Doubly Indexed Infinite Series

Transcription

1 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

2 Dedication To My Parents, To My Brothers, and To My Sisters.

3 Contents Acknowledgements Abstract iv v Introduction 1 1 Double Sequences Double Sequences and Their Limits Monotone Double Sequences Cauchy Double Sequences Double Series Nonnegative Double Series Summing by Curves Cauchy s Product Divergent Double Series Transformation of Double Sequences Regularity of Double Linear Transformation Double Absolute Summability Factor Theorem References 68 ii

4 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

5 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

6 Introduction We 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) n1 m1 x m,n, (ii) m1 n1 x m,n, (iii) m,n x m,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 x m,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 summability 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 several examples and we have proved the Cauchy theorem which says that you can interchange 1

7 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

8 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. lim n,m x n,m, 2. lim n (lim m x n,m ), 3. lim m (lim n x n,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 (x n,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

9 Definition [4] A double sequence of complex numbers is a function x : N N C. We shall use the notation (x(n, m)) or simply (x n,m ). We say that a double sequence (x n,m ) converges to a C and we write lim n,m x n,m a, if the following condition is satisfied: For every ɛ > 0, there exists N N(ɛ) N such that x n,m a < ɛ number n, m N. The a is called the double limit of the double sequence (x n,m ). If no such a exists, we say that the sequence (x n,m ) diverges. Definition [4] Let (x n,m ) be a double sequence of real numbers. (i) We say that (x n,m ) tends to, and we write lim n,m x n,m, if for every α R, there exists K K(α) N such that if n, m K, then x n,m > α. (ii) We say that (x n,m ) tends to, and we write lim n,m x n,m, if for every β R, there exists K K(β) N such that if n, m K, then x n,m < β. We say that (x n,m ) is properly divergent in case we have lim n,m x n,m or lim n,m x n,m. In case (x n,m ) does not converge to a R and also it does not diverge properly, then we say that (x n,m ) oscillates finitely or infinitely according as (x n,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 (a) For the double sequence x n,m 1, we have n+m lim x n,m 0. n,m To see this, given ɛ > 0, choose N N such that N > 2. Then n, m N, we have ɛ 1, 1 1, which implies that n m N x n,m 0 (b) The double sequence x n,m 1 n + m < 1 n + 1 m < 1 N + 1 N 2 N < ɛ. n is divergent. Indeed, for all sufficiently large n, m N n+m with n m, we have x n,m 1, whereas for all sufficiently large n, m N with n 2m, 2 4

10 we have x n,m 2 3. It follows that x n,m does not converge to a for any a R as n, m. (c) The double sequence x n,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 x n,m 1 n m is properly divergent to. Indeed, given β R, there exists K N such that K > β + 1. Then n, m K n, m < 2 2 β n m < β. Theorem [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 (x n,m ). Then given ɛ > 0, there exist natural numbers N 1, N 2 such that n, m N 1 x n,m a < ɛ 2 (1.1) and such that n, m N 2 x n,m a < ɛ 2. (1.2) Let N : max{n 1, N 2 }. Then for all n, m N, implications (1.1) and (1.2) yield 0 a a a x n,m + x n,m a x n,m a + x n,m a < ɛ 2 + ɛ 2 ɛ. It follows that a a 0, and hence the limit is unique whenever it exists. Definition [4] A double sequence (x n,m ) is called bounded if there exists a real number M > 0 such that x n,m M n, m N. Theorem [4] A convergent double sequence of complex numbers is bounded. Proof. Suppose x n,m a and let ɛ 1. Then there exists N N such that n, m N x n,m a < 1. 5

11 This and the triangle inequality yield that x n,m < 1 + a n, m N. Let M : max{ x 1,1, x 1,2, x 2,1,..., x N 1,N 1, a + 1}. Clearly, x n,m M n, m N. Definition For a double sequence (x n,m ), the limits are called iterated limits. lim ( lim x n,m), and lim ( lim x n,m) n m m n Outrightly there is no reason to suppose the equality of the above two iterated limits whenever they exist, as the following example shows. Example Consider the sequence x n,m m N, lim n x n,m 1 and hence n m+n of Example 1.1.1(b). Then for every lim ( lim x n,m) 1. m n While for every n N, lim m x n,m 0 and hence lim ( lim x n,m) 0. n m 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 Consider the sequence x n,m ( 1) n+m ( 1 n + 1 m ). Clearly, lim n,m x n,m 0. In fact, given ɛ > 0, choose N N such that 1 < ɛ. Then we N 2 have n, m N ( 1) n+m ( 1 n + 1 m ) 1 n + 1 m 2 N < ɛ. 6

