1.17. LIM SUP AND LIM INF 31

Proof: First note that

sup{ak : k ≥ n} ≥ inf{ak : k ≥ n}

and so,

lim supn→∞

an ≡ limn→∞

sup{ak : k ≥ n} ≥ limn→∞

inf{ak : k ≥ n} ≡ lim infn→∞

an.

Suppose first that limn→∞ an exists and is a real number a. Then from the definition of alimit, there exists N corresponding to ε/6 in the definition. Hence, if m,n≥ N, then

|an−am| ≤ |an−a|+ |a−an|<ε

6+

ε

6=

ε

3.

From the definition of sup{ak : k ≥ N} , there exists n1 ≥ N such that

sup{ak : k ≥ N} ≤ an1 + ε/3.

Similarly, there exists n2 ≥ N such that

inf{ak : k ≥ N} ≥ an2 − ε/3.

It follows that

sup{ak : k ≥ N}− inf{ak : k ≥ N} ≤ |an1 −an2 |+2ε

3< ε.

Since the sequence, {sup{ak : k ≥ N}}∞

N=1 is decreasing and {inf{ak : k ≥ N}}∞

N=1 is in-creasing, it follows that

0≤ limN→∞

sup{ak : k ≥ N}− limN→∞

inf{ak : k ≥ N} ≤ ε

Since ε is arbitrary, this shows

limN→∞

sup{ak : k ≥ N}= limN→∞

inf{ak : k ≥ N} (1.3)

Next suppose 1.3 and both equal a ∈ R. Then

limN→∞

(sup{ak : k ≥ N}− inf{ak : k ≥ N}) = 0

Since sup{ak : k ≥ N}≥ inf{ak : k ≥ N} it follows that for every ε > 0, there exists N suchthat

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 limn→∞ an = a. In case

a = ∞ = limN→∞

sup{ak : k ≥ N}= limN→∞

inf{ak : k ≥ N}

then if r ∈ R is given, there exists N such that inf{ak : k ≥ N} > r which is to say thatlimn→∞ an = ∞. 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 con-tained in the following theorem which follows quickly from the definition.