤
1. So
|a
n
| ≤ |l| + 1. So a
n
is eventually bounded, and therefore bounded.
Lemma (Product of sequences). Let a
n
→ a and b
n
→ b. Then a
n
b
n
→ ab.
Proof. Let a
n
= a + ε
n
. Then a
n
b
n
= (a + ε
n
)b
n
= ab
n
+ ε
n
b
n
.
Since
b
n
→ b
,
ab
n
→ ab
. Since
ε
n
→
0 and
b
n
is bounded,
ε
n
b
n
→
0. So
a
n
b
n
→ ab.
Proof.
(alternative) Observe that
a
n
b
n
−ab
= (
a
n
−a
)
b
n
+(
b
n
−b
)
a
. We know that
a
n
−a →
0 and
b
n
−b →
0. Since (
b
n
) is bounded, so (
a
n
−a
)
b
n
+ (
b
n
−b
)
a →
0.
So a
n
b
n
→ ab.
Note that in this proof, we no longer write “Let
ε >
0”. In the beginning, we
have no lemmas proven. So we must prove everything from first principles and
use the definition. However, after we have proven the lemmas, we can simply
use them instead of using first principles. This is similar to in calculus, where
we use first principles to prove the product rule and chain rule, then no longer
use first principles afterwards.
Lemma (Quotient of sequences). Let (
a
n
) be a sequence such that (
∀n
)
a
n
6
= 0.
Suppose that a
n
→ a and a 6= 0. Then 1/a
n
→ 1/a.
Proof. We have
1
a
n
−
1
a
=
a − a
n
aa
n
.
We want to show that this
→
0. Since
a −a
n
→
0, we have to show that 1
/
(
aa
n
)
is bounded.
Since
a
n
→ a
,
∃N
such that
∀n ≥ N
,
|a
n
− a| ≤ a/
2. Then
∀n ≥ N
,
|a
n
| ≥
|a|/
2. Then
|
1
/
(
a
n
a
)
| ≤
2
/|a|
2
. So 1
/
(
a
n
a
) is bounded. So (
a − a
n
)
/
(
aa
n
)
→
0
and the result follows.
Corollary. If a
n
→ a, b
n
→ b, b
n
, b 6= 0, then a
n
/b
n
= a/b.
Proof.
We know that 1
/b
n
→
1
/b
. So the result follows by the product rule.
Lemma (Sandwich rule). Let (
a
n
) and (
b
n
) be sequences that both converge to
a limit x. Suppose that a
n
≤ c
n
≤ b
n
for every n. Then c
n
→ x.
Proof.
Let
ε >
0. We can find
N
such that
∀n ≥ N
,
|a
n
−x| < ε
and
|b
n
−x| < ε
.
Then ∀n ≥ N, we have x − ε < a
n
≤ c
n
≤ b
n
< x + ε. So |c
n
− x| < ε.
Example. 1
/
2
n
→
0. For every
n
,
n <
2
n
. So 0
<
1
/
2
n
<
1
/n
. The result
follows from the sandwich rule.
Example. We want to show that
n
2
+ 3
(n + 5)(2n − 1)
→
1
2
.
We can obtain this by
n
2
+ 3
(n + 5)(2n − 1)
=
1 + 3/n
2
(1 + 5/n)(2 − 1/n)
→
1
2
,
by sum rule, sandwich rule, Archimedean property, product rule and quotient
rule.
Example. Let k ∈ N and let δ > 0. Then
n
k
(1 + δ)
n
→ 0.
This can be summarized as “exponential growth beats polynomial growth even-
tually”.
By the binomial theorem,
(1 + δ)
n
≥
n
k + 1
δ
k+1
.
Also, for n ≥ 2k,
n
k + 1
=
n(n − 1) ···(n − k)
(k + 1)!
≥
(n/2)
k+1
(k + 1)!
.
So for sufficiently large n,
n
k
(1 + δ)
n
≤
n
k
2
k+1
(k + 1)!
n
k+1
δ
k+1
=
2
k+1
(k + 1)!
δ
k+1
·
1
n
→ 0.