12 But lim n (lim m x n,m ) does not exist, since does not exist, and also lim m x n,m does not exist, since lim ( lim x n,m) m n does not exist. lim n x n,m It should be noted that, in general, the existence and the values of the iterated and double limits of a double sequence (x n,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 and 1.1.3, the following examples shed some light on these cases. Example (a) For the sequence x n,m 1 + 1, note first that the double limit n m lim n,m x n,m 0. Indeed, given ɛ > 0, N N such that 1 N < ɛ 2. Then, n, m N 1 n + 1 m 1 n + 1 m 2 N < ɛ. Moreover, since lim n x n,m 1 and lim m m x n,m 1, it follows that the iterated n limits also exist and lim ( lim x n,m) lim ( lim x n,m) 0. n m m n (b) Consider the sequence x n,m ( 1) m ( ). Clearly, by an argument similar to that n m given in part (a), we have lim n,m x n,m 0. Also, the iterated limit lim ( lim x n,m) 0, m n 7

13 since ( lim n x n,m ) ( 1)m m. But the other iterated limit lim n (lim m x n,m ) does not exist, since, lim m x n,m does not exist. (c) For the sequence x n,m iterated limits exist. ( 1) n+m, it is clear that neither the double limit nor the The next result gives a necessary and sufficient condition for the existence of an iterated limit of a convergent double sequence. Theorem [4] Let lim n,m x n,m a. Then lim m (lim n x n,m ) a if and only if lim n x n,m exists for each m N. Proof. The necessity is obvious. As for sufficiency, assume lim n x n,m c m for each m N. We need to show that c m a as m. Let ɛ > 0 be given. Since x n,m a as n, m, there exists N 1 N such that n, m N 1 x n,m a < ɛ 2, and since for each m N, x n,m c m as n, there exists N 2 N such that n N 2 x n,m c m < ɛ 2. Now choose n max{n 1, N 2 }. Then m N 1, we have c m a c m x n,m + x n,m a c m x n,m + x n,m a < ɛ 2 + ɛ 2 ɛ. Hence, c m a as m. It should be noted that the following theorem similar to Theorem with an interchange of the n and m symbols. 8

14 Theorem [4] Let lim n,m x n,m a. Then lim n (lim m x n,m ) a if and only if lim m x n,m exists for each n N. Theorem [4] Let lim n,m x n,m a. Then the iterated limits exist and both are equal to a if and only if (i) lim m x n,m exists for each n N, and (ii) lim n x n,m exists for each m N. lim ( lim x n,m) and lim ( lim x n,m) n m m n Proof. By combining Theorems and 1.1.4, we obtain the result. true. The following example shows that the converse of Theorem or Theorem is not Example Consider the sequence x n,m nm. Clearly, n 2 +m 2 for each m N, lim n x n,m 0 and hence lim m (lim n x n,m ) 0. But x n,m 1 2 when n m and x n,m 2 5 when n 2m, and hence it follows that the double limit lim n,m x n,m cannot exist in this case. The next result can be viewed as a partial converse of Theorem But, first, we give some definitions and theorems. Definition [3] Suppose S R n and that for each n N, f n is a function from S to R. The sequence f n converges uniformly to the function f on S if an only if given ɛ > 0, there exists N n(ɛ) such that n > N(ɛ) f n (x) f(x) < ɛ, for all x S. For S a subset of R n, let F S be the set of bounded functions from S to R. For two functions f, g F S, define f g sup x S f(x) g(x). 9

15 Theorem [3]Let S be any subset of R n, Let (f n ) be a sequence of bounded functions from S tor, and let f be a bounded function from S to R. Then (f n ) converges uniformly to f on S if and only if for all ɛ > 0 there exists some N N(ɛ) such that n > N sup x S f(x) g(x) < ɛ. Proof. (f n ) converges uniformly to f on S if and only of for any ɛ > 0 there is some N N(ɛ) such that n > N(ɛ), x S f n (x) f(x) < ɛ/2 if and only if for n > N(ɛ), sup x S f n (x) f(x) ɛ/2 < ɛ. Theorem [4] If (x n,m ) is a double sequence such that (i) the iterated limit lim m (lim n x n,m ) a, and (ii) the limit lim n f n (m) : x n,m exists uniformly in m N, then the double limit lim n,m x n,m a. Proof. For each n N, define a function f n on N by f n (m) : x n,m m N. Then, by hypothesis (ii), f n f uniformly on N, where f(m) : lim n x n,m. So given ɛ > 0, there exists N 1 N such that n N 1 x n,m f(m) < ɛ 2 m N. Since, by hypothesis (i), lim m f(m) lim m (lim n x n,m ) a, then for the same ɛ, there exists N 2 N such that m N 2 f(m) a < ɛ 2. Now, letting N : max{n 1, N 2 }, we have n, m N x n,m a x n,m f(m) + f(m) a < ɛ 2 + ɛ 2 ɛ, 10

