Sometimes it is required to consider double series which are of the form
∞ ∞ ∞ ( ∞ )
∑ ∑ a ≡ ∑ ( ∑ a ) .
k=m j=m jk k=m j=m jk
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
∑∞ ∞∑ ∑∞ ∑∞
ajk = ajk.
k=mj=m j=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=m j=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 10.5.1Consider the following picturewhich depicts some of the orderedpairs
(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, anda_{mn} = −c for all
(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=1 n=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 10.5.2Let f
(a,b)
∈
[− ∞,∞ ]
for a ∈ A and b ∈ B whereA,B are 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 largerthan every 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)
where S_{b}≡
{f (a,b) : a ∈ A}
.
Unlike limits, you can take the sup in different orders.
Lemma 10.5.3Let f
(a,b)
∈
[− ∞, ∞ ]
for a ∈ A and b ∈ B where A,B are sets.Then
sup sup f (a,b) = supsup f (a,b).
a∈A b∈B b∈Ba∈A
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,
sup sup f (a,b) ≤ supsup f (a,b) .
a∈A b∈B b∈Ba∈A
Repeat the same argument interchanging a and b, to get the conclusion of the lemma.
■
Theorem 10.5.4Let 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
One of the most important applications of this theorem is to the problem of
multiplication of series.
Definition 10.5.6Let∑_{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.