Sometimes it is required to consider double series which are of the form
( )
∑∞ ∞∑ ∑∞ ∞∑
ajk ≡ ( ajk) .
k=mj=m k=m j=m
In other words, first sum on j yielding something which depends on k and then sum these.
The major consideration for these double series is the question of when
∞ ∞ ∞ ∞
∑ ∑ ∑ ∑
j=m ajk = j=m ajk.
k=m k=m
In other words, when does it make no difference which subscript is summed over first? In the
case of finite sums there is no issue here. You can always write
∑M N∑ ∑N ∑M
ajk = ajk
k=mj=m j=m k=m
because addition is commutative. However, there are limits involved with infinite sums and
the 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 of thumb
is this: If something involves changing the order in which two limits are taken, you may not
do it without agonizing over the question. In general, limits foul up algebra and also introduce
things which are counter intuitive. Here is an example. This example is a little technical.
It is placed here just to prove conclusively there is a question which needs to be
considered.
Example 5.4.1Consider the following picturewhich depicts some of the ordered pairs
(m, n)
where m,n are positive integers.
PICT
The numbers next to the point are the values of a_{mn}. You see a_{nn} = 0 for all n,a_{21} = a,a_{12} = b,a_{mn} = c for
(m,n)
on the line y = 1 + x whenever m > 1, and a_{mn} = −c forall
(m, n)
on the line y = x − 1 whenever m > 2.
Then ∑_{m=1}^{∞}a_{mn} = a if n = 1, ∑_{m=1}^{∞}a_{mn} = b−c if n = 2 and if n > 2,∑_{m=1}^{∞}a_{mn} = 0.
Therefore,
∞∑ ∑∞
amn = a + b− c.
n=1m=1
Next observe that ∑_{n=1}^{∞}a_{mn} = b if m = 1,∑_{n=1}^{∞}a_{mn} = a + c if m = 2, and
∑_{n=1}^{∞}a_{mn} = 0 if m > 2. Therefore,
∑∞ ∞∑
amn = b+ a+ c
m=1n=1
and so the two sums are different. Moreover, you can see that by assigning different values of
a,b, and c, you can get an example for any two different numbers desired.
Don’t become upset by this. It happens because, as indicated above, limits are taken in
two different orders. An infinite sum always involves a limit and this illustrates why
you must always remember this. This example in no way violates the commutative
law of addition which has nothing to do with limits. However, it turns out that if
a_{ij}≥ 0 for all i,j, then you can always interchange the order of summation. This is
shown next and is based on the following lemma. First, some notation should be
discussed.
Definition 5.4.2Let f
(a,b)
∈
[− ∞, ∞ ]
for a ∈ A and b ∈ B where A,Bare setswhich means that f
(a,b)
is either a number, ∞, or −∞. The symbol, +∞is interpreted as a point out at the end of the number line which is larger thanevery real number. Of course there is no such number. That is why it is called ∞.The symbol, −∞ is interpreted similarly. Then sup_{a∈A}f
(a,b)
means sup
(Sb)
whereS_{b}≡
{f (a,b) : a ∈ A}
.
Unlike limits, you can take the sup in different orders.
Lemma 5.4.3Let f
(a,b)
∈
[− ∞,∞ ]
for a ∈ A and b ∈ B where A,B are sets.Then
sau∈pAsbu∈pB f (a,b) = sb∈uBpsau∈pAf (a,b).
Proof: Note that for all a,b, f
(a,b)
≤ sup_{b∈B} sup_{a∈A}f
(a,b)
and therefore, for all a,
sup_{b∈B}f
(a,b)
≤ sup_{b∈B} sup_{a∈A}f
(a,b)
. Therefore,
sau∈pAsbu∈pB f (a,b) ≤ sb∈uBpsau∈pAf (a,b).
Repeat the same argument interchanging a and b, to get the conclusion of the lemma.