16 which means that lim n,m x n,m a. It should be noted that the hypothesis that lim n x n,m exists uniformly in m N in the theorem above cannot be weakened to lim n x n,m exists for every m N. Indeed, reconsider the double sequence x n,m nm n 2 +m 2 (1) (x n,m ) is bounded, since x n,m 1 n, m N. (2) For each m N, lim n x n,m 0. of Example It can be easily seen that (3) lim n x n,m 0 uniformly in m N. Indeed, if for each n N we let then we obtain f n (m) : x n,m nm n 2 + m 2, m N, f n 0 : sup m { f n (m) 0 : m N} nm sup m { 0 : m N} n 2 + m2 nm sup m { n 2 + m : m N} (put) m n n N. which implies that lim n f n 0 0. (4) lim m (lim n x n,m ) 0. (5) lim n,m x n,m does not exist, as has been shown in Example We conclude this section with the following remark: In Theorem 1.1.3, the assumption that the limit lim n x n,m exists for each m N does not follow from the assumption that the double limit lim n,m x n,m exists, as the following example shows. Example Consider the double sequence x n,m ( 1)n. Note, first, that m lim x n,m n,m lim n,m ( 1) n m 0. Indeed, given ɛ > 0, choose N N such that 1 N ( 1)n m 0 1 m 1 N < ɛ. < ɛ. Then for all n, m N, we have 11

17 ( 1) On the other hand, lim n n does not exist for each fixed m N, since lim n ( 1) n does not exist. m 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 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 Let (x n,m ) be a double sequence of real numbers. (i) If x n,m x j,k (ii) If x n,m x j,k (n, m) (j, k) in N N, we say the sequence is increasing. (n, m) (j, k) in N N, we say the sequence is decreasing. (ii) If (x n,m ) is either increasing or decreasing, then we say it is monotone. Theorem (Monotone Convergence Theorem).[4] A monotone double sequence of real numbers is convergent if and only if it is bounded. Further: (a) If (x n,m ) is increasing and bounded above, then lim ( lim x n,m) lim ( lim x m n n n,m) lim x n,m m n,m sup{x n,m : n, m N}. 12

18 (b) If (x n,m ) is decreasing and bounded below, then lim ( lim x n,m) lim ( lim x m n n n,m) lim x n,m m n,m inf{x n,m : n, m N}. Proof. It was shown in Theorem that a convergent sequence must be bounded. Conversely, let (x n,m ) be a bounded monotone sequence. Then (x n,m ) is increasing or decreasing. (a) We first treat the case that (x n,m ) is increasing and bounded above. By the supremum principle of real numbers, the supremum a : sup{x n,m : n, m N} exists. We shall show that the double and iterated limits of (x n,m ) exist and are equal to a. If ɛ > 0 is given, then a ɛ is not an upper bound for the set {x n,m : n, m N}; hence there exists natural numbers K(ɛ) and J(ɛ) such that a ɛ < x K,J. But since (x n,m ) is increasing, it follows that a ɛ < x K,J x n,m a < a + ɛ (n, m) (K, J), and hence x n,m a < ɛ (n, m) (K, J). Since ɛ > 0 was arbitrary, it follows that (x n,m ) converges to a. Next, to show that lim ( lim x n,m) lim x n,m a, m n n,m note that since (x n,m ) is bounded above, then, for each fixed m N, the single sequence {x n,m : n N} is bounded above and increasing, so, by Monotone Convergence Theorem of [3] for single sequences, we have lim x n,m sup{x n,m : n N} : l m m N. n Hence, by Theorem 1.1.3, the iterated limit lim m (lim n x n,m ) exists and lim ( lim x n,m) lim x n,m a. m n n,m 13

