The Islamic University of Gaza
Deanery of Higher studies
Faculty of Science
Department of Mathematics
Doubly Indexed Infinite Series
Presented By
Ahed Khaleel Abu ALees
Supervisor
Professor Eissa D. Habil
SUBMITTED IN PARTIAL FULFILMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
MASTER OF MATHEMATICS
2008
Dedication
To My Parents,
To My Brothers,
and
To My Sisters.
Contents
Acknowledgements iv
Abstract v
Introduction 1
1 Double Sequences 3
1.1 Double Sequences and Their Limits . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Monotone Double Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.3 Cauchy Double Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Double Series 16
2.1 Nonnegative Double Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2 Summing by Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 Cauchy’s Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3 Divergent Double Series 32
3.1 Transformation of Double Sequences . . . . . . . . . . . . . . . . . . . . . . 32
3.2 Regularity of Double Linear Transformation . . . . . . . . . . . . . . . . . . 35
3.3 Double Absolute Summability Factor Theorem . . . . . . . . . . . . . . . . 52
References 68
ii
Acknowledgements
My thanks go (after God) to my advisor Prof. Dr. Eissa D. Habil for suggesting the topic of
the thesis, for his encouragement, invaluable discussions and kind help during the preparation
of the thesis and offering me advice and assistance at every stage of my thesis. Special thanks
go to my family members for their encouragement.
iii
Abstract
As we know the theory of double series has been widely used in mathematics and entered
in many applications. So we discuss important cases in the theory of double series. The
first case is that when can we interchange the order of summations in a double sum without
altering the sum? We Present several results. We also present the concept of Cauchy’s
product and we give the related theorems.
We also discuss the case of divergent double series and produce several results on the
method of summability of double series, such as transformation of double sequences and series
and the concept of double absolute summability factor theorem which is a generalization of
absolute summability factor for single series.
iv
IntroductionWe have important cases which are important in the theory of double series and have
several applications.
The first case is when the following are equal:
(i)∑∞
n=1
∑∞m=1 xm,n,
(ii)∑∞
m=1
∑∞n=1 xm,n,
(iii)∑
m,n xm,n.
That is when can we interchange the order of summations? And when the iterated sums
in (i), (ii) are equal and equal the double sum in (iii). In [4], this question has been carefully
discussed by Habil in the complex case; i.e., when the xm,n’s are complex numbers.
Also we discuss some results about the method of summability of double series such as
transformation of double sequences and series and the concept of double absolute summabil-
ity factor theorem which is a generalization of absolute summability factor for single series
which have been discussed in [7].
Our thesis comes into 3 chapters. In the first chapter, we give the definitions of double
sequences of complex numbers. We define the convergence and divergence of the double
sequences. We have proved the uniqueness of the limit of the double sequence. We have
proved that a convergent double sequence of complex numbers is bounded. We have defined
the iterated limits of double sequences and we presented conditions on when they are equal.
We have defined monotone double sequences, and we have proved a monotone convergence
theorem for double sequences (see[4]).
In the second chapter, we have defined the iterated sums, double sums, and we have
proved the double-series theorem for double series with non-negative terms which says that we
can interchange the order of summations in the nonnegative case without any conditions(see
[5]). And we have discussed Alternating Nonnegative double series theorem (see [6]).
Also in this chapter we discuss the concept of summing by curves and we have given sev-
eral examples and we have proved the Cauchy theorem which says that you can interchange
1
the order of summation for the arbitrary case but under certain conditions (see [6]).
In the third chapter, we introduce the concept of linear transformation of a double series
and give the definition of the regular linear transformation and we present some theorems
about the subject (see[1],[7]). We also present the concept of double absolute summability
factor theorem and we give the conditions which are important to the series to be summable
|A|k for some triangular transformation A (see [8]).
Throughout this thesis, the symbols R, C, Z and N denote, respectively, the set of all
real numbers, all complex numbers, all integers, and all natural numbers. The notation :=
means “equals by definition”.
2
Chapter 1
Double Sequences
The theory of double sequences and double series is an extension of the theory of single or
ordinary sequences and series. To each double sequence x : N×N −→ C, there corresponds
three important limits; namely:
1. limn,m→∞ xn,m,
2. limn→∞(limm→∞ xn,m),
3. limm→∞(limn→∞ xn,m).
The important question that is usually considered in this regard is the question of when
can we interchange the order of the limit for a double sequence (xn,m); that is, when the
limit (2) above equals the limit (3) above. These equations were answered in [4] as we will
see in this chapter.
1.1 Double Sequences and Their Limits
In this section, we introduce double sequences of complex numbers and we shall give the
definition of their convergence, divergence and oscillation. Then we study the relationship
between double and iterated limits of double sequences.
3
Definition 1.1.1. [4] A double sequence of complex numbers is a function x : N×N −→ C.
We shall use the notation (x(n, m)) or simply (xn,m). We say that a double sequence (xn,m)
converges to a ∈ C and we write limn,m→∞ xn,m = a, if the following condition is satisfied:
For every � > 0, there exists N = N(�) ∈ N such that |xn,m − a| < � ∀n, m ≥ N. The
number a is called the double limit of the double sequence (xn,m). If no such a exists, we
say that the sequence (xn,m) diverges.
Definition 1.1.2. [4] Let (xn,m) be a double sequence of real numbers.
(i) We say that (xn,m) tends to ∞, and we write limn,m→∞ xn,m = ∞, if for every α ∈ R,
there exists K = K(α) ∈ N such that if n, m ≥ K, then xn,m > α.
(ii) We say that (xn,m) tends to −∞, and we write limn,m→∞ xn,m = −∞, if for every
β ∈ R, there exists K = K(β) ∈ N such that if n,m ≥ K, then xn,m < β.
We say that (xn,m) is properly divergent in case we have
limn,m→∞ xn,m = ∞ or limn,m→∞ xn,m = −∞. In case (xn,m) does not converge
to a ∈ R and also it does not diverge properly, then we say that (xn,m) oscillates finitely or
infinitely according as (xn,m) is also bounded or not. For example, the sequence ((−1)n+m)
oscillates finitely, while the sequence ((−1)n+m(n + m)) oscillates infinitely.
Example 1.1.1.
(a) For the double sequence xn,m =1
n+m, we have
limn,m→∞
xn,m = 0.
To see this, given � > 0, choose N ∈ N such that N > 2�. Then ∀n,m ≥ N , we have
1n, 1
m≤ 1
N, which implies that
| xn,m − 0 |=|1
n + m|< 1
n+
1
m<
1
N+
1
N=
2
N< �.
(b) The double sequence xn,m =n
n+mis divergent. Indeed, for all sufficiently large n, m ∈ N
with n = m, we have xn,m =12, whereas for all sufficiently large n, m ∈ N with n = 2m,
4
we have xn,m =23. It follows that xn,m does not converge to a for any a ∈ R as
n, m →∞.
(c) The double sequence xn,m = n + m is properly divergent to ∞. Indeed, given α ∈ R,
there exists K ∈ N such that K > α. Then n, m ≥ K ⇒ n + m > α.
(d) The double sequence xn,m = 1 − n − m is properly divergent to −∞. Indeed, given
β ∈ R, there exists K ∈ N such that K > −β2
+ 12. Then n, m ≥ K ⇒ −n,−m <
β2− 1
2⇒ 1− n−m < β.
Theorem 1.1.1. [4] (Uniqueness of Double Limits). A double sequence of complex
numbers can have at most one limit.
Proof. Suppose that a, a′ are both limits of (xn,m). Then given � > 0, there exist natural
numbers N1, N2 such that
n, m ≥ N1 ⇒ | xn,m − a |<�
2(1.1)
and such that
n,m ≥ N2 ⇒ | xn,m − a′ |<�
2. (1.2)
Let N := max{N1, N2}. Then for all n, m ≥ N , implications (1.1) and (1.2) yield
0 ≤ |a− a′| = |a− xn,m + xn,m − a′|
≤ |xn,m − a|+ |xn,m − a′|
<�
2+
�
2= �.
It follows that a− a′ = 0, and hence the limit is unique whenever it exists. �
Definition 1.1.3. [4] A double sequence (xn,m) is called bounded if there exists a real number
M > 0 such that |xn,m| ≤ M ∀n, m ∈ N.
Theorem 1.1.2. [4] A convergent double sequence of complex numbers is bounded.
Proof. Suppose xn,m → a and let � = 1. Then there exists N ∈ N such that
n,m ≥ N ⇒ |xn,m − a| < 1.
5
This and the triangle inequality yield that |xn,m| < 1 + |a| ∀n, m ≥ N . Let
M := max{|x1,1|, |x1,2|, |x2,1|, . . . , |xN−1,N−1|, |a|+ 1}.
Clearly, |xn,m| ≤ M ∀n, m ∈ N. �
Definition 1.1.4. For a double sequence (xn,m), the limits
limn→∞
( limm→∞
xn,m), and limm→∞
( limn→∞
xn,m)
are called iterated limits.
Outrightly there is no reason to suppose the equality of the above two iterated limits
whenever they exist, as the following example shows.
Example 1.1.2. Consider the sequence xn,m =n
m+nof Example 1.1.1(b). Then for every
m ∈ N, limn→∞ xn,m = 1 and hence
limm→∞
( limn→∞
xn,m) = 1.
While for every n ∈ N, limm→∞ xn,m = 0 and hence
limn→∞
( limm→∞
xn,m) = 0.
Note that the double limit of this sequence does not exist, as has been shown in Example
1.1.1(b).
In the theory of double sequences, one of the most interesting questions is the following:
For a convergent double sequence, is it always the case that the iterated limits exist? The
answer to this question is no, as the following example shows.
Example 1.1.3. Consider the sequence xn,m = (−1)n+m( 1n +1m
).
Clearly, limn,m→∞ xn,m = 0. In fact, given � > 0, choose N ∈ N such that 1N <�2. Then we
have
n,m ≥ N ⇒ |(−1)n+m( 1n
+1
m)| ≤ 1
n+
1
m≤ 2
N< �.
6
But limn→∞(limm→∞ xn,m) does not exist,
since
limm→∞
xn,m
does not exist, and also
limm→∞
( limn→∞
xn,m)
does not exist, since
limn→∞
xn,m
does not exist.
It should be noted that, in general, the existence and the values of the iterated and double
limits of a double sequence (xn,m) depend on its form. While one of these limits exists, the
other may or may not exist and even if these exist, their values may differ. Besides Examples
1.1.2 and 1.1.3, the following examples shed some light on these cases.
Example 1.1.4.
(a) For the sequence xn,m =1n
+ 1m
, note first that the double limit
limn,m→∞ xn,m = 0. Indeed, given � > 0, ∃N ∈ N such that 1N <�2. Then,
n, m ≥ N ⇒ |1n
+1
m| = 1
n+
1
m≤ 2
N< �.
Moreover, since limn→∞ xn,m =1m
and limm→∞ xn,m =1n, it follows that the iterated
limits also exist and
limn→∞
( limm→∞
xn,m) = limm→∞
( limn→∞
xn,m) = 0.
(b) Consider the sequence xn,m = (−1)m( 1n +1m
). Clearly, by an argument similar to that
given in part (a), we have limn,m→∞ xn,m = 0. Also, the iterated limit
limm→∞
( limn→∞
xn,m) = 0,
7
since
( limn→∞
xn,m) =(−1)m
m.
But the other iterated limit limn→∞(limm→∞ xn,m) does not exist, since,
limm→∞ xn,m does not exist.
(c) For the sequence xn,m = (−1)n+m, it is clear that neither the double limit nor the
iterated limits exist.
The next result gives a necessary and sufficient condition for the existence of an iterated
limit of a convergent double sequence.
Theorem 1.1.3. [4] Let limn,m→∞ xn,m = a. Then limm→∞(limn→∞ xn,m) = a if and only
if limn→∞ xn,m exists for each m ∈ N.
