30 CHAPTER 1. SOME PREREQUISITE TOPICS

Before discussing limsup and liminf, here is a very useful observation about doublesums.

Theorem 1.17.2 Let ai j ≥ 0. Then

∞

∑i=1

∞

∑j=1

ai j =∞

∑j=1

∞

∑i=1

ai j.

Proof: First note there is no trouble in defining these sums because the ai j are allnonnegative. If a sum diverges, it only diverges to ∞ and so ∞ is the value of the sum. Nextnote that

∞

∑j=r

∞

∑i=r

ai j ≥ supn

∞

∑j=r

n

∑i=r

ai j

because ∑∞j=r ∑

∞i=r ai j ≥ ∑

∞j=r ∑

ni=r ai j for each n. Therefore,

∞

∑j=r

∞

∑i=r

ai j ≥ supn

∞

∑j=r

n

∑i=r

ai j = supn

limm→∞

m

∑j=r

n

∑i=r

ai j

= supn

limm→∞

n

∑i=r

m

∑j=r

ai j = supn

n

∑i=r

limm→∞

m

∑j=r

ai j

= supn

n

∑i=r

∞

∑j=r

ai j = limn→∞

n

∑i=r

∞

∑j=r

ai j =∞

∑i=r

∞

∑j=r

ai j

Interchanging the i and j in the above argument proves the theorem. ■

Lemma 1.17.3 Let {an} be a sequence of real numbers and Un ≡ sup{ak : k ≥ n} . Then{Un} is a decreasing sequence. Also if Ln ≡ inf{ak : k ≥ n} , then {Ln} is an increasingsequence. Therefore, limn→∞ Ln and limn→∞ Un both exist.

Proof: Let Wn be an upper bound for {ak : k ≥ n} . Then since these sets are gettingsmaller, it follows that for m < n, Wm is an upper bound for {ak : k ≥ n} . In particular ifWm =Um, then Um is an upper bound for {ak : k ≥ n} and so Um is at least as large as Un,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 1.17.4 Let {an} be any sequence of points of [−∞,∞]

lim supn→∞

an ≡ limn→∞

sup{ak : k ≥ n}

lim infn→∞

an ≡ limn→∞

inf{ak : k ≥ n} .

Now the following shows the relation of liminf and limsup to the limit.

Theorem 1.17.5 Suppose {an} is a sequence of real numbers and that limsupn→∞ an andliminfn→∞ an are both real numbers. Then limn→∞ an exists if and only if liminfn→∞ an =limsupn→∞ an and in this case,

limn→∞

an = lim infn→∞

an = lim supn→∞

an.