19 Similarly, it can be shown that lim ( lim x n,m) lim x n,m a. n m n,m (b) If (x n,m ) is decreasing and bounded below, then the sequence ( x n,m ) is increasing and bounded above. Hence, by part (a), we obtain Therefore it follows that lim ( lim x n,m) lim ( lim x m n n n,m) lim x n,m m n,m sup{ x n,m : n, m N} inf{x n,m : n, m N}. lim ( lim x n,m) lim ( lim x m n n n,m) lim x n,m m n,m inf{x n,m : n, m N}. 1.3 Cauchy Double Sequences We present in this section the important Cauchy Criterion for convergence of double sequences. Definition A double sequence (x n,m ) of complex numbers is called a Cauchy sequence if and only if for every ɛ > 0, there exists a natural number N N(ɛ) such that x p,q x n,m < ɛ p n N and q m N. Theorem (Cauchy Convergence Criterion for Double Sequences).[4] A double sequence (x n,m ) of complex numbers converges if and only if it is a Cauchy sequence. Proof. ( ) : Assume that x n,m a as n, m. Then given ɛ > 0, there exists N N such that x n,m a < ɛ 2 n, m N. Hence, p n N and q m N we have x p,q x n,m x p,q a + a x n,m x p,q a + x n,m a < ɛ 2 + ɛ 2 ɛ; 14

20 that is, (x n,m ) is a Cauchy sequence. ( ) : Assume that (x n,m ) is a Cauchy sequence, and let ɛ > 0 be given. Taking m n and writing x n,n b n, we see that there exists K N such that b p b n < ɛ p n K. Therefore, by Cauchy s Criterion for single sequences, the sequence (b n ) converges, say to a C. Hence, there exists N 1 N such that b n a < ɛ 2 n N 1. (1.3) Since (x n,m ) is a Cauchy sequence, there exists N 2 N such that x p,q b n < ɛ 2 p, q n N 2. (1.4) Let N : max{n 1, N 2 } and choose n N. Then, by (1.3) and (1.4), we have x p,q a x p,q b n + b n a < ɛ 2 + ɛ 2 ɛ p, q N. Hence, (x n,m ) converges to a. 15

21 Chapter 2 Double series A double series has terms with two subscripts instead of one. think of the terms x m,n arranged in a rectangular array, with If the terms are x m,n we x m,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 m0 ( n0 a m,n ) (add by rows) and n0 ( m0 a m,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 [5] The partial sums for the first (add by rows) series are M ( ) M R M : a m,n r m, where r m : a m,n, m0 n0 n0 n0 16

22 and the partial sums for the second (add by columns) series are N ( ) N C N : a m,n c n, where c n : a m,n. n0 m0 n0 m0 The sum r m is the sum along row m. Each row is named by the column index, here m, which remains constant in that row. Similarly, c n is the the sum along column n. We define convergence, for each way of summing, in the usual way-by requiring the appropriate (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 ( ) ( x m,n a m,n ); (2.1) m0 n0 n0 m0 i.e., can we switch the order of summation? the answer is No! and here is an example to see this: Example [5] Let a k,k 1, a k,k+1 1 for all k N 0 : N {0}, and let a m,n 0 otherwise. Then r m 0 for all m and c 0 1, while c n 0, when n > 0. Thus ( ( ) 0 r m a m,n )and a m,n c n 1. m0 m0 n0 n0 m0 n0 Thus both double sums are finite but ( ) ( ) a m,n a m,n m0 n0 n0 m0 Theorem (The Double-Series Theorem for Double Series with Non-negative Terms)[5] If p m,n 0 for every pair (m, n) N 0 N 0, then ( ) ( p m,n p m,n ). n0 m0 m0 n0 Remark: The equation in the statement of the theorem is true whether or not the double sums are finite. 17

23 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: ( ) p m,n < +. n0 m0 Now p M,n p m,n c n, for each M N 0 and n N 0, since p M,n is a term in the sum m0 p m,n. These inequalities, for each fixed m0 (now putting M m), p m,n c n and c n converges, so r m n0 M N 0, and the comparison test yield that p m,n c n < + for each m N 0. In other words, each row sum is finite (convergent) because the series summed by columns m0 is convergent. Now we have to prove that r m < (converges). Since the terms r m are nonnegative, all we have to do is to show that the partial sums R M n0 n0 M r m 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 : c n p m,n < +. (2.2) n0 n0 m0 M We actually only show that for every ɛ > 0 and for every M N 0, R M r m < M C + ɛ. But then we will know that r m < C + ɛ ɛ > 0, and hence r m < C. m0 m0 m0 First we allow M to be a given natural number, no matter how large, and we allow ɛ > 0 to be given, no matter how small. Since r m p m,n < + m N 0, we can find (for each m N 0 ) a natural N m such that ( Nm ) r m p m,n < p m,n + ɛ/(m + 1). n0 n0 This only needs to be done for 0 m M. Now we define N ɛ : max{n 0, N 1,, N M }. Then because p m,n 0, we can replace each N m by the larger quantity N ɛ : 18 n0 m0

