Sometimes the limit of a sequence does not exist. For example, if a_{n} =
(− 1)
^{n}, then lim_{n→∞}a_{n} does not
exist. This is because the terms of the sequence are a distance of 1 apart. Therefore there can’t exist a
single number such that all the terms of the sequence are ultimately within 1∕4 of that number. The nice
thing about limsup and liminf is that they always exist. First here is a simple lemma and
definition.
Definition 1.14.4Denote by
[− ∞, ∞]
the real line along with symbols ∞ and −∞. It is understood that∞ is larger than every real number and −∞ is smaller than every real number. Then if
{A }
n
is anincreasing sequence of points of
[− ∞, ∞ ]
, lim_{n→∞}A_{n}equals ∞ if the only upper bound of the set
{A }
n
is∞. If
{A }
n
is bounded above by a real number, then lim_{n→∞}A_{n}is defined in the usual way and equals theleast upper bound of
{A }
n
. If
{A }
n
is a decreasing sequence of points of
[− ∞, ∞ ]
, lim_{n→∞}A_{n}equals −∞if the only lower bound of the sequence
{A }
n
is −∞. If
{A }
n
is bounded below by a real number, then
lim_{n→∞}A_{n}is defined in the usual way and equals the greatest lower bound of
{A }
n
. More simply, if
{A }
n
is increasing,
lim An ≡ sup{An}
n→ ∞
and if
{An }
is decreasing then
lim A ≡ inf{A }.
n→ ∞ n n
Lemma 1.14.5Let
{an}
be a sequence of real numbers and let U_{n}≡ sup
{ak : k ≥ n}
. Then
{Un}
is a decreasing sequence. Also if L_{n}≡ inf
{ak : k ≥ n }
, then
{Ln}
is an increasing sequence.Therefore, lim_{n→∞}L_{n}and lim_{n→∞}U_{n}both exist.
Proof: Let W_{n} be an upper bound for
{ak : k ≥ n}
. Then since these sets are getting
smaller, it follows that for m < n, W_{m} is an upper bound for
{ak : k ≥ n}
. In particular if
W_{m} = U_{m}, then U_{m} is an upper bound for
{ak : k ≥ n}
and so U_{m} is at least as large as
U_{n}, the least upper bound for
{ak : k ≥ n}
. The claim that
{Ln}
is decreasing is similar.
■
From the lemma, the following definition makes sense.
Definition 1.14.6Let
{an}
be any sequence of points of
[− ∞, ∞]
lim nsu→p∞an ≡ nli→m∞sup {ak : k ≥ n}
lim infan ≡ lim inf{ak : k ≥ n}.
n→ ∞ n→ ∞
Theorem 1.14.7Suppose
{an}
is a sequenceof real numbers and that
limns→up∞ an
and
lim inf an
n→ ∞
are both real numbers. Then lim_{n→∞}a_{n}exists if and only if
lim inf a = lim sup a
n→ ∞ n n→ ∞ n
and in this case,
nli→m∞ an = lim nin→f∞ an = lim sn→up∞ an.
Proof: First note that
sup {ak : k ≥ n} ≥ inf{ak : k ≥ n}
and so,
lim snu→p∞ an ≡ nl→im∞ sup {ak : k ≥ n}
≥ lim inf{a : k ≥ n }
n→ ∞ k
≡ lim ni→nf∞ an.
Suppose first that lim_{n→∞}a_{n} exists and is a real number a. Then from the definition of a limit, there exists
N corresponding to ε∕6 in the definition. Hence, if m,n ≥ N, then
ε ε ε
|an − am | ≤ |an − a|+ |a− an| < 6 + 6 = 3.
From the definition of sup
{ak : k ≥ N}
, there exists n_{1}≥ N such that
sup {ak : k ≥ N } ≤ an1 + ε∕3.
Similarly, there exists n_{2}≥ N such that
inf {ak : k ≥ N} ≥ an2 − ε∕3.
It follows that
2ε
sup{ak : k ≥ N } − inf{ak : k ≥ N } ≤ |an1 − an2|+ 3 < ε.
Since the sequence,
{sup{ak : k ≥ N }}
_{N=1}^{∞} is decreasing and
{inf{ak : k ≥ N}}
_{N=1}^{∞} is increasing, it
follows that
0 ≤ lim sup{ak : k ≥ N} − lim inf{ak : k ≥ N } ≤ ε
N→∞ N→∞
Since ε is arbitrary, this shows
Nlim→∞ sup{ak : k ≥ N } = Nlim→∞ inf{ak : k ≥ N } (1.1)
it follows that for every ε > 0, there exists N such
that
sup{ak : k ≥ N }− inf{ak : k ≥ N } < ε,
and for every N,
inf{ak : k ≥ N } ≤ a ≤ sup {ak : k ≥ N }
Thus if n ≥ N,
|a− an| < ε
which implies that lim_{n→∞}a_{n} = a. In case
a = ∞ = Nlim→∞ sup{ak : k ≥ N } = Nlim→∞ inf{ak : k ≥ N }
then if r ∈ ℝ is given, there exists N such that inf
{ak : k ≥ N}
> r which is to say that lim_{n→∞}a_{n} = ∞.
The case where a = −∞ is similar except you use sup
{ak : k ≥ N }
. ■
The significance of limsup and liminf, in addition to what was just discussed, is contained in the
following theorem which follows quickly from the definition.
Theorem 1.14.8Suppose
{an}
is a sequence of points of
[− ∞,∞ ]
. Let
λ = lim nsu→p∞ an.
Then if b > λ, it follows there exists N such that whenever n ≥ N,
an ≤ b.
If c < λ, then a_{n}> c for infinitely many values of n. Let
γ = lim inf an.
n→∞
Then if d < γ, it follows there exists N such that whenever n ≥ N,
an ≥ d.
If e > γ, it follows a_{n}< e for infinitely many values of n.
The proof of this theorem is left as an exercise for you. It follows directly from the definition and it is
the sort of thing you must do yourself. Here is one other simple proposition.
Proposition 1.14.9Let lim_{n→∞}a_{n} = a > 0. Then
lim sup anbn = a lim sup bn.
n→∞ n→∞
Proof: This follows from the definition. Let λ_{n} = sup
{akbk : k ≥ n}
. For all n large enough, a_{n}> a−ε
where ε is small enough that a − ε > 0. Therefore,
λn ≥ sup {bk : k ≥ n}(a− ε)
for all n large enough. Then
lim sup anbn = lim λn ≡ lim sup anbn
n→ ∞ n→∞ n→∞
≥ lnim→∞ (sup {bk : k ≥ n}(a− ε))
= (a − ε)lim sup b
n→∞ n
Similar reasoning shows
lim sup anbn ≤ (a +ε)lim sup bn
n→∞ n→ ∞
Now since ε > 0 is arbitrary, the conclusion follows. ■