Proof. The necessity is obvious. As for sufficiency, assume limn→∞ xn,m = cm for each
m ∈ N. We need to show that cm → a as m → ∞. Let � > 0 be given. Since xn,m → a as
n, m →∞, there exists N1 ∈ N such that
n, m ≥ N1 ⇒ |xn,m − a| <�
2,
and since for each m ∈ N, xn,m → cm as n →∞, there exists N2 ∈ N such that
n ≥ N2 ⇒ |xn,m − cm| <�
2.
Now choose n ≥ max{N1, N2}. Then ∀m ≥ N1, we have
|cm − a| = |cm − xn,m + xn,m − a|
≤ |cm − xn,m|+ |xn,m − a|
<�
2+
�
2= �.
Hence, cm → a as m →∞. �
It should be noted that the following theorem similar to Theorem 1.1.3 with an interchange
of the n and m symbols.
8
Theorem 1.1.4. [4] Let limn,m→∞ xn,m = a. Then limn→∞(limm→∞ xn,m) = a if and only
if limm→∞ xn,m exists for each n ∈ N.
Theorem 1.1.5. [4] Let limn,m→∞ xn,m = a. Then the iterated limits
limn→∞
( limm→∞
xn,m) and limm→∞
( limn→∞
xn,m)
exist and both are equal to a if and only if
(i) limm→∞ xn,m exists for each n ∈ N, and
(ii) limn→∞ xn,m exists for each m ∈ N.
Proof. By combining Theorems 1.1.3 and 1.1.4, we obtain the result.
The following example shows that the converse of Theorem 1.1.3 or Theorem 1.1.4 is not
true.
Example 1.1.5. Consider the sequence xn,m =nm
n2+m2. Clearly,
for each m ∈ N, limn→∞ xn,m = 0 and hence limm→∞(limn→∞ xn,m) = 0. But xn,m = 12 when
n = m and xn,m =25
when n = 2m, and hence it follows that the double limit limn,m→∞ xn,m
cannot exist in this case.
The next result can be viewed as a partial converse of Theorem 1.1.3. But, first, we give
some definitions and theorems.
Definition 1.1.5.[3] Suppose S ⊂ Rn and that for each n ∈ N, fn is a function from
S to R. The sequence fn converges uniformly to the function f on S if an only if given
� > 0, there exists N = n(�) such that
n > N(�) ⇒ |fn(x)− f(x)| < �, for all x ∈ S.
For S a subset of Rn, let FS be the set of bounded functions from S to R.
For two functions f, g ∈ FS, define ‖ f − g ‖= supx∈S |f(x)− g(x)|.
9
Theorem 1.1.6. [3]Let S be any subset of Rn, Let (fn) be a sequence of bounded functions
from S toR, and let f be a bounded function from S to R. Then (fn) converges uniformly
to f on S if and only if for all � > 0 there exists some N = N(�) such that n > N ⇒
supx∈S |f(x)− g(x)| < �.
Proof. (fn) converges uniformly to f on S if and only of for any � > 0 there is some N = N(�)
such that n > N(�), x ∈ S ⇒ |fn(x)−f(x)| < �/2 if and only if for n > N(�), supx∈S |fn(x)−
f(x)| ≤ �/2 < �.
Theorem 1.1.6. [4] If (xn,m) is a double sequence such that
(i) the iterated limit limm→∞(limn→∞ xn,m) = a, and
(ii) the limit limn→∞ fn(m) := xn,m exists uniformly in m ∈ N,
then the double limit limn,m→∞ xn,m = a.
Proof. For each n ∈ N, define a function fn on N by
fn(m) := xn,m ∀ m ∈ N.
Then, by hypothesis (ii), fn → f uniformly on N, where f(m) := limn→∞ xn,m. So given
� > 0, there exists N1 ∈ N such that
n ≥ N1 ⇒ |xn,m − f(m)| <�
2∀ m ∈ N.
Since, by hypothesis (i), limm→∞ f(m) = limm→∞(limn→∞ xn,m) = a, then for the same �,
there exists N2 ∈ N such that
m ≥ N2 ⇒ |f(m)− a| <�
2.
Now, letting N := max{N1, N2}, we have
n, m ≥ N ⇒ |xn,m − a| ≤ |xn,m − f(m)|+ |f(m)− a|
<�
2+
�
2= �,
10
which means that limn,m→∞ xn,m = a. �
It should be noted that the hypothesis that limn→∞ xn,m exists uniformly in m ∈ N in
the theorem above cannot be weakened to limn→∞ xn,m exists for every m ∈ N. Indeed,
reconsider the double sequence xn,m =nm
n2+m2of Example 1.1.5. It can be easily seen that
(1) (xn,m) is bounded, since |xn,m| ≤ 1 ∀n, m ∈ N.
(2) For each m ∈ N, limn→∞ xn,m = 0.
(3) limn→∞ xn,m 6= 0 uniformly in m ∈ N. Indeed, if for each n ∈ N we let
fn(m) := xn,m =nm
n2 + m2, m ∈ N,
then we obtain
‖ fn − 0 ‖ := supm{|fn(m)− 0| : m ∈ N}
= supm{|nm
n2 + m2− 0| : m ∈ N}
= supm{nm
n2 + m2: m ∈ N} (put) m = n
=1
2∀n ∈ N.
which implies that limn→∞ ‖ fn − 0 ‖6= 0.
(4) limm→∞(limn→∞ xn,m) = 0.
(5) limn,m→∞ xn,m does not exist, as has been shown in Example 1.1.5.
We conclude this section with the following remark: In Theorem 1.1.3, the assumption
that the limit limn→∞ xn,m exists for each m ∈ N does not follow from the assumption that
the double limit limn,m→∞ xn,m exists, as the following example shows.
Example 1.1.6. Consider the double sequence xn,m =(−1)n
m. Note, first, that
limn,m→∞
xn,m = limn,m→∞
(−1)n
m= 0.
Indeed, given � > 0, choose N ∈ N such that 1N
< �. Then for all n, m ≥ N , we have
| (−1)n
m− 0| = 1
m≤ 1
N< �.
11
On the other hand, limn→∞(−1)n
mdoes not exist for each fixed m ∈ N, since limn→∞ (−1)n
does not exist.
1.2 Monotone Double Sequences
In this section, we define increasing and decreasing double sequences of real numbers and
we prove a monotone convergence theorem for such sequences that are parallel to their
counterparts for single sequences.
Definition 1.2.1. We define a relation ≤ on N×N to be the lexicographic ordering, that is,
(a, b) ≤ (c, d)
if and only if
a < c
or
(a = c and b ≤ d)
Definition 1.2.2. Let (xn,m) be a double sequence of real numbers.
(i) If xn,m ≤ xj,k ∀ (n,m) ≤ (j, k) in N× N, we say the sequence is increasing.
(ii) If xn,m ≥ xj,k ∀ (n,m) ≤ (j, k) in N× N, we say the sequence is decreasing.
(ii) If (xn,m) is either increasing or decreasing, then we say it is monotone.
Theorem 1.2.1 (Monotone Convergence Theorem).[4] A monotone double sequence
of real numbers is convergent if and only if it is bounded. Further:
(a) If (xn,m) is increasing and bounded above, then
limm→∞
( limn→∞
xn,m) = limn→∞
( limm→∞
xn,m) = limn,m→∞
xn,m
= sup{xn,m : n,m ∈ N}.
12
(b) If (xn,m) is decreasing and bounded below, then
limm→∞
( limn→∞
xn,m) = limn→∞
( limm→∞
xn,m) = limn,m→∞
xn,m
= inf{xn,m : n, m ∈ N}.
Proof. It was shown in Theorem 1.1.2 that a convergent sequence must be bounded.
Conversely, let (xn,m) be a bounded monotone sequence. Then (xn,m) is increasing or
decreasing.
(a) We first treat the case that (xn,m) is increasing and bounded above. By the supremum
principle of real numbers, the supremum a∗ := sup{xn,m : n, m ∈ N} exists. We shall show
that the double and iterated limits of (xn,m) exist and are equal to a∗. If � > 0 is given,
then a∗ − � is not an upper bound for the set {xn,m : n, m ∈ N}; hence there exists natural
numbers K(�) and J(�) such that a∗ − � < xK,J . But since (xn,m) is increasing, it follows
that
a∗ − � < xK,J ≤ xn,m ≤ a∗ < a∗ + � ∀ (n,m) ≥ (K, J),
and hence
|xn,m − a∗| < � ∀ (n, m) ≥ (K, J).
Since � > 0 was arbitrary, it follows that (xn,m) converges to a∗.
Next, to show that
limm→∞
( limn→∞
xn,m) = limn,m→∞
xn,m = a∗,
note that since (xn,m) is bounded above, then, for each fixed m ∈ N, the single sequence
{xn,m : n ∈ N} is bounded above and increasing, so, by Monotone Convergence Theorem
3.3.2 of [3] for single sequences, we have
limn→∞
xn,m = sup{xn,m : n ∈ N} =: lm ∀ m ∈ N.
Hence, by Theorem 1.1.3, the iterated limit limm→∞(limn→∞ xn,m) exists and
limm→∞
( limn→∞
xn,m) = limn,m→∞
xn,m = a∗.
13
Similarly, it can be shown that
limn→∞
( limm→∞
xn,m) = limn,m→∞
xn,m = a∗.
(b) If (xn,m) is decreasing and bounded below, then the sequence (−xn,m) is increasing
and bounded above. Hence, by part (a), we obtain
limm→∞
( limn→∞
−xn,m) = limn→∞
( limm→∞
−xn,m) = limn,m→∞
−xn,m
= sup{−xn,m : n,m ∈ N}
= − inf{xn,m : n,m ∈ N}.
Therefore it follows that
limm→∞
( limn→∞
xn,m) = limn→∞
( limm→∞
xn,m) = limn,m→∞
xn,m
= inf{xn,m : n, m ∈ N}. �
1.3 Cauchy Double Sequences
We present in this section the important Cauchy Criterion for convergence of double se-
quences.
Definition 1.2.1. A double sequence (xn,m) of complex numbers is called a Cauchy sequence if
and only if for every � > 0, there exists a natural number N = N(�) such that
|xp,q − xn,m| < � ∀ p ≥ n ≥ N and q ≥ m ≥ N.
Theorem 1.2.1 (Cauchy Convergence Criterion for Double Sequences).[4] A double
sequence (xn,m) of complex numbers converges if and only if it is a Cauchy sequence.
Proof. (⇒) : Assume that xn,m → a as n,m → ∞. Then given � > 0, there exists
N ∈ N such that |xn,m − a| < �2 ∀n, m ≥ N. Hence, ∀p ≥ n ≥ N and ∀q ≥ m ≥ N we have
|xp,q − xn,m| = |xp,q − a + a− xn,m|
≤ |xp,q − a|+ |xn,m − a|
<�
2+
�
2= �;
14
that is, (xn,m) is a Cauchy sequence.
(⇐) : Assume that (xn,m) is a Cauchy sequence, and let � > 0 be given.
Taking m = n and writing xn,n = bn, we see that there exists K ∈ N such that
|bp − bn| < � ∀p ≥ n ≥ K.
Therefore, by Cauchy’s Criterion for single sequences, the sequence (bn) converges, say to
a ∈ C. Hence, there exists N1 ∈ N such that
|bn − a| <�
2∀n ≥ N1. (1.3)
Since (xn,m) is a Cauchy sequence, there exists N2 ∈ N such that
|xp,q − bn| <�
2∀p, q ≥ n ≥ N2. (1.4)
Let N := max{N1, N2} and choose n ≥ N. Then, by (1.3) and (1.4), we have
|xp,q − a| ≤ |xp,q − bn|+ |bn − a|
<�
2+
�
2= � ∀p, q ≥ N.
Hence, (xn,m) converges to a.
15
Chapter 2
Double series
A double series has terms with two subscripts instead of one. If the terms are xm,n we
think of the terms xm,n arranged in a rectangular array, with xm,n assigned to the point
(m,n). We regard the first subscript (here m) as the “row index”, the second subscript (here
n) as the “column index”.
For a double series we choose whether to “add by rows” or “add by columns.” That is, we
choose between∑∞m=0
(∑∞n=0 am,n
)(add by rows) and
∑∞n=0
(∑∞m=0 am,n
)(add by columns).