■
Lemma 5.4.4If
{An }
is an increasingsequence in
[− ∞,∞ ]
, then
sup{An } = nl→im∞ An.
Proof: Let sup
({An : n ∈ ℕ})
= r. In the first case, suppose r < ∞. Then letting ε > 0
be given, there exists n such that A_{n}∈ (r −ε,r]. Since
{An}
is increasing, it follows if m > n,
then r −ε < A_{n}≤ A_{m}≤ r and so lim_{n→∞}A_{n} = r as claimed. In the case where r = ∞, then
if a is a real number, there exists n such that A_{n}> a. Since
{Ak}
is increasing, it follows that
if m > n, A_{m}> a. But this is what is meant by lim_{n→∞}A_{n} = ∞. The other case
is that r = −∞. But in this case, A_{n} = −∞ for all n and so lim_{n→∞}A_{n} = −∞.■
Theorem 5.4.5Let a_{ij}≥ 0. Then
∑∞ ∑∞ ∑∞ ∑∞
aij = aij.
i=1j=1 j=1i=1
Proof: First note there is no trouble in defining these sums because the a_{ij} are all
nonnegative. If a sum diverges, it only diverges to ∞ and so ∞ is the value of the sum. Next
note that
both exist. It only remains to verify they are equal. By similar reasoning you can replace a_{ij}
with Rea_{ij} or with Ima_{ij} in the above and the two sums will exist.
The real part of a finite sum of complex numbers equals the sum of the real parts. Then
passing to a limit, it follows
∑∞ ∑∞ ∑∞ ∑∞
Re aij = Re aij
j=ri=r j=ri=r
and similarly,
∑∞ ∑∞ ∑∞ ∑∞
Im aij = Im aij
i=rj=r i=rj=r
Note 0 ≤
(|aij|+ Re aij)
≤ 2
|aij|
. Therefore, by Theorem 5.4.5 and Theorem 5.1.5 on Page
197
∑∞ ∑∞ ∑∞ ∑∞ ∞∑ ∞∑
|aij|+ Re aij = (|aij|+ Re aij)
j=ri=r j=ri=r j=∞r i=∞r
∑ ∑
= (|aij|+ Re aij)
i=∞r j=∞r ∞ ∞
= ∑ ∑ |a |+∑ ∑ Rea
i=r j=r ij i=r j=r ij
∞ ∞ ∞ ∞
= ∑ ∑ |aij|+∑ ∑ Reaij
j=r i=r i=r j=r
and so
∑∞ ∞∑ ∑∞ ∞∑ ∑∞ ∑∞ ∞∑ ∞∑
Re aij = Reaij = Re aij = Re aij
j=r i=r j=r i=r i=rj=r i=r j=r
Similar reasoning applies to the imaginary parts. Since the real and imaginary parts of the
two series are equal, it follows the two series are equal. ■
One of the most important applications of this theorem is to the problem of multiplication
of series.
Definition 5.4.7Let∑_{i=r}^{∞}a_{i}and∑_{i=r}^{∞}b_{i}be two series. For n ≥ r,define
∑n
cn ≡ akbn−k+r.
k=r
The series∑_{n=r}^{∞}c_{n}is called theCauchy product of the two series.
It isn’t hard to see where this comes from. Formally write the following in the case
r = 0:
(a0 + a1 + a2 + a3⋅⋅⋅)(b0 + b1 + b2 + b3⋅⋅⋅)
and start multiplying in the usual way. This yields
a0b0 +(a0b1 + b0a1)+ (a0b2 + a1b1 + a2b0)+ ⋅⋅⋅
and you see the expressions in parentheses above are just the c_{n} for n = 0,1,2,
⋅⋅⋅
. Therefore,
it is reasonable to conjecture that
∑∞ ∑∞ ∞∑
ai bj = cn
i=r j=r n=r
and of course there would be no problem with this in the case of finite sums but in the case of
infinite sums, it is necessary to prove a theorem. The following is a special case of Merten’s
theorem.