24 r m p m,n < ( Nɛ n0 n0 m0 ) p m,n + ɛ/(m + 1), 0 m M. Therefore (We will drop the parentheses about the sum in n now) ( M M Nɛ ) R M r m < p m,n + ɛ/(m + 1) m0 n0 M N ɛ p m,n + m0 n0 M N ɛ p m,n + ɛ m0 n0 N ɛ M n0 m0 p m,n + ɛ M ɛ/(m + 1) m0 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, R M < N ɛ n0 m0 N ɛ n0 m0 N ɛ n0 M p m,n + ɛ p m,n + ɛ c n + ɛ C Nɛ + ɛ C + ɛ. Thus R M < C + ɛ, so R : ( ) p m,n m0 n0 m0 r m lim M R M C + ɛ. Since ɛ > 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

25 Recall that if (a n ) is a sequence of complex numbers, then we say that a n converges if the sequence (s n ) converges, where s n : n k1 a k. By analogy, we define a double series of complex numbers as follows. Let (a m,n ) be a double sequences of complex numbers and let s m,n : called the mn th partial sum of a m,n. m i1 n a i,j, j1 We say that the double series a m,n converges if the double sequences (s m,n ) of partial sums converges. If a m,n exists, we can ask whether or not am,n a m,n n1 m1 a m,n? (2.3) Here, with s m,n m i1 n j1 a i,j the iterated series on the right are defined as a m,n : lim m lim n s m,n and n1 m1 a m,n : lim lim s n m,n. m Theorem (Alternative Nonnegative double series theorem). [6] If a m,n converges where a m,n 0 for all m,n, then both iterated series converge, and m,n a m,n a m,n a m,n. m,n n1 m1 Moreover, given ɛ > 0 there is an N such that k > N a i,j < ɛ i1 jk and a i,j < ɛ. ik j1 Proof. Assume that the series a m,n converges and let ɛ > 0. Since a m,n converges, setting s : a m,n and s m,n : m n a i,j i1 j1 n m a i,j, (2.4) j1 i1 by definition of convergence we can choose N such that m, n > N s s m,n < ɛ/2. 20

26 Given i N, choose m i such that m > N and let n > N. Then in view of (2.4) we have n m n a ij a i,j s m,n < s + ɛ/2. j1 i1 j1 Therefore, the partial sums of j1 a i,j are bounded above by a fixed constant and hence (by the nonnegative test), for any i N, the sum j1 a i,j exists. In particular, lim s m,n lim n n m i1 n a i,j j1 m i1 a i,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 m > N s a i,j ɛ 2 < ɛ. Since ɛ > 0 was arbitrary, by definition of convergence, i1 j1 a i,j exists with sum s m,n a m,n : a m,n Similarly, using the second form of m,n show that a m,n j1 i1 a i,j. i1 i1 j1 j1 a i,j j1 s m,n in (2.4), one can use an analogous argument to k N, We now prove the last statement of our theorem. To this end, we observe that for any we have s i1 which implies that a i,j j1 i1 jk+1 ( k a i,j + i1 a i,j s j1 i1 k j1 jk+1 a i,j ) i1 k a i,j + j1 i1 jk+1 a i,j, a i,j s lim m s m,k. (2.5) By the first part of this proof, we know that s lim lim s k m,k, so taking k in (2.5), m we see that the left-hand side of (2.5) tends to zero as k, so it follows that for some N 1 N, k > N 1 21 i1 a i,j < ɛ. jk

27 A similar argument shows that there is an N 2 N such that k > N 2 a i,j < ɛ. ik j1 Setting N as the largest of N 1 and N 2 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 a m,n be an absolutely convergent series, and consider the following pictures. Figure 2.1 Summing by squares and Summing by triangles Example [6] Let S k {(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) S k a m,n the sum of those a m,n s within the k k square S k. Explicitly, a m,n (m,n) S k The sum by curves theorem implies that am,n lim k (m,n) S k a m,n lim k k a m,n. k k 22 k a m,n. (2.6)

28 As we already noted, (m,n) S k a m,n involves summing the a m,n s within a k k square; for this reason, (2.6) is referred as summing by squares. Example [6] Let S k T 1 T k, where T l {(m, n) : m + n l + 1}. Notice that T l {(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, T 3 {(1, 3), (2, 2), (3, 1)} is the third diagonal in Figure 2.1. Then (m,n) S k a m,n k a m,n l1 (m,n) T l is the sum of the a m,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) S k a m,n lim k or using that T k {(1, k), (2, k 1),, (k, 1)} we have am,n k l1 (m,n) T l a m,n, (a 1,k + a 2,k a k,1 ). (2.7) k1 We refer to (2.7) as summing by triangles. Theorem (Sum by Curves Theorem)[6]. An absolutely convergent series a m,n itself converges. Moreover, if S 1 S 2 S 3 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} S k S k+1 S k+2, (2.8) then the sequence (s k ) converges, where s k is the finite sum and furthermore, s k : (m,n) S k a m,n, am,n lim s k. 23