What we have here is a pair of series consisting of terms that are themselves series.
2.1 Nonnegative Double Series
In this section we discuss the case of nonnegative series firstly by adding by rows and columns
and secondly by using the mn-th partial sum of the double series as follows.
Definition 2.1.1. [5]
The partial sums for the first (add by rows) series are
RM :=M∑
m=0
( ∞∑n=0
am,n
)=
M∑n=0
rm, where rm :=∞∑
n=0
am,n,
16
and the partial sums for the second (add by columns) series are
CN :=N∑
n=0
( ∞∑m=0
am,n
)=
N∑n=0
cn, where cn :=∞∑
m=0
am,n.
The sum rm is the “sum” along row m. Each row is named by the column index, here m,
which remains constant in that row. Similarly, cn is the the “sum” along column n.
We define convergence, for each way of summing, in the usual way-by requiring the ap-
propriate (row or column) sequences of partial sums to converge. Otherwise, we say the
double series diverges by rows or by columns.
We ask is it true that∞∑
m=0
( ∞∑n=0
xm,n
)=
∞∑n=0
( ∞∑m=0
am,n
); (2.1)
i.e., can we switch the order of summation? the answer is “No!” and here is an example to
see this:
Example 2.1.1.[5] Let ak,k = 1, ak,k+1 = −1 for all k ∈ N0 := N ∪ {0}, and let am,n =
0 otherwise. Then rm = 0 for all m and c0 = 1, while cn = 0, when n > 0. Thus
0 =∞∑
m=0
rm =∞∑
m=0
( ∞∑n=0
am,n
)and
∞∑n=0
( ∞∑m=0
am,n
)=
∞∑n=0
cn = 1.
Thus both double sums are finite but
∞∑m=0
( ∞∑n=0
am,n
)6=
∞∑n=0
( ∞∑m=0
am,n
)Theorem 2.1.1 (The Double-Series Theorem for Double Series with Non-negative
Terms)[5]
If pm,n ≥ 0 for every pair (m,n) ∈ N0 × N0, then∞∑
n=0
( ∞∑m=0
pm,n
)=
∞∑m=0
( ∞∑n=0
pm,n
).
Remark: The equation in the statement of the theorem is true whether or not the double
sums are finite.
17
Proof. If both series diverge to +∞, there is nothing to show. Suppose that one of the sums
is finite. The first case is that the series added by columns is finite. We are assuming:
∞∑n=0
( ∞∑m=0
pm,n
)< +∞.
Now pM,n ≤∞∑
m=0
pm,n = cn, for each M ∈ N0 and n ∈ N0, since pM,n is a term in the sum
∞∑m=0
pm,n. These inequalities, for each fixed M ∈ N0, and the comparison test yield that
(now putting M = m),
pm,n ≤ cn and∞∑
n=0
cn converges, so rm =∞∑
n=0
pm,n ≤∞∑
n=0
cn < +∞ for each m ∈ N0.
In other words, each row sum is finite (convergent) because the series summed by columns
is convergent.
Now we have to prove that∞∑
m=0
rm < ∞ (converges). Since the terms rm are non-
negative, all we have to do is to show that the partial sums RM =M∑
m=0
rm are bounded
above. It will be to our advantage if we show that the partial sums are bounded above by
the sum by columns, namely
C :=∞∑
n=0
cn =∞∑
n=0
( ∞∑m=0
pm,n
)< +∞. (2.2)
We actually only show that for every � > 0 and for every M ∈ N0, RM =M∑
m=0
rm <
C + �. But then we will know thatM∑
m=0
rm < C + � ∀ � > 0, and hence∞∑
m=0
rm < C.
First we allow M to be a given natural number, no matter how large, and we allow
� > 0 to be given, no matter how small. Since rm =∞∑
n=0
pm,n < +∞ ∀m ∈ N0, we can find
(for each m ∈ N0) a natural Nm such that
rm =∞∑
n=0
pm,n <
(Nm∑n=0
pm,n
)+ �/(M + 1).
This only needs to be done for 0 ≤ m ≤ M. Now we define N� := max{N0, N1, · · · , NM}. Then
because pm,n ≥ 0, we can replace each Nm by the larger quantity N� :
18
rm =∞∑
n=0
pm,n <
(N�∑
n=0
pm,n
)+ �/(M + 1), 0 ≤ m ≤ M.
Therefore (We will drop the parentheses about the sum in n now)
RM =M∑
m=0
rm <
M∑m=0
(N�∑
n=0
pm,n + �/(M + 1)
)
=M∑
m=0
N�∑n=0
pm,n +M∑
m=0
�/(M + 1)
=M∑
m=0
N�∑n=0
pm,n + �
=N�∑
n=0
M∑m=0
pm,n + �
At last, we have used the fact that we can rearrange finite sums. Therefore in the inner sum
we can replace M by + ∞, because that can only increase the terms. This gives, from
what we just did,
RM <N�∑
n=0
M∑m=0
pm,n + �
≤N�∑
n=0
∞∑m=0
pm,n + �
=N�∑
n=0
cn + � = CN� + � ≤ C + �.
Thus RM < C + �, so
R :=∞∑
m=0
(∞∑
n=0
pm,n
)=
∞∑m=0
rm = limM→∞
RM ≤ C + �.
Since � > 0 is arbitrary, R ≤ C. We have shown that if C < ∞, then R < ∞, and we
have even shown that R ≤ C. But now that we know R < ∞, we can redo the argument,
with the rows and columns reversed and obtain the estimate C ≤ R. Thus the case that
one sum be infinite and the other be finite cannot occur. This completes the proof.
19
Recall that if (an) is a sequence of complex numbers, then we say that∑
an converges
if the sequence (sn) converges, where sn :=∑n
k=1 ak. By analogy, we define a double series
of complex numbers as follows. Let (am,n) be a double sequences of complex numbers and
let
sm,n :=m∑
i=1
n∑j=1
ai,j,
called the mn−th partial sum of∑
am,n. We say that the double series∑
am,n converges if
the double sequences (sm,n) of partial sums converges. If∑
am,n exists, we can ask whether
or not ∑am,n =
∞∑m=1
∞∑n=1
am,n =∞∑
n=1
∞∑m=1
am,n? (2.3)
Here, with sm,n =∑m
i=1
∑nj=1 ai,j the iterated series on the right are defined as
∞∑m=1
∞∑n=1
am,n := limm→∞
limn→∞
sm,n and∞∑
n=1
∞∑m=1
am,n := limn→∞
limm→∞
sm,n.
Theorem 2.1.2.(Alternative Nonnegative double series theorem). [6]
If∑m,n
am,n converges where am,n ≥ 0 for all m,n, then both iterated series converge, and
∑m,n
am,n =∞∑
m=1
∞∑n=1
am,n =∞∑
n=1
∞∑m=1
am,n.
Moreover, given � > 0 there is an N such that
k > N ⇒∞∑i=1
∞∑j=k
ai,j < � and
∞∑i=k
∞∑j=1
ai,j < �.
Proof. Assume that the series∑
am,n converges and let � > 0. Since∑
am,n converges,
setting s :=∑
am,n and
sm,n :=m∑
i=1
n∑j=1
ai,j =n∑
j=1
m∑i=1
ai,j, (2.4)
by definition of convergence we can choose N such that
m, n > N ⇒ |s− sm,n| < �/2.
20
Given i ∈ N, choose m ≥ i such that m > N and let n > N. Then in view of (2.4) we haven∑
j=1
aij ≤m∑
i=1
n∑j=1
ai,j = sm,n < s + �/2.
Therefore, the partial sums of∑∞
j=1 ai,j are bounded above by a fixed constant and hence
(by the nonnegative test), for any i ∈ N, the sum∑∞
j=1 ai,j exists. In particular,
limn→∞
sm,n = limn→∞
m∑i=1
n∑j=1
ai,j =m∑
i=1
∞∑j=1
ai,j;
here, we can interchange the limit with the first sum because the first sum is finite.
Now taking n →∞ in (2.4), we see that
m > N ⇒
∣∣∣∣∣s−m∑
i=1
∞∑j=1
ai,j
∣∣∣∣∣ ≤ �2 < �.Since � > 0 was arbitrary, by definition of convergence,
∑∞i=1
∑∞j=1 ai,j exists with sum s =∑
m,n am,n : ∑m,n
am,n =∞∑i=1
∞∑j=1
ai,j
Similarly, using the second form of sm,n in (2.4), one can use an analogous argument to
show that∑
am,n =∑∞
j=1
∑∞i=1 ai,j.
We now prove the last statement of our theorem. To this end, we observe that for any
k ∈ N, we have
s =∞∑i=1
∞∑j=1
ai,j =∞∑i=1
(k∑
j=1
ai,j +∞∑
j=k+1
ai,j
)=
∞∑i=1
k∑j=1
ai,j +∞∑i=1
∞∑j=k+1
ai,j,
which implies that
∞∑i=1
∞∑j=k+1
ai,j = s−∞∑i=1
k∑j=1
ai,j = s− limm→∞
sm,k. (2.5)
By the first part of this proof, we know that s = limk→∞
limm→∞
sm,k, so taking k →∞ in (2.5),
we see that the left-hand side of (2.5) tends to zero as k → ∞, so it follows that for some
N1 ∈ N,
k > N1 ⇒∞∑i=1
∞∑j=k
ai,j < �.
21
A similar argument shows that there is an N2 ∈ N such that
k > N2 ⇒∞∑
i=k
∞∑j=1
ai,j < �.
Setting N as the largest of N1 and N2 completes the proof.
2.2 Summing by curves
Before presenting the “sum by curves theorem” it might be helpful to give a couple examples
of this theorem to help in understanding what it says. Let∑
am,n be an absolutely conver-
gent series, and consider the following pictures.
Figure 2.1 “Summing by squares” and “Summing by triangles”
Example 2.2.1.[6] Let
Sk = {(m, n) : 1 ≤ m ≤ k, 1 ≤ n ≤ k},
which represents a k× k square of numbers; see the left-hand picture in Figure (2.1) for 1×
1, 2× 2, 3× 3, and 4× 4 examples. We denote by∑
(m,n)∈Sk am,n the sum of those am,n’s
within the k × k square Sk. Explicitly,∑(m,n)∈Sk
am,n =k∑
m=1
k∑n=1
am,n.
The sum by curves theorem implies that
∑am,n = lim
k→∞
∑(m,n)∈Sk
am,n = limk→∞
k∑m=1
k∑n=1
am,n. (2.6)
22
As we already noted,∑
(m,n)∈Sk am,n involves summing the am,n’s within a k × k square;
for this reason, (2.6) is referred as “summing by squares”.
Example 2.2.2.[6] Let
Sk = T1 ∪ · · · ∪ Tk, where Tl = {(m, n) : m + n = l + 1}.
Notice that Tl = {(m, n) : m + n = l + 1} = {(1, l), (2, l− 1), · · · , (l, 1)} represents the l−th
diagonal in the right-hand picture in Figure 2.1; for instance, T3 = {(1, 3), (2, 2), (3, 1)} is
the third diagonal in Figure 2.1. Then∑(m,n)∈Sk
am,n =k∑
l=1
∑(m,n)∈Tl
am,n
is the sum of the am,n’s that are within the triangle consisting of the first k diagonals.
The sum by curves theorem implies that∑am,n = lim
k→∞
∑(m,n)∈Sk
am,n = limk→∞
k∑l=1
∑(m,n)∈Tl
am,n,
or using that Tk = {(1, k), (2, k − 1), · · · , (k, 1)} we have
∑am,n =
∞∑k=1
(a1,k + a2,k−1 + · · ·+ ak,1). (2.7)
We refer to (2.7) as “summing by triangles”.
