64 CHAPTER 2. SOME BASIC TOPICS

Proof: Let sup({An : n ∈ N}) = r. In the first case, suppose r < ∞. Then letting ε > 0be given, there exists n such that An ∈ (r− ε,r]. Since {An} is increasing, it follows ifm > n, then r− ε < An ≤ Am ≤ r and so limn→∞ An = r as claimed. In the case wherer = ∞, then if a is a real number, there exists n such that An > a. Since {Ak} is increasing,it follows that if m > n, Am > a. But this is what is meant by limn→∞ An = ∞. The othercase is that r =−∞. But in this case, An =−∞ for all n and so limn→∞ An =−∞. The casewhere An is decreasing is entirely similar. ■

2.5 Double SeriesDouble series are of the form ∑

∞k=m ∑

∞j=m a jk ≡∑

∞k=m

(∑

∞j=m a jk

). In other words, first sum

on j yielding something which depends on k and then sum these. The major considerationfor these double series is the question of when ∑

∞k=m ∑

∞j=m a jk = ∑

∞j=m ∑

∞k=m a jk In other

words, when does it make no difference which subscript is summed over first? In the caseof finite sums there is no issue here. You can always write ∑

Mk=m ∑

Nj=m a jk = ∑

Nj=m ∑

Mk=m a jk

because addition is commutative. However, there are limits involved with infinite sums andthe interchange in order of summation involves taking limits in a different order. Therefore,it is not always true that it is permissible to interchange the two sums. A general rule ofthumb is this: If something involves changing the order in which two limits are taken, youmay not do it without agonizing over the question. In general, limits foul up algebra andalso introduce things which are counter intuitive. Here is an example. This example is alittle technical. It is placed here just to prove conclusively there is a question which needsto be considered.

Example 2.5.1 Consider the following picture which depicts some of the ordered pairs(m,n) where m,n are positive integers.

...0 0 c 0 −c0 c 0 −c 0b 0 −c 0 00 a 0 0 0

· · ·

The a,b,c are the values of amn. Thus ann = 0 for all n≥ 1, a21 = a,a12 = b,am(m+1) =−cwhenever m > 1, and am(m−1) = c whenever m > 2. The numbers next to the point are thevalues of amn. You see ann = 0 for all n, a21 = a,a12 = b,amn = c for (m,n) on the liney = 1+ x whenever m > 1, and amn = −c for all (m,n) on the line y = x− 1 wheneverm > 2.

Then ∑∞m=1 amn = a if n = 1, ∑

∞m=1 amn = b− c if n = 2 and if n > 2,∑∞

m=1 amn = 0.Therefore,

∞

∑n=1

∞

∑m=1

amn = a+b− c.

Next observe that ∑∞n=1 amn = b if m = 1,∑∞

n=1 amn = a+ c if m = 2, and ∑∞n=1 amn = 0 if

m > 2. Therefore,∞

∑m=1

∞

∑n=1

amn = b+a+ c