29 Proof. We first prove that the sequence (s k ) converges by showing that the sequence is Cauchy. Indeed, let ɛ > 0 be given. By assumption, a m,n converges, so, by Theorem 2.1.2, we can choose N N such that k > N a i,j < ɛ and a i,j < ɛ. (2.9) i1 jk ik j1 By the property (2.8) of the sets {S k } there is an N such that {1, 2,, N} {1, 2,, N} S N S N +1 S N +2 (2.10) Let k > l > N. Then, since S l S k, we have s k s l a i,j (i,j) S k (i,j) S l a i,j a i,j a i,j. (i,j) (S k \S l ) (i,j) (S k \S l ) Since l > N, by (2.10), S l contains {1, 2,, N} {1, 2,, N}. Hence, S k \S l is a subset of N {N +1, N +2, } or {N +1, N +2, } N, (to show that suppose on contrary that S k \S l 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 S k \S l 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} S l 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 s k s l a i,j a i,j < ɛ. (i,j) (S k \S l ) i1 j This shows that (s k ) is Cauchy and hence converges. We now show that a m,n converges with sum equals to lim s k. Let ɛ > 0 be given and choose N such that (2.9) holds with ɛ replaced by ɛ/2. Fix natural numbers m, n > N. By the property (2.8) and the fact that s k s : lim s k we can choose a k > N such that {1, 2,, m} {1, 2,, n} S k 24

30 and s k s < ɛ/2. Now observe that s k s m,n a i,j (i,j) S k (i,j) {1,,m} {1,,n} a i,j a i,j. (i,j) (S k \({1,,m} {1,,n})) Notice that S k \({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 ɛ replaced with ɛ/2), we have Hence, s k s m,n a i,j < a i,j < ɛ/2. (i,j) (S k \({1,,m} {1,,n})) i1 jn+1 s m,n s s m,n s k + s k s < ɛ 2 + ɛ 2 ɛ. This proves that a m,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 (Cauchy s Double Series Theorem).[6] A series a m,n is absolutely convergent if and only if a m,n < or a m,n <, n1 m1 in which case, m1,n1 a m,n a m,n n1 m1 in the sense that both iterated sums converge and are equal to the sum of the series. Proof. Assume that the sum a m,n converges absolutely. Then, by Theorem 2.1.2, we a m,n know that, m1,n1 a m,n a m,n n1 m1 a m,n. 25

31 We shall prove that the iterated sums a m,n and a m,n converge and equal n1 m1 s : a m,n, which exists by the sum by curves theorem. Let s m,n denote the partial sums of a m,n. Let ɛ > 0 be given and choose a natural number N such that m, n > N s s m,n < ɛ 2. (2.11) Since a m,n <, this implies, in particular, that for any m N, the sum n1 a m,n converges, and hence for any m N, n1 a m,n lim n s m,n converges. Thus, letting n in (2.11), we obtain m > N s lim n s m,n ɛ 2 < ɛ. But this means that s lim m lim n s m,n ; that is, s a m,n. A similar argument gives this equality with the sums reversed. Now assume that a m,n t <. We will show that a m,n is absolutely convergent; a similar proof shows that if a m,n <, n1 m1 then a m,n is absolutely convergent. Let ɛ > 0 be given. Then the fact that ( a i,j ) < i1 j1 implies, by the Cauchy criterion for single series, there is an N such that for m > N, ( ) m > n a i,j < ɛ 2. im+1 j1 26

32 Let m, n > N. Then for any k > m, we have k m k a i,j a i,j i1 j1 i1 j1 im+1 j1 Letting k shows that for all m, n > N, m t n a i,j ɛ 2 < ɛ, i1 j1 a i,j a i,j < ɛ 2. im+1 j1 which proves that a m,n converges, and completes the proof of the theorem. Example [6] Consider the sum m,n 1/(mp n q ) where p, q R. Since in this case, ( 1 m p n 1 ) q m. 1, p n q n1 it follows that ( ) ( 1 m p n ) q m p n q m1 n1 Therefore, by Cauchy s double series theorem and the p test, 1/(m p n q ) converges if and only if both p, q > 1. Example [6] Consider the sum m,n n1 1/(m 4 + n 4 ). Observe that (m 2 n 2 ) 2 0 m 4 + n 4 2m 2 n 2 0 Then by the comparison test for single series we have n N. But since by the previous example m,n too. m1 m1 1 m 4 + n 1 4 2m 2 n, 2 m1 1/(m 2 n 2 ) converges, by we see that m,n 1 m 4 + n 1 4 2m 2 n. 2 1, converges for some m 4 + n4 1/(m 4 +n 4 ) converges 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 z m+n m0 n0 z m 1. 1 z 1 (1 z ) <, 2 m0 27