Theorem 2.2.1 (Sum by Curves Theorem)[6]. An absolutely convergent series∑
am,n itself
converges. Moreover, if S1 ⊆ S2 ⊆ S3 ⊆ · · · ⊆ N × N is a nondecreasing sequence of finite
sets having the property that for any m,n there is a k such that
{1, 2, · · · , m}×{1, 2, · · · , n} ⊆ Sk ⊆ Sk+1 ⊆ Sk+2 ⊆ · · · , (2.8)
then the sequence (sk) converges, where sk is the finite sum
sk :=∑
(m,n)∈Sk
am,n,
and furthermore, ∑am,n = lim sk.
23
Proof. We first prove that the sequence (sk) converges by showing that the sequence is
Cauchy. Indeed, let � > 0 be given. By assumption,∑
|am,n| converges, so, by Theorem
2.1.2, we can choose N ∈ N such that
k > N ⇒∞∑i=1
∞∑j=k
|ai,j| < � and∞∑
i=k
∞∑j=1
|ai,j| < �. (2.9)
By the property (2.8) of the sets {Sk} there is an N ′ such that
{1, 2, · · · , N}× {1, 2, · · · , N} ⊆ SN ′ ⊆ SN ′+1 ⊆ SN ′+2 ⊆ · · · (2.10)
Let k > l > N ′. Then, since Sl ⊆ Sk, we have
|sk − sl| =
∣∣∣∣∣ ∑(i,j)∈Sk
ai,j −∑
(i,j)∈Sl
ai,j
∣∣∣∣∣ =∣∣∣∣∣ ∑
(i,j)∈(Sk\Sl)
ai,j
∣∣∣∣∣ ≤ ∑(i,j)∈(Sk\Sl)
|ai,j|.
Since l > N ′, by (2.10), Sl contains {1, 2, · · · , N} × {1, 2, · · · , N}. Hence,
Sk\Sl is a subset of N×{N +1, N +2, · · · } or {N +1, N +2, · · · }×N, (to show that suppose
on contrary that Sk\Sl is neither a subset of N× {N + 1, N + 2, · · · } nor a subset of {N +
1, N + 2, · · · } × N, so there is some element (a, b) in Sk\Sl which is neither in N × {N +
1, N +2, · · · } nor in {N +1, N +2, · · · }×N, which implies that a, b not in {N +1, N +2, · · · },
hence a, b ∈ {1, 2, · · · , N} ⇒ (a, b) ∈ {1, 2, · · · , N} × {1, 2, · · · , N} ⊂ Sl a contradiction.)
For concreteness, assume that the first case holds; the second case can be dealt with in a
similar way. In this case, by the property (2.9), we have
|sk − sl| ≤∑
(i,j)∈(Sk\Sl)
|ai,j| ≤∞∑i=1
∞∑j=N+1
|ai,j| < �.
This shows that (sk) is Cauchy and hence converges.
We now show that∑
am,n converges with sum equals to lim sk. Let � > 0 be given and
choose N such that (2.9) holds with � replaced by �/2. Fix natural numbers m, n > N. By
the property (2.8) and the fact that sk → s := lim sk we can choose a k > N such that
{1, 2, · · · , m} × {1, 2, · · · , n} ⊆ Sk
24
and |sk − s| < �/2. Now observe that
|sk − sm,n| =
∣∣∣∣∣ ∑(i,j)∈Sk
ai,j −∑
(i,j)∈{1,··· ,m}×{1,··· ,n}
ai,j
∣∣∣∣∣ ≤ ∑(i,j)∈(Sk\({1,··· ,m}×{1,··· ,n}))
|ai,j|.
Notice that Sk\({1, · · · , m}×{1, · · · , n}) is a subset of N×{n+1, n+2, · · · } or{m+1, m+
2, · · · } ×N. For concreteness, assume that the first case holds; the second case can be dealt
with in a similar manner. In this case, by the property (2.9) (with � replaced with �/2), we
have
|sk − sm,n| ≤∑
(i,j)∈(Sk\({1,··· ,m}×{1,··· ,n}))
|ai,j| <∞∑i=1
∞∑j=n+1
|ai,j| < �/2.
Hence,
|sm,n − s| ≤ |sm,n − sk|+ |sk − s| <�
2+
�
2= �.
This proves that∑
am,n = s and completes the proof.
We now come to Cauchy’s double series theorem, the most important result of this section.
Instead of summing by curves, in many applications we are interested in summing by rows
or by columns.
Theorem 2.2.2 (Cauchy’s Double Series Theorem).[6]
A series∑
am,n is absolutely convergent if and only if
∞∑m=1
∞∑n=1
|am,n| < ∞ or∞∑
n=1
∞∑m=1
|am,n| < ∞,
in which case∞,∞∑
m=1,n=1
am,n =∞∑
m=1
∞∑n=1
am,n =∞∑
n=1
∞∑m=1
am,n
in the sense that both iterated sums converge and are equal to the sum of the series.
Proof. Assume that the sum∑
am,n converges absolutely. Then, by Theorem 2.1.2, we
know that∞,∞∑
m=1,n=1
|am,n| =∞∑
m=1
∞∑n=1
|am,n| =∞∑
n=1
∞∑m=1
|am,n|.
25
We shall prove that the iterated sums∞∑
m=1
∞∑n=1
am,n and∞∑
n=1
∞∑m=1
am,n converge and equal
s :=∑
am,n, which exists by the sum by curves theorem. Let sm,n denote the partial sums
of∑
am,n. Let � > 0 be given and choose a natural number N such that
m,n > N ⇒ |s− sm,n| <�
2. (2.11)
Since∞∑
m=1
∞∑n=1
|am,n| < ∞,
this implies, in particular, that for any m ∈ N, the sum∑∞
n=1 |am,n| converges, and hence
for any m ∈ N,∑∞
n=1 am,n = limn→∞ sm,n converges. Thus, letting n → ∞ in (2.11), we
obtain
m > N ⇒ |s− limn→∞
sm,n| ≤�
2< �.
But this means that s = limm→∞ limn→∞ sm,n; that is,
s =∞∑
m=1
∞∑n=1
am,n.
A similar argument gives this equality with the sums reversed.
Now assume that∞∑
m=1
∞∑n=1
|am,n| = t < ∞.
We will show that∑
am,n is absolutely convergent; a similar proof shows that if
∞∑n=1
∞∑m=1
|am,n| < ∞,
then∑
am,n is absolutely convergent. Let � > 0 be given. Then the fact that
∞∑i=1
(∞∑
j=1
|ai,j|) < ∞
implies, by the Cauchy criterion for single series, there is an N such that for m > N,
m > n ⇒∞∑
i=m+1
(∞∑
j=1
|ai,j|
)<
�
2.
26
Let m,n > N. Then for any k > m, we have∣∣∣∣∣k∑
i=1
∞∑j=1
|ai,j| −m∑
i=1
∞∑j=1
|ai,j|
∣∣∣∣∣ ≤k∑
i=m+1
∞∑j=1
|ai,j| ≤∞∑
i=m+1
∞∑j=1
|ai,j| <�
2.
Letting k →∞ shows that for all m,n > N,∣∣∣∣∣t−m∑
i=1
n∑j=1
|ai,j|
∣∣∣∣∣ ≤ �2 < �,which proves that
∑|am,n| converges, and completes the proof of the theorem.
Example 2.2.3.[6] Consider the sum∑
m,n 1/(mpnq) where p, q ∈ R. Since in this case,
∞∑n=1
1
mpnq=
1
mp.
(∞∑
n=1
1
nq
),
it follows that∞∑
m=1
∞∑n=1
1
mpnq=
(∞∑
m=1
1
mp
).
(∞∑
n=1
1
nq
).
Therefore, by Cauchy’s double series theorem and the p−test,∑
1/(mpnq) converges if and
only if both p, q > 1.
Example 2.2.4.[6] Consider the sum∑m,n
1/(m4 + n4). Observe that
(m2 − n2)2 ≥ 0 ⇒ m4 + n4 − 2m2n2 ≥ 0 ⇒ 1m4 + n4
≤ 12m2n2
.
Then by the comparison test for single series we have∞∑
m=1
1
m4 + n4, converges for some
n ∈ N. But∞∑
m=1
1
m4 + n4≤
∞∑m=1
1
2m2n2,
since by the previous example∑m,n
1/(m2n2) converges, by we see that∑m,n
1/(m4+n4) converges
too.
Example 2.2.5[6] For an application of Cauchy’s double series theorem and the sum by
curves theorem, we look at the double sum∑
m,n zm+n for |z| < 1. For such z, this sum
converges absolutely because∞∑
m=0
∞∑n=0
|z|m+n =∞∑
m=0
|z|m. 11− |z|
=1
(1− |z|)2< ∞,
27
where we used the geometric series test (twice): If |r| < 1, then∑∞
k=0 rk =
1
1− r. So∑
zm+n converges absolutely by Cauchy’s double series theorem, and∑zm+n =
∞∑m=0
∞∑n=0
zm+n =∞∑
m=0
zm.1
1− z=
1
(1− z)2.
On the other hand by our sum by curves theorem, we can determine∑
zm+n by summing
over curves; we shall choose to sum over triangles. Thus, if we set
Sk = T0 ∪ T1 ∪ T2 ∪ · · · ∪ Tk , where Tl = {(m, n) : m + n = l, m, n ≥ 0},
then ∑zm+n = lim
k→∞
∑(m,n)∈Sk
zm+n = limk→∞
k∑l=0
∑(m,n)∈Tl
zm+n.
Since Tl = {(m, n) : m + n = l, m, n ≥ 0} = {(0, l), (1, l − 1), · · · , (l, 0)}, we have∑(m,n)∈Tl
zm+n = z0+l + z1+(l−1) + z2+(l−2) + · · ·+ zl+0 = (l + 1)zl.
Thus,∑
zm+n =∑∞
k=0(k+1)zk. However, we already proved that
∑zm+n = 1/(1−z)2, so
1
(1− z)2=
∞∑n=1
nzn−1. (2.12)
2.3 Cauchy’s Product
Definition 2.3.1(Cauchy’s Product).[6] Given two series∑∞
n=0 an and∑∞
n=0 bn, their
Cauchy product is the series∑∞
n=0 cn, where
cn = a0bn + a1bn−1 + · · ·+ anb0 =n∑
k=0
akbn−k.
A natural question to ask is if∑∞
n=0 an and∑∞
n=0 bn converge, then is it true that( ∞∑n=0
an
)( ∞∑n=0
bn
)=
∞∑n=0
cn?
The answer is “no.”
Example 2.3.1[6] Let us consider the example (∑∞
n=1(−1)n−1√
n)(∑∞
n=1(−1)n−1√
n),
which is due to Cauchy. That is, let a0 = b0 = 0 and
28
an = bn = (−1)n−11√n
, n = 1, 2, 3, · · · .
We know, by the alternating series test, that∑∞
n=1(−1)n−1√
nconverges, and hence
(∞∑
n=1
(−1)n−1√n
)(∞∑
n=1
(−1)n−1√n
).
converges. However, we shall see that the cauchy product does not converge. Indeed,
c0 = a0b0 = 0, c1 = a0b1 + a1b0 = 0,
and for n ≥ 2,
cn =n∑
k=0
akbn−k =n−1∑k=1
(−1)k(−1)n−k√k√
n− k= (−1)n
n−1∑k=1
1√k√
n− k.
Since for 1 ≤ k ≤ n− 1, we have
k(n− k) ≤ (n− 1)(n− 1) = (n− 1)2 ⇒ 1n− 1
≤ 1√k(n− k)
,
we see that
(−1)ncn =n−1∑k=1
1√k(n− k)
≥n−1∑k=1
1
n− 1=
1
n− 1
n−1∑k=1
1 = 1.
Thus, the terms cn do not tend to zero as n → ∞, so by the n-th term test, the series∑∞n=0 cn does not converge.
The problem with this example is that the series∑∞
n=1(−1)n−1√
ndoes not converge ab-
solutely. However, for absolutely convergent series, there is no problem as the following
theorem shows.
Theorem 2.3.1.(Mertens’ multiplication theorem).[6] If at least one of two convergent
series∑
an = A and∑
bn = B converges absolutely, then their Cauchy product converges
with sum equals AB.
