2Convergence of sequences
IA Analysis I
2.5 Limit supremum and infimum
Here we will define the limit supremum and infimum. While these are technically
not part of the course, eventually some lecturers will magically assume students
know this definition. So we might as well learn it here.
Definition (Limit supremum/infimum). Let (
a
n
) be a bounded sequence. We
define the limit supremum as
lim sup
n→∞
a
n
= lim
n→∞
sup
m≥n
a
m
.
To see that this exists, set
b
n
=
sup
m≥n
a
m
. Then (
b
n
) is decreasing since we
are taking the supremum of fewer and fewer things, and is bounded below by
any lower bound for (a
n
) since b
n
≥ a
n
. So it converges.
Similarly, we define the limit infimum as
lim inf
n→∞
a
n
= lim
n→∞
inf
m≥n
a
m
.
Example. Take the sequence
2, −1,
3
2
, −
1
2
,
4
3
, −
1
3
, ···
Then the limit supremum is 1 and the limit infimum is 0.
Lemma. Let (
a
n
) be a sequence. The following two statements are equivalent:
– a
n
→ a
– lim sup a
n
= lim inf a
n
= a.
Proof. If a
n
→ a, then let ε > 0. Then we can find an n such that
a −ε ≤ a
m
≤ a + ε for all m ≥ n
It follows that
a − ε ≤ inf
m≥n
a
m
≤ sup
m≥n
a
m
≤ a + ε.
Since ε was arbitrary, it follows that
lim inf a
n
= lim sup a
n
= a.
Conversely, if
lim inf a
n
=
lim sup a
n
=
a
, then let
ε >
0. Then we can find
n
such that
inf
m≥n
a
m
> a − ε and sup
m≥n
a
m
< a + ε.
It follows that ∀m ≥ n, we have |a
m
− a| < ε.