33 where we used the geometric series test (twice): If r < 1, then k0 rk 1 1 r. So z m+n converges absolutely by Cauchy s double series theorem, and z m+n m0 n0 z m+n z m 1. 1 z 1 (1 z). 2 On the other hand by our sum by curves theorem, we can determine z m+n by summing m0 over curves; we shall choose to sum over triangles. Thus, if we set S k T 0 T 1 T 2 T k, where T l {(m, n) : m + n l, m, n 0}, then z m+n lim k (m,n) S k z m+n lim k k l0 (m,n) T l z m+n. Since T l {(m, n) : m + n l, m, n 0} {(0, l), (1, l 1),, (l, 0)}, we have (m,n) T l z m+n z 0+l + z 1+(l 1) + z 2+(l 2) + + z l+0 (l + 1)z l. Thus, z m+n k0 (k +1)zk. However, we already proved that z m+n 1/(1 z) 2, so 2.3 Cauchy s Product 1 (1 z) nz n 1. (2.12) 2 n1 Definition 2.3.1(Cauchy s Product).[6] Given two series n0 a n and n0 b n, their Cauchy product is the series n0 c n, where A natural question to ask is if The answer is no. c n a 0 b n + a 1 b n a n b 0 n0 n a k b n k. k0 a n and n0 b n converge, then is it true that ( )( ) a n b n n0 n0 Example 2.3.1[6] Let us consider the example ( n1 which is due to Cauchy. That is, let a 0 b 0 0 and 28 c n? n0 ( 1) n 1 n )( ( 1) n 1 n1 n ),

34 a n b n ( 1) n 1 1 n, n 1, 2, 3,. We know, by the alternating series test, that ( 1) n 1 n1 n ( n1 ( 1) n 1 )( n n1 ( 1) n 1 n ). converges, and hence converges. However, we shall see that the cauchy product does not converge. Indeed, c 0 a 0 b 0 0, c 1 a 0 b 1 + a 1 b 0 0, and for n 2, c n n a k b n k k0 n 1 k1 ( 1) k ( 1) n k k n k n 1 ( 1) n 1. k n k k1 Since for 1 k n 1, we have k(n k) (n 1)(n 1) (n 1) 2 1 n 1 1, k(n k) we see that ( 1) n c n n 1 k1 1 n 1 k(n k) k1 1 n 1 1 n 1 n 1 k Thus, the terms c n do not tend to zero as n, so by the n-th term test, the series n0 c n does not converge. The problem with this example is that the series ( 1) n 1 n1 n does not converge absolutely. theorem shows. However, for absolutely convergent series, there is no problem as the following Theorem (Mertens multiplication theorem).[6] If at least one of two convergent series a n A and b n B converges absolutely, then their Cauchy product converges with sum equals AB. 29

35 Proof. Consider the partial sums of the Cauchy product: C n c 0 + c c n a 0 b 0 + (a 0 b 1 + a 1 b 0 ) + + (a 0 b n + a 1 b n a n b 0 ) (2.13) a 0 (b b n ) + a 1 (b b n 1 ) + + a n b 0. We need to show that C n tends to AB as n. Because our notation is symmetric in A and B, we may assume that the sum a n is absolutely convergent. If A n denotes the n th partial sum of a n and B n that of b n, then from (2.13), we have C n a 0 B n + a 1 B n a n B 0. Set β k : B k B. Since B k B we have β k 0. Now we write C n a 0 (B + β n ) + a 1 (B + β n 1 ) + + a n (B + β 0 ) A n B + (a 0 β n + a 1 β n a n β 0 ) Since A n A, then A n B AB. Thus, we just need to show that the term in parenthesis tends to zero as n. To see this, let ɛ > 0 be given. Putting α a n and using that β n 0, we can choose a natural number N such that for all n > N, we have β n < ɛ. (2α) Also, since β n 0, we can choose a constant C such that β n < C for every n. Then for n > N, a 0 β n + a 1 β n a n β 0 a 0 β n + a 1 β n a n β +a n N β N + + a n β 0 a 0 β n + a 1 β n a n β + a n N β N + + a n β 0 ( ) < a 0 + a a n. ɛ ( ) 2α + a n N + + a n.c α. ɛ 2α + C( a n N + + a n ) ɛ ) ( a 2 + C n N + + a n. 30