29
Proof. Consider the partial sums of the Cauchy product:
Cn = c0 + c1 + · · ·+ cn
= a0b0 + (a0b1 + a1b0) + · · ·+ (a0bn + a1bn−1 + · · ·+ anb0) (2.13)
= a0(b0 + · · ·+ bn) + a1(b0 + · · ·+ bn−1) + · · ·+ anb0.
We need to show that Cn tends to AB as n → ∞. Because our notation is symmetric in
A and B, we may assume that the sum∑
an is absolutely convergent. If An denotes the
n−th partial sum of∑
an and Bn that of∑
bn, then from (2.13), we have
Cn = a0Bn + a1Bn−1 + · · ·+ anB0.
Set βk := Bk −B. Since Bk → B we have βk → 0. Now we write
Cn = a0(B + βn) + a1(B + βn−1) + · · ·+ an(B + β0)
= AnB + (a0βn + a1βn−1 + · · ·+ anβ0)
Since An → A, then AnB → AB. Thus, we just need to show that the term in parenthesis
tends to zero as n →∞. To see this, let � > 0 be given. Putting α =∑|an| and using that
βn → 0, we can choose a natural number N such that for all n > N, we have |βn| < �(2α) .
Also, since βn → 0, we can choose a constant C such that |βn| < C for every n. Then for
n > N,
|a0βn + a1βn−1 + · · ·+ anβ0| = |a0βn + a1βn−1 + · · ·+ an−N+1βN+1
+an−NβN + · · ·+ anβ0|
≤ |a0βn + a1βn−1 + · · ·+ an−N+1βN+1|+ |an−NβN + · · ·+ anβ0|
<
(|a0|+ |a1|+ · · ·+ |an−N+1|
).
�
2α+
(|an−N |+ · · ·+ |an|
).C
≤ α. �2α
+ C(|an−N |+ · · ·+ |an|
)=
�
2+ C
(|an−N |+ · · ·+ |an|
).
30
Since∑|an| < ∞, by the Cauchy criterion for single series, we choose N ′ > N such that
n > N ′ ⇒ |an−N |+ · · ·+ |an| <�
2C.
Then for n > N ′, we see that
|a0βn + a1βn−1 + · · ·+ anβ0| <�
2+
�
2= �.
Since � > 0 was arbitrary, this completes the proof.
Theorem 2.3.2 (Cauchy’s Multiplication Theorem).[6]
If two series∑
an = A and∑
bn = B converge absolutely, then the double series∑m,n ambn converges absolutely and has the value AB.
Proof. Since
∞∑m=0
∞∑n=0
|ambn| =∞∑
m=0
|am|∞∑
n=0
|bn| =
(∞∑
m=0
|am|
)(∞∑
n=0
|bn|
)< ∞,
by Cauchy’s double series theorem, the double series∑
ambn converges absolutely, and we
can iterate the sums:
∑ambn =
∞∑m=0
∞∑n=0
ambn =∞∑
m=0
am
∞∑n=0
bn =
(∞∑
m=0
am
)(∞∑
n=0
bn
)= AB.
31
Chapter 3
Divergent Double series
3.1 Transformation of Double Sequences
There are two types of linear transformation defined by an infinite matrix on numbers.
T :
a1,1
a2,1 a2,2
a3,1 a3,2 a3,3
a4,1 a4,2 a4,3 a4,4...
......
......
S :
a1,1 a1,2 a1,3 . . .
a2,1 a2,2 a2,3 . . .
a3,1 a3,2 a3,3 . . ....
......
One by a triangular matrix T , the other by a square matrix S. For any sequence (xn) a
new sequence (yn) is defined as follows:
yn =∑n
k=1 an,k xk, for the matrix T .
yn =∑∞
k=1 an,k xk, for the matrix S,
provided in the latter case yn has a meaning. If to any matrix of type T we adjoint the
elements an,k = 0, k > n ∀n ∈ N, we obtain a matrix of type S.
32
Definition 3.1.1.[7] If for either transformation limn→∞ yn exists, the limit is called the
generalized value of the sequence (xn) by the transformation.
Definition 3.1.2.[7] The transformation is said to be regular if whenever xn converges, yn converges
to the same value.
The criterion for regularity of these transformations is stated as follows:
Theorem 3.1.1.[7] A necessary and sufficient condition that the transformation T be regular
is that:
(1) limn→∞ an,k = 0 for every k,
(2) limn→∞∑n
k=1 an,k = 1,
(3)∑n
k=1 |an,k| < A for all n and some A > 0,
Proof. For a proof see [7]
Theorem 3.1.2.[7] A necessary and sufficient condition that the transformation S be regular
is that:
(1) limn→∞ an,k = 0 for every k,
(2)∑∞
k=1 |an,k| converges for each n,
(3)∑∞
k=1 |an,k| < A for all n and some A > 0,
(4) limn→∞∑∞
k=1 an,k = 1.
Proof. For a proof see [7]
Corresponding to these definitions of summability for single series, we have the following
definitions for giving a value to a divergent double series. Let the given series∑
um,n be
represented as follows:
u1,1 + u1,2 + u1,3 + u1,4 + u1,5 + · · ·
+ u2,1 + u2,2 + u2,3 + u2,4 + u2,5 + · · ·
33
+ u3,1 + u3,2 + u3,3 + u3,4 + · · ·
+ u4,1 + · · · · · · · · · · · · · · · · · · · · · ;
then the corresponding double sequence (xm,n) of partial sums for this series is given by the
following equality:
xm,n =
m,n∑k=1,l=1
uk,l. (3.1)
Thus we have also
um,n = xm,n + xm−1,n−1 − xm−1,n − xm,n−1, m > 1, n > 1;
u1,n = x1,n − x1,n−1, n > 1;
um,1 = xm,1 − xm−1,1, m > 1;
u1,1 = x1,1.
We define a new double sequence (ym,n) by the relation
ym,n =
m,n∑k=1,l=1
am,n,k,l xk,l.
We shall call this transformation and its matrix A = (am,n,k,l) a double A−transformation of
the type T ; here k ≤ m; l ≤ n. We may write
ym,n =
∞,∞∑k=1,l=1
am,n,k,l xk,l,
provided ym,n has a meaning. We shall call this transformation and its matrix A = (am,n,k,l) a
double A− transformation of the type S; here k, l ≥ 0. Any double A-transformation of
type T may be considered as a special case of a transformation of type S; by adding the
elements
am,n,k,l = 0, m < k, n < l, ∀ m and n,
am,n,k,l = 0, 1 ≤ k ≤ m, n < l, ∀ m and n,
am,n,k,l = 0, m < k, 1 ≤ l ≤ n, ∀ m and n,
to any matrix of the type T we obtain a matrix of type S such that the resulting transfor-
mation is identical with the original one.
Definition 3.1.3.[7] If for either transformation the double sequence (ym,n) possesses a
34
limit, the limit is called the generalized value of the sequence (xm,n) by the transformation.
It is a well-known fact that if a single series converges, the corresponding sequence is
bounded. This need not hold for a double series. Thus consider the series
u1,n = 1,
u2,n = −1,
um,n = 0, m ≥ 3, n ≥ 1.
This series converges, but the corresponding sequences (xm,n) as defined by (3.1) is not
bounded. Thus convergent double series may be divided into two classes according to whether
the corresponding sequences are bounded or not. The following definition for regularity of a
transformation is constructed with regard to a convergent bounded sequence; thus even if a
transformation is regular it need not give to an unbounded convergent sequence the value to
which it is convergent.
Definition 3.1.4.[7] If whenever xm,n is bounded convergent sequence, ym,n converges to
the same value, then the transformation is said to be regular.
Definition 3.1.5.[7] A regular transformation of real numbers (am,n,k,l) is said to be totally
regular, provided when applied to a sequence of real numbers (xm,n), which has the following
properties,
(1) xm,n is bounded for each m,
(2) xm,n is bounded for each n,
(3) limm,n→∞ xm,n = +∞,
it transforms this sequence into a sequences which has its limit +∞.
3.2 Regularity of Double Linear Transformation
The criterion of regularity of the transformations of type T is given in the following theorem.
35
Theorem 3.2.1.[7] The necessary and sufficient conditions that any transformation of type
T be regular are
(a) limm,n→∞
am,n,k,l = 0, for each k and l,
(b) limm,n→∞
m,n∑k=1,l=1
am,n,k,l = 1,
(c) limm,n→∞
m∑k=1
|am,n,k,l| = 0, for each l,
(d) limm,n→∞
n∑l=1
|am,n,k,l| = 0, for each k, and
(e)
m,n∑k=1,l=1
|am,n,k,l| ≤ K, for all m, n,
where K is some constant.
Proof of necessity. (a) Define a sequence (xm,n) as follows:
xm,n =
1, if m = p, n = q;0, otherwise.Then lim
m,n→∞xm,n = 0, ym,n = am,n,p,q. Hence in order that lim
m,n→∞ym,n = 0, it is necessary
that limm,n→∞
am,n,p,q = 0 for each p and q. Thus condition (a) is necessary.
(b) Consider the sequence (xm,n) defined as follows : xm,n = 1 ∀ m,n. Then
limm,n→∞
xm,n = 1, ym,n =
m,n∑k=1, l=1
am,n,k,l.
Since by regularity of T limm,n→∞
ym,n = 1, condition (b) is necessary.
(c) To show the necessity of condition (c) we assume that condition (a) is satisfied and
that (c) is not, and obtain a contradiction. Since we are assuming that for l = l0 (some
fixed integer) the sequencem∑
k=1
|am,n,k,l0| 9 0,
36
for some preassigned constant h > 0 there must exist a sub-sequence of this sequence, such
that each element of it is greater than h. Choose m1 and n1 such that
m1∑k=1
|am1,n1,k,l0| > h.
Choose m2 > m1; n2 > n1 and such that
m1∑k=1
|am2,n2,k,l0| ≤h
2,
m2∑k=1
|am2,n2,k,l0| > h,
and in general choose mp > mp−1; np > np−1 and such that
mp−1∑k=1
|amp,np,k,l0| <h
2p−1,
mp∑k=1
|amp,np,k,l0| > h. (3.2)
From (3.2) we have
mp∑k=mp−1+1
|amp,np,k,l0| > h−h
2p−1= h
(1− 1
2p−1
). (3.3)
Define a sequence (xm,n) as follows :
xm,n =
0, if n 6= l0;
sgn am1,n1,k,l0 , m ≤ m1; (3.4)
sgn am2,n2,k,l0 , m1 < m ≤ m2;
· · · · · · · · · · · ·
sgn amp,np,k,l0 , mp−1 < m ≤ mp;
· · · · · · · · · · · ·
Here limm,n→∞
xm,n = 0 since limm,n→∞ xm,n = limm→∞(limn→∞ xm,n) = limm→∞(0) = 0. For
this sequence (xm,n) we have
ymp,np =
mp∑k=1
amp,np,k,l0 xk,l0
=
mp−1∑k=1
amp,np,k,l0 xk,l0 +
mp∑k=mp−1+1
amp,np,k,l0 xk,l0 .
Since
mp−1∑k=1
amp,np,k,l0 sgn amp,np,k,l0 =
mp−1∑k=1
|amp,np,k,l0|, so we have from (3.2), (3.3), and (3.4)
it follows that
37
∣∣∣∣∣mp−1∑k=1
amp,np,k,l0 xk,l0
∣∣∣∣∣ =∣∣∣∣∣mp−1∑k=1
amp,np,k,l0 sgn amp,np,k,l0
∣∣∣∣∣≤
mp−1∑k=1
∣∣amp,np,k,l0∣∣ ≤ h2p−1 ,mp∑
k=mp−1+1
amp,np,k,l0 xk,l0 =
mp∑k=mp−1+1
∣∣amp,np,k,l0∣∣ ≥ h(1− 12p−1)
.
Hence
ymp,np ≥ h(
1− 12p−1
)− h
2p−1= h
(1− 1
2p−2
).
Thus ym,n does not have the limit zero, from which follows the necessity of condition (c).
(d) The above proof can be used for showing the necessity of condition (d) by simply
interchanging the roles of rows and columns.
(e) Assume conditions (a) and (b) are satisfied and that (e) does not hold. Choose
m1 and n1 such thatm1,n1∑
k=1,l=1
∣∣∣∣am1,n1,k,l∣∣∣∣ ≥ 1.Choose m2 > m1, n2 > n1, and such that
m1,n1∑k=1,l=1
∣∣∣∣am2,n2,k,l∣∣∣∣ ≤ 2,m2,n2∑
k=1,l=1
∣∣∣∣am2,n2,k,l∣∣∣∣ ≥ 24,and, in general, choose mp > mp−1, np > np−1, such that
mp−1,np−1∑k=1,l=1
∣∣∣∣amp,np,k,l∣∣∣∣ ≤ 2p−1, mp,np∑k=1, l=1
∣∣∣∣amp,np,k,l∣∣∣∣ ≥ 22p. (3.5)From (3.5) we have
mp−1,np∑k=1, l=np−1+1
∣∣amp,np,k,l∣∣+ mp,np−1∑k=mp−1+1,l=1
∣∣amp,np,k,l∣∣+ mp,np∑k=mp−1+1, l=np−1+1
∣∣amp,np,k,l∣∣≥ 22p − 2p−1 ≥ 22p − 22p−1 = 22p−1.
38
We now have two sequences of integers
m1 < m2 < m3 < m4 · · · ,
n1 < n2 < n3 < n4 · · · ,
such that
mp−1,np−1∑k=1,l=1
∣∣amp,np,k,l∣∣ ≤ 2p−1, p > 1, (3.6)mp−1, np∑
k=1,l=np−1+1
∣∣amp,np,k,l∣∣+ mp, np−1∑k=mp−1+1,l=1
∣∣amp,np,k,l∣∣+ mp,np∑k=mp−1+1, l=np−1+1
∣∣amp,np,k,l∣∣ ≥ 22p−1. (3.7)Define a sequence (xm,n) as follows:
xm,n = sgn am1,n1,k,l, k ≤ m1, l ≤ n1,
xm,n =12sgn am2,n2,k,l
m1 < k ≤ m2, 1 ≤ l ≤ n11 ≤ k ≤ m1, m1 < l ≤ n2m1 < k ≤ m2, n1 < l ≤ n2
· · · · · · · · · · · · · · · · · · (3.8)
xm,n =1
2p−1sgn amp,np,k,l
mp−1 < k ≤ mp, 1 ≤ l ≤ np−11 ≤ k ≤ mp−1, np−1 < l ≤ npmp−1 < k ≤ mp, np−1 < l ≤ np
· · · · · · · · · · · · · · · · · ·
Here limm,n→∞
xm,n = 0. Consider
ymp,np =
mp,np∑k=1, l=1
amp,np,k,l xk,l =
mp−1,np−1∑k=1, l=1
amp,np,k,l xk,l +
mp−1,np∑k=1, l= np−1+1
am,n,k,l xk,l
+
mp,np−1∑k=mp−1+1, l=1
amp,np,k,l xk,l +
mp,np∑k=mp−1+1, l=np−1+1
amp,np,k,l xk,l.
From (3.7) and (3.8) we have∣∣∣∣∣mp−1,np−1∑k=1, l=1
amp,np,k,l xk,l∣∣ ≤ mp−1,np−1∑
k=1, l=1
∣∣amp,np,k,l∣∣∣∣∣ ≤ 2p−1,
39
mp−1,np∑k=1, l=np−1+1
amp,np,k,l xk,l +
mp,np−1∑k=mp−1+1, l=1
amp,np,k,l xk,l +
mp,np∑k=mp−1+1, l=np−1+1
amp,np,k,l xk,l
=1
2p−1
[mp−1,np∑
k=1, l=np−1+1
∣∣amp,np,k,l∣∣ + mp,np−1∑k=mp−1+1, l=1
∣∣amp,np,k,l∣∣ + mp,np∑k=mp−1+1, l=np−1+1
∣∣amp,np,k,l∣∣]
≥ 12p−1
22p−1 = 2p.
Hence
|ymp,nP | ≥ 2p − 2p−1 = 2p−1.
Thus
limp→∞
|ymp,nP | = ∞.
Since the subsequence of the (ym,n) sequence does not converge, the sequence (ym,n) has
no limit and thus condition (e) is necessary.
Proof of sufficiency. Let the limit of the convergent sequence (xm,n) be x; then
ym,n − x =m,n∑
k=1, l=1
am,n,k,l xk,l − x.
From condition (b) we may write
m,n∑k=1, l=1
am,n,k,l + rm,n = 1, (3.9)
where rm,n is any sequence such that
limm,n→∞
rm,n = 0. (3.10)
Therefore, from (3.9), we have
ym,n − x =m,n∑
k=1, l=1
am,n,k,l(xk,l − x)− rm,nx;
|ym,n − x| ≤
∣∣∣∣∣p,q∑
k=1, l=1
am,n,k,l (xk,l − x)
∣∣∣∣∣+∣∣∣∣∣
p,n∑k=1, l=q+1
am,n,k,l (xk,l − x)
∣∣∣∣∣+
∣∣∣∣∣m,q∑
k=p+1, l=1
am,n,k,l (xk,l − x)
∣∣∣∣∣ +∣∣∣∣∣
m,n∑k=p+1, l=q+1
am,n,k,l (xk,l − x)
∣∣∣∣∣ + |rm,nx|. (3.11)40
Since xm,n → x, we can choose p and q so large that for any preassigned small constant �,
|xk,l − x| <�
5A, whenever k ≥ p, l ≥ q.
Let L = max{|xk,l − x| : ∀k, l}. Now choose M and N such that whenever m ≥ M, n ≥
N, the following inequalities are satisfied:
(i)
p,q∑k=1, l=1
|am,n,k,l| <�
5pqL(from condition (a)),
(ii) |rm,n| <�
5|x|(from equation (3.10)),
(iii)m∑
k=1
|am,n,k,l| <�
5qL, l = 1, 2, · · · , q (from condition (c)),
(iv)n∑
l=1
|am,n,k,l| <�
5pL, k = 1, 2, · · · , p (from condition (d)).
Hence whenever m ≥ M, n ≥ N we have from ( 3.11) that
|ym,n − x| ≤�
5pqLpqL +
�
5pLpL +
�
5qLqL +
�
5AA +
�
5|x||x| = �. Thus
limm,n→∞
ym,n − x = 0,
or ym,n → x, which proves that the transformation is regular, and hence the theorem. �
Example 3.2.1.[7] We define a transformation of type T as follows:
am,n,k,l =1
2kn, m, n, k, l ∈ N.
We have
(a) limm,n→∞
am,n,k,l = 0,
(b) limm,n→∞
m,n∑k=1, l=1
am,n,k,l = limm,n→∞
n∑k=1
1
2k= 1,
(c) limm,n→∞
m∑k=1
am,n,k,l ≤ limm,n→∞
1
n= 0,
41
(d) limm,n→∞
n∑l=1
am,n,k,l = limm,n→∞
1
2k=
1
2k6= 0,
(e)
m,n∑k=1, l=1
|am,n,k,l| =m,n∑
k=1, l=1
am,n,k,l ≤ 1.
Before considering the regularity of a transformation of type S, we will prove certain
lemmas.
Lemma 3.2.1.[7] If am,n ≥ 0 ∀m, n,∞,∞∑
m=1, n=1
am,n diverges, it is possible to find a bounded
sequence �m,n such that
(i) �m,n ≥ 0,
(ii) limm,n→∞
�m,n = 0,
(iii)
∞,∞∑k=1, l=1
ak,l �k,l diverges.
Proof. Let Sm,n =
m,n∑k=1, l=1
ak,l. Then Sm+1,n+1 ≥ Sm,n+1 ≥ Sm,n, Sm+1,n+1 ≥ Sm+1,n ≥
Sm,n, hence limm,n→∞
Sm,n = +∞. We define
�m,n =
1Sm,n , if Sm,n 6= 0,0, if Sm,n = 0.Then
(i) �m,n is bounded and ≥ 0 ∀m, n,
(ii) limm,n→∞
�m,n = 0.
From some value of m, n onward,
�m,n ≥ �m+1,n ≥ �m+1,n+1, �m,n ≥ �m,n+1 ≥ �m+1,n+1.
For a fixed m, n, where �m,n 6= 0, we have
42
p,q∑k=m+1, l=n+1
ak,l �k,l +
m,q∑k=1, l=n+1
ak,l �k,l +
p,m∑k=m+1, l=1
ak,l �k,l
≥ �m+p, n+q
[p,q∑
k=m+1, l=n+1
ak,l +
m,q∑k=1, l=n+1
ak,l +
p,n∑k=m+1, l=1
ak,l
]= �m+p, n+q (Sm+q,n+q−Sm,n). (3.12)
In the above summation �m+p,n+q is in the smallest term except those which are zero, but
these terms drop out of the summation. Now choose p and q so large that
Sm,n ≤1
2Sm+p,n+q.
Then inequality (3.12) becomesp,q∑
k=m+1, l=n+1
ak,l �k,l +
m,q∑k=1, l=n+1
ak,l �k,l +
p,n∑k=m+1, l=1
ak,l �k,l
≥ 12�m+p,n+pSm+p,n+q =
1
2.
Setting m1 = p, n1 = q, this process can be repeated. Since this can be carried out an
infinite number of times, we have∞,∞∑
m=1, n=1
am,n �m,n = ∞.
Lemma 3.2.2.[7] If
∞,∞∑m=1, n=1
am,n is not absolutely convergent, it is possible to choose a
bounded sequence xm,n such that xm,n → 0 and∞,∞∑
m=1, n=1
am,n xm,n diverges to +∞.
Proof. Under the hypothesis,
∞,∞∑m=1, n=1
|am,n| diverges. Now choose �m,n as in the preceding
lemma with regard to the series
∞,∞∑m=1, n=1
|am,n|. We define xm,n = �m,n sgn am,n; then
am,n xm,n = am,n�m,n sgn am,n = �m,n|am,n|. By the preceding lemma∞,∞∑
m=1, n=1
am,n xm,n diverges,
as wished to prove.
We now proceed to consider transformations of type S.
43
Theorem 3.2.2.[7] In order that whenever a bounded sequence (xm,n) possesses a limit
x,
∞,∞∑k=1, l=1
am,n,k,l xk,l shall converge and limm,n→∞
∞,∞∑k=1, l=1
am,n,k,l xk,l = x, it is necessary and
sufficient that
(a) limm,n→∞
am,n,k,l = 0 for each k and l,
(b)
∞,∞∑k=1, l=1
|am,n,k,l| converges for each m and n,
(c) limm,n→∞
∞,∞∑k=1, l=1
am,n,k,l = 1,
(d) limm,n→∞
∞∑k=1
|am,n,k,l| = 0 for each l,
(e) limm,n→∞
∞∑l=1
|am,n,k,l| = 0 for each k,
(f)
∞,∞∑k=1, l=1
|am,n,k,l| ≤ A for all m and n, and for some A > 0.
Proof of necessity. (a) Define a sequence (xm,n) as follows:
xm,n =
1, if m = p, n = q,0, otherwise.Here ym,n = am,n,p,q. Hence condition (a) is necessary.
(b) Choose any fixed m and n and assume
∞,∞∑k=1, l=1
|am,n,k,l| diverges; then there exists
by the preceding Lemma 3.2.2 a bounded sequence xm,n having the limit zero, and such
that
∞,∞∑k=1, l=1
am,n,k,l xk,l diverges. This contradicts our hypothesis; hence condition (b) is
necessary.
(c) Consider the sequence (xm,n) defined as follows:
xm,n = 1, ∀ m, n ∈ N. Then ym,n =∞,∞∑
k=1, l=1
am,n,k,l; and thus condition (c) is necessary.
(d) To show the necessity of condition (d) we assume that conditions (a) and (b) are
satisfied but (d) does not hold; and obtain a contradiction. Since the double sequence
44
∞∑k=1
|am,n,k,l0| (l0 being a fixed integer) does not approach zero, then for some preassigned
small constant h > 0 there must exist a subsequence of this sequence such that every element
of it is greater than h.
Choose m1, n1 and r1 at random.
Choose m2 > m1, n2 > n1 such that
r1∑k=1
|am2,n2,k,l0| ≤h
8from (a),
∞∑k=1
|am2,n2,k,l0| ≥ h;
and r2 > r1 such that
∞∑k=r2+1
|am2,n2,k,l0| ≤h
8from (b).
In general choose mp > mp−1, np > np−1 such that
rp−1∑k=1
|amp,np,k,l0| ≤h
8from (a),
∞∑k=1
|amp,np,k,l0| ≥ h
(3.13)and rp > rp−1 such that
∞∑k= rp+1
|amP ,np,k,l0| ≤h
8, from (b). (3.14)
From (3.13) and (3.14) we haverp∑
k= rp−1+1
|amp,np,k,l0| ≥3
4h. (3.15)
Define a sequence (xk,l) as follows:
xk,l =
0, l 6= l0sgn am1,n1,k,l0 , k ≤ r1;
sgn am2,n2,k,l0 , r1 < k ≤ r2;
· · · · · · · · · · · ·
sgn amp,np,k,l0 , rp−1 < k ≤ rp;
· · · · · · · · · · · ·
(3.16)
45
Here limk,l→∞ xk,l = 0. From (3.13) and (3.14),∣∣∣∣∣rp−1∑k=1
amp,np,k,l0 xk,l0
∣∣∣∣∣ ≤rp−1∑k=1
|amp,np,k,l0| ≤h
8,
and from (3.15) and (3.16),∣∣∣∣∣rp∑
k=rp−1+1
amp,np,k,l0 xk,l0
∣∣∣∣∣ =rp∑
k=rp−1+1
|amp,np,k,l0| ≥3
4h.
Hence ∀p,
|ymp,np | =
∣∣∣∣∣∞∑
k=1
amp,np,k,l0 xk,l0
∣∣∣∣∣≥
∣∣∣∣∣rp∑
k=rp−1+1
amp,np,k,l0 xk,l0
∣∣∣∣∣−∣∣∣∣∣
rp−1∑k=1
amp,np,k,l0 xk,l0
∣∣∣∣∣−∣∣∣∣∣
∞∑k=rp+1
amp,np,k,l0 xk,l0
∣∣∣∣∣≥ 3
4h− 2h
8=
h
2.
Thus limm,n→∞ ym,n is not zero, hence (d) is necessary.
(e) In a similar manner we can show the necessity of (e).
(f) Assume conditions (a) and (b) are satisfied and that (f) does not hold. Choose any
m1, m2; r1, s1 at random.
Choose m2 > m1; n2 > n1 such that
m1,n1∑k=1,l=1
∣∣∣∣am2,n2,k,l∣∣∣∣ ≤ 2, from (a),m2,n2∑
k=1,l=1
∣∣∣∣am2,n2,k,l∣∣∣∣ ≥ 24.Choose r2 > r1; s2 > s1 such that
∞,∞∑k=r2+1, l=s2+1
∣∣∣∣am2,n2,k,l∣∣∣∣+ r2,∞∑k=1,l=s2+1
∣∣∣∣am2,n2,k,l∣∣∣∣ + ∞,s2∑k=r2+1,l=1
∣∣∣∣am2,n2,k,l∣∣∣∣ ≤ 22, from (b).Choose m3 > m2; n3 > n2 such that
m2,n2∑k=1, l=1
|am3,n3,k,l| ≤ 22,
46
∞,∞∑k=1, l=1
|am3,n3,k,l| ≥ 26.
Choose r3 > r2, s3 > s2 such that
∞,∞∑k=r3+1, l=s3+1
|am3,n3,k,l|+∞,s3∑
k=r3+1, l=1
|am3,n3,k,l|+r3,∞∑
k=1, l=s3+1
|am3,n3,k,l| ≤ 24.
In general, choose mp > mp−1, np > np−1, such that
mp−1,np−1∑k=1, l=1
|amp,np,k,l| ≤ 2p−1,
∞,∞∑k=1, l=1
|amp,np,k,l| ≥ 22p;
(3.17)and rp > rp−1, sp > sp−1 such that
∞,∞∑k=rp−1+1, l=sp−1+1
|amp,np,k,l|+∞,sp∑
k=rp−1+1, l=1
|amp,np,k,l|+rp,∞∑
k=1, l=sp−1+1
|amp,np,k,l| ≤ 22p−2. (3.18)
From these inequalities we have
rp,sp∑k=rp−1, l=sp−1+1
|amp,np,k,l|+rp−1,sp∑
r=1, l= sp−1+1
|amp,np,k,l|+sp,sp−1∑
k=rp−1+1, l= 1
|amp,np,k,l|
≥ 22p − 22p−2 − 2p−1 = 2p−1[2p+1 − 2p−1 − 1];
≥ 2p−1[2p+1 − 2p−1 − 2p−1] = 22p−1.
(3.19)
Define a sequence (xm,n) as follows:
xk,l = sgn am1,n1,k,l, k ≤ m1, l ≤ n1;
xk,l =12sgn am2,n2,k,l
1 ≤ k ≤ m1, n1 < l ≤ n2m1 < k ≤ m2, 1 ≤ l ≤ n1m1 < k ≤ m2, n1 < l ≤ n2
· · · · · · · · · · · · · · · · · · (3.20)
xk,l =1
2p−1sgn amp,np,k,l
1 ≤ k ≤ mp−1, np−1 < l ≤ npmp−1 < k ≤ mp, 1 < l ≤ np−1mp−1 < k ≤ mp, np−1 < l ≤ np
47
Here limm,n→∞
xm,n = 0. From the preceding inequalities (3.16),(3.17), (3.18) and (3.19),∣∣∣∣∣mp−1, np−1∑
k=1, l=1
amp,np,k,l xk,l
∣∣∣∣∣ ≤mp−1, np−1∑
k=1, l=1
|amp,np,k,l| ≤ 2p−1; (3.21)
mp, np∑k=mp−1+1, l=np−1+1
amp,np,k,l xk,l +
mp−1, np∑k=1, l=np−1+1
amp,np,k,l xk,l +
mp, np−1∑k=mp−1+1, l=1
amp,np,k,l xk,l
=1
2p−1
[mp, np∑
k=mp−1+1, l=np−1+1
|amp,np,k,l|+mp−1, np∑
k=1, l=np−1+1
|amp,np,k,l|+mp, np∑
k=mp−1+1, l=1
|amp,np,k,l|
]≥ 1
2p−122p−1 = 2p; (3.22)
∣∣∣∣∣∞, ∞∑
k=mp+1, l=np+1
amp,np,k,l xk,l +
mp, ∞∑k=1, l=np+1
amp,np,k,lxk,l +
∞, np∑k=mp+1, l=1
amp,np,k,l xk,l
∣∣∣∣∣≤ 1
2p
[ ∞,∞∑k=mp+1, l=np+1
|amp,np,k,l|+mp, ∞∑
k=1, l=np+1
|amp,np,k,l|+∞, np∑
k=mp+1, l=1
|amp,np,k,l|
]
≤ 12p
22p−2 = 2p−2. (3.23)
Consider
|ymp, np | =
∣∣∣∣∣∞,∞∑
k=1, l=1
amp,np,k,l xk,l
∣∣∣∣∣≥
∣∣∣∣∣mp, np∑
k=mp−1+1, l=np−1+1
amp,np,k,l xk,l +
mp−1, np∑k=1, l=np−1+1
amp,np,k,l xk,l
+
mp, np−1∑k=mp−1+1, l=1
amp,np,k,l xk,l
∣∣∣∣∣−∣∣∣∣∣
∞,∞∑k=mp+1, l=np+1
amp,np,k,l xk,l
+
mp,∞∑k=1, l=np+1
amp,np,k,l xk,l +
∞, np∑k=mp+1, l=1
amp,np,k,l xk,l
∣∣∣∣∣−
∣∣∣∣∣mp, np∑
k=1, l=1
amp,np,k,l xk,l
∣∣∣∣∣ ≥ 2p − 2p−1 − 2p−2 = 2p−2[4− 2− 1]= 2p−2, from (3.21),(3.22),(3.23).
Hence limp→∞ |ymp,np | = +∞. Thus the sequence (ym,n) has no finite limit, hence (e) is
necessary.
48
Proof of sufficiency. From the definition of ym,n we write
ym,n − x =∞,∞∑
k=1, l=1
am,n,k,l xk,l − x.
From condition (c) we have
∞,∞∑k=1, l=1
am,n,k,l + rm,n = 1,
where limm,n→∞ rm,n = 0. Hence
ym,n − x =∞,∞∑
k=1, l=1
am,n,k,l (xk,l − x) + rm,nx;
|ym,n − x| ≤
∣∣∣∣∣p,q∑
k=1, l=1
am,n,k,l xk,l
∣∣∣∣∣+∣∣∣∣∣
p,∞∑k=1, l=q+1
amp,np,k,l xk,l
∣∣∣∣∣+
∣∣∣∣∣∞,q∑
k=p+1, l=1
am,n,k,l xk,l
∣∣∣∣∣+∣∣∣∣∣
∞,∞∑k=p+1, l=q+1
amp,np,k,l xk,l
∣∣∣∣∣+ |rm,nx|.Given � > 0, we can choose p and q so large such that
|xk,l − x| ≤�
5A, when k > p, l > q.
Let L := max{|xk,l − x| : ∀ k, l}. Using conditions (a), (d), and (e) we can choose two
integers M and N such that whenever m ≥ M, n ≥ N,
(1)
p,q∑k=1, l=1
|amp,np,k,l| <�
5pqL;
(ii)∞∑l=1
|amp,nP ,k,l| <�
5pL; (k = 1, 2, 3, · · · , p);
(iii)∞∑
k=1
|amp,np,k,l| <�
5qL; (l = 1, 2, 3, · · · , q);
(iv) |rm,n| <�
5|x|.
We have, whenever m > M, n > N,
|ym,n − x| ≤�
5LpqLpq +
�
5LpLp +
�
5LqLq
49
+�
5|x||x|+ �
5AA = �.
Hence limm,n→∞
ym,n = x. Thus the theorem is proved. �
From the equation
ym,n =
∞,∞∑k=1, l=1
am,n,k,l xk,l,
we have, taking absolute values,
|ym,n| ≤
∣∣∣∣∣∞,∞∑
k=1, l=1
am,n,k,l xk,l
∣∣∣∣∣ ≤∞,∞∑
k=1, l=1
|am,n,k,l| K ≤ AK,
where |xk,l| ≤ K. Hence we have.
Corollary 3.2.2.[7] A bounded sequence (xm,n) is transformed by a regular transformation
of type S into a bounded sequence (ym,n).
Definition 3.2.6. [1] A double A−transformation will be said to be the “product” of two
simple A−transformations, A′ and A′′, whenever we have
am,n,k,l = a′m,ka
′′nl, (m,n, k, l = 1, 2, 3, · · · );
we shall then write A = A′ � A′′.
Theorem 3.2.9. [1] Let A′ and A′′ be any two regular simple A-transformations. Then the
double A-transformation A = A′ �A′′ is regular for the class of double sequences of which
each row is transformed by A′′, and each column by A′, into a bounded sequence.
Proof. It follows at once from the definition that every double A − transformation is
distributive. Moreover, since A′ and A′′ are regular and theorem 3.1.1 is consequently
satisfied by each, it is seen that any constant sequence, (xm,n) = (c), is transformed by
the double A−transformation A = A′ � A′′ into a sequence converging to c. Hence we
may, without loss of generality, confine ourselves to proving that if (xm,n) converges to
50
zero, (ym,n) also converges to zero; that is, given any � > 0, there exists positive numbers
M, N such that
|ym,n| =
∣∣∣∣∣m,n∑
k=1, l=1
am,n,k,l xk,l
∣∣∣∣∣ < �, (3.24)for m > M, n > N. Let any � > 0 be assigned. A′ and A′′ being regular, a′m,k and a
′′n,l
satisfy conditions 2,3 of theorem 3.1.1 where one K may serve for both transformations. And
since (xm,n) converges to zero, there exists positive integers p, q such that we have
|xm,n| < �/(4K2), (3.25)
for m > p, n > q. Let p, q be held fast; henceforth we restrict ourselves to values of
m > p and of n > q. Now
|ym,n| =
∣∣∣∣∣m,n∑
k=1, l=1
am,n,k,l xk,l
∣∣∣∣∣≤
∣∣∣∣∣p, q∑
k=1, l=1
am,n,k,l xk,l
∣∣∣∣∣+∣∣∣∣∣
m, n∑k=p+1, l=q+1
am,n,k,l xk,l
∣∣∣∣∣+
∣∣∣∣∣m, q∑
k=p+1, l=1
am,n,k,l xk,l
∣∣∣∣∣+∣∣∣∣∣
p, n∑k=1, l=q+1
am,n,k,l xk,l
∣∣∣∣∣. (3.26)We consider the four terms of (3.26) seriatim.
Let |xk,l| ≤ D, k = 1, 2, 3, · · · , p; l = 1, 2, 3, · · · , q.
By Theorem 3.1.1 we have
limm,n→∞
am,n,k,l = limm→∞
a′m,k limn→∞
a′′n,l = 0.
Hence there exist numbers M1, N1 such that we have, for m > M1, n > N1,
|am,n,k,l| < �/(4pqD), (k = 1, 2, · · · , p; l = 1, 2, · · · , q);
and the first term of (3.26) is at most equal to
p, q∑k=1, l=1
� |xk,l|/(4pqD) ≤ �/4 for m > M1, n > N1.
51
By (3.25) and Theorem 3.1.1, the second term of (2.26) is less than or equal to
m,n∑k=p+1, l=q+1
|am,n,k,l| �/(4K2) =�
4K2
m∑k=p+1
|a′m,n|n∑
l=q+1
|a′′n,l| ≤�
4K2.K.K = �/4.
The third term of (3.26) is less than or equal to
q∑l=1
∣∣∣∣∣m∑
k=p+1
am,n,k,l xk,l
∣∣∣∣∣ =q∑
l=1
|a′′n,l|.
∣∣∣∣∣m∑
k=p+1
a′m,k xk,l
∣∣∣∣∣≤
q∑l=1
|a′′n,l|.
[ ∣∣∣∣∣m∑
k=1
a′m,kxk,l
∣∣∣∣∣+∣∣∣∣∣
p∑k=1
a′m,k xk,l
∣∣∣∣∣]
≤q∑
l=1
|a′′n,l|.
[Bl + D
p∑k=1
|a′m,k|
],
where Bl > 0 denotes a bound for the A′-transform of the lth column of (xm,n).
Let B ≥ max{Bl : l = 1, 2, · · · , q}; then by Theorem 3.1.1, we infer the existence of numbers
M2, N2 such that we have
|a′m,k| < B/(pD), (for m > M2; k = 1, 2, · · · , p);
|a′′n,l| < �/(8qB), (for n > N2; l = 1, 2, · · · , q).
Thus the third term of (3.26) is less than �/4 for m > M2, n > N2. Similarly there exist
numbers M3, N3 such that the fourth term of (3.26) is less than �/4 for
m > M3, n > N3. Hence if M = max{M1, M2, M3}, and N = max{N1, N2, N3}, the
inequality (3.24) will be satisfied. This completes the proof.
3.3 Double Absolute Summability Factor Theorem
Suppose that A = (am,n,j,k) 3 am,n,j,k = 0 for j > m or k > n. The mn−th term of the
A−transform of a double sequence (sm,n) is defined by
Tm,n :=m∑
µ=0
n∑v=0
am,n,µ,vsµ,v.
For any double sequence um,n, ∆11 is defined by
∆11um,n = um,n − um+1,n − um,n+1 + um+1,n+1.
52
For any fourfold sequence vm,n,i,j,
∆11vm,n,i,j := vm,n,i,j − vm+1,n,i,j − vm,n+1,i,j + vm+1,n+1,i,j
∆ijvm,n,i,j := vm,n,i,j − vm,n,i+1,j − vm,n,i,j+1 + vm,n,i+1,j+1 (3.27)
∆0jvm,n,i,j := vm,n,i,j − vm,n,i,j+1
∆i0vm,n,i,j := vm,n,i,j − vm,n,i+1,j
Definition 3.3.1(The Big Oh Notation). [9]
Let f, g : N → R be functions. We say that f(n) is O(g(n)) if there is an n0 ∈ N and a c > 0
in R such that for all n ≥ n0 we have
f(n) ≤ c.g(n).
O(g(n)) can be seen as the set of all functions f(n) that are
O(g(n)) = {f(n) : ∃c, n0,∀n ≥ n0, f(n) ≤ c.g(n)}.
Then The statement f(x) is O(g(x)) as defined above is usually written as f(x) = O(g(x)).
We have the following hierarchy of big-Oh classes:
O(1) ⊂ O(√
n) ⊂ O(nk) ⊂ O(2n)
such that:
O(1) : functions bounded above by a constant.
O(√
n) :square root of n.
O(nk), for some integer k > 1 :polynomials.
There is properties for the big oh notation for example:
1. The product : f1 = O(g1) and f2 = O(g2) ⇒ f1.f2 = O(g1g2).
f.O(g) = O(fg)
2. The sum : f1 = O(g1) and f2 = O(g2) ⇒ f1 + f2 = O(g1 + g2).
3. Multiplication by a constant: let k be a positive real number then, O(kg) = O(g)
53
and f = O(g) ⇒ k.f = O(g)
Definition 3.3.2. [8] A series∑∑
bm,n, with partial sums sm,n is said to be summable
|A|k, k ≥ 1 if∞∑
m=1
∞∑n=1
(mn)k−1|∆11Tm−1,n−1|k < ∞. (3.28)
Definition 3.3.3. [8] We may associate with A = (am,n,j,k) two doubly triangular matrices
A and  as follows:
am,n,i,j =m∑
µ=i
n∑v=j
am,n,µ,v, m, n = 0, 1, 2, · · · ,
and
âm,n,i,j = ∆11am−1,n−1,i,j m,n = 0, 1, 2, 3, · · · (3.29)
Note that â0,0,0,0 = a0,0,0,0 = a0,0,0,0.
Let ym,n denote the mn−th term of the A transform of∑m
µ=0
∑nv=0 bµ,vλµv. Then we may
write
ym,n =m∑
µ=0
n∑v=0
am,n,µ,v
µ∑i=0
v∑j=0
bi,jλi,j
=m∑
i=0
n∑j=0
bi,jλi,j
m∑µ=i
n∑v=j
am,n,µ,v
=m∑
i=0
n∑j=0
bi,jλi,jam,n,i,j.
54
It then follows that
∆11ym−1,n−1 = ym−1,n−1 − ym,n−1 − ym−1,n + ym,n
=m−1∑i=0
n−1∑j=0
bi,jλi,jam−1,n−1,i,j −m∑
i=0
n−1∑j=0
bi,jλi,jam,n−1,i,j
−m−1∑i=0
n∑j=0
bi,jλi,jam−1,n,i,j +m∑
i=0
n∑j=0
bi,jλi,jam,n,i,j
=m∑
i=0
n∑j=0
bi,jλi,j âm,n,i,j −n∑
j=0
bm,jλm,jam−1,n−1,m,j
−m−1∑i=0
bi,nλi,nam−1,n−1,i,n +m∑
i=0
bi,nλi,nam,n−1,i,n +n∑
j=0
bm,nλm,jam−1,n,m,j
=m∑
i=0
n∑j=0
bi,jλi,j âm,n,i,j,
since A doubly triangle matrices, so that
am−1,n−1,m,j = am−1,n−1,i,n = am,n−1,i,n = am−1,n,m,n = 0.
But bm,n = sm−1,n−1 − sm−1,n − sm,n−1 + sm,n, so
∆11ym−1,n−1 =m∑
i=0
n∑j=0
âm,n,i,jλi,j(si−1,j−1 − si−1,j − si,j−1 + sij)
=m−1∑i=0
n−1∑j=0
âm,n,i+1,j+1λi+1,j+1si,j −m−1∑i=0
n∑j=0
âm,n,i+1,jλi+1,jsi,j
−m∑
i=0
n−1∑j=0
âm,n,i,j+1λi,j+1si,j +m∑
i=0
n∑j=0
âm,n,i,jλi,jsi,j
=m−1∑i=0
n−1∑j=0
∆ij(âm,n,i,jλi,j)si,j −m−1∑i=0
âm,n,i+1,nλi+1,nsi,n
−n−1∑j=0
âm,n,i,j+1λm,j+1sm,j +n∑
j=0
âm,n,m,jλm,jsm,j +m−1∑i=0
âm,n,i,nλi,nsi,n
=m−1∑i=0
n−1∑j=0
∆ij(âm,n,i,jλi,j)si,j +m−1∑i=0
(∆i0âm,n,i,nλi,n)si,n
+n−1∑j=0
(∆0j âm,n,m,jλm,j)sm,j + âm,n,m,nλm,nsm,n. (3.30)
55
Note that we may write
∆i0âm,n,i,nλi,n = λi,n∆i0âm,n,i,n + âm,n,i+1,n∆i0λi,n,
and
∆0j âm,n,m,jλm,j = λm,j∆0j âm,n,m,j + âm,n,m,j+1∆0jλm,j,
so that
m−1∑i=0
(∆i0âm,n,i,nλi,n)si,n +n−1∑j=0
(∆0j âm,n,m,jλm,j)sm,j =m−1∑i=0
[λi,n∆i0âm,n,i,n + âm,n,i+1,n∆i0λin]si,n
+n−1∑j=0
[λmj∆0j âm,n,m,j + âm,n,m,j+1∆0jλmj]sm,j. (3.31)
Lemma 3.3.1 (Hölder Inequality). [8] Let a1, a2, · · · , an, b1, b2, · · · , bn be nonnegative
numbers. Let p,q > 1 be real numbers with the property 1p
+ 1q
= 1. Then:
n∑j=1
ajbj ≤
(n∑
j=1
apj
) 1p(
n∑j=1
bqj
) 1q
.
Lemma 3.3.2 (Minkowski’s Inequality). [8] Let a1, a2, · · · , an, b1, b2, · · · , bn be nonneg-
ative numbers. Let p > 1 be a real number. Then:(n∑
j=1
|aj + bj|p) 1
p
≤
(n∑
j=1
|aj|p) 1
p
+
(n∑
j=1
|bj|p) 1
p
We shall need the following lemma, which is the two-dimensional formula for the first
difference of a product of two sequences.
Lemma 3.3.3. [8]
Let (ui,j), (vi,j) be two double sequences. Then
∆ij(ui,jvi,j) = vi,j∆ijui,j +(∆0jui+1,j)(∆i0vi,j)+(∆i0ui,j+1)(∆0jvi,j)+ui+1,j+1∆ijvi,j. (3.32)
Proof. Simply expand the right hand side of (3.32).
Theorem 3.3.1. [7]
Let A= (am,n,j,k) be a doubly triangular matrix with non-negative entries satisfying
56
(i) ∆11am−1,n−1,i,j ≥ 0,
(ii)∑n
v=0 am,n,i,v =∑n−1
v=0 am,n−1,i,v = b(m, i),∑mµ=0 am,n,µ,j =
∑m−1µ=0 am−1,n,µ,j = a(n, j),
(iii) mn am,n,m,n = O(1),
(iv) am,n,i,j ≥ max{am,n+1,i,j, am+1,n,i,j} for m ≥ i, n ≥ j, and i, j = 0, 1, · · · ,
(v)∑m
i=0
∑nj=0 am,n,i,j = O(1).
Let (Xm,n) be a given double sequence of positive numbers and suppose that sm,n = O(Xm,n) as
m, n →∞. If (λm,n) is a double sequence of complex numbers satisfying :