36 Since a n <, by the Cauchy criterion for single series, we choose N > N such that Then for n > N, we see that n > N a n N + + a n < ɛ 2C. a 0 β n + a 1 β n a n β 0 < ɛ 2 + ɛ 2 ɛ. Since ɛ > 0 was arbitrary, this completes the proof. Theorem (Cauchy s Multiplication Theorem).[6] If two series a n A and b n B converge absolutely, then the double series m,n a mb n converges absolutely and has the value AB. Proof. Since ( )( ) a m b n a m b n a m b n <, m0 n0 m0 n0 m0 n0 by Cauchy s double series theorem, the double series a m b n converges absolutely, and we can iterate the sums: am b n a m b n a m ( )( ) b n a m b n AB. m0 n0 m0 n0 m0 n0 31

37 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. a 1,1 T : a 2,1 a 2,2 a 3,1 a 3,2 a 3,3 a 4,1 a 4,2 a 4,3 a 4, S : a 1,1 a 1,2 a 1,3... a 2,1 a 2,2 a 2,3... a 3,1 a 3,2 a 3,3.... One by a triangular matrix T, the other by a square matrix S. For any sequence new sequence (y n ) is defined as follows: y n n k1 a n,k x k, for the matrix T. y n k1 a n,k x k, for the matrix S,.. (x n ) a provided in the latter case y n has a meaning. If to any matrix of type T we adjoint the elements a n,k 0, k > n n N, we obtain a matrix of type S. 32

38 Definition [7] If for either transformation lim n y n exists, the limit is called the generalized value of the sequence (x n ) by the transformation. Definition [7] The transformation is said to be regular if whenever x n converges, y n converges to the same value. The criterion for regularity of these transformations is stated as follows: Theorem [7] A necessary and sufficient condition that the transformation T be regular is that: (1) lim n a n,k 0 for every k, (2) lim n n k1 a n,k 1, (3) n k1 a n,k < A for all n and some A > 0, Proof. For a proof see [7] Theorem [7] A necessary and sufficient condition that the transformation S be regular is that: (1) lim n a n,k 0 for every k, (2) k1 a n,k converges for each n, (3) k1 a n,k < A for all n and some A > 0, (4) lim n k1 a n,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 u m,n be represented as follows: u 1,1 + u 1,2 + u 1,3 + u 1,4 + u 1,5 + + u 2,1 + u 2,2 + u 2,3 + u 2,4 + u 2,5 + 33

39 + u 3,1 + u 3,2 + u 3,3 + u 3,4 + + u 4,1 + ; then the corresponding double sequence (x m,n ) of partial sums for this series is given by the following equality: Thus we have also x m,n m,n k1,l1 u k,l. (3.1) u m,n x m,n + x,n 1 x,n x m,n 1, m > 1, n > 1; u 1,n x 1,n x 1,n 1, n > 1; u m,1 x m,1 x,1, m > 1; u 1,1 x 1,1. We define a new double sequence (y m,n ) by the relation y m,n m,n k1,l1 a m,n,k,l x k,l. We shall call this transformation and its matrix A (a m,n,k,l ) a double A transformation of the type T ; here k m; l n. We may write y m,n, k1,l1 a m,n,k,l x k,l, provided y m,n has a meaning. We shall call this transformation and its matrix A (a m,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 a m,n,k,l 0, m < k, n < l, m and n, a m,n,k,l 0, 1 k m, n < l, m and n, a m,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 transformation is identical with the original one. Definition [7] If for either transformation the double sequence (y m,n ) possesses a 34

40 limit, the limit is called the generalized value of the sequence (x m,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 u 1,n 1, u 2,n 1, u m,n 0, m 3, n 1. This series converges, but the corresponding sequences (x m,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 [7] If whenever x m,n is bounded convergent sequence, y m,n converges to the same value, then the transformation is said to be regular. Definition [7] A regular transformation of real numbers (a m,n,k,l ) is said to be totally regular, provided when applied to a sequence of real numbers (x m,n ), which has the following properties, (1) x m,n is bounded for each m, (2) x m,n is bounded for each n, (3) lim m,n x m,n +, it transforms this sequence into a sequences which has its limit Regularity of Double Linear Transformation The criterion of regularity of the transformations of type T is given in the following theorem. 35

41 Theorem [7] The necessary and sufficient conditions that any transformation of type T be regular are (a) lim a m,n,k,l 0, for each k and l, m,n (b) lim m,n m,n k1,l1 a m,n,k,l 1, (c) (d) lim m,n lim m,n m a m,n,k,l 0, for each l, k1 n a m,n,k,l 0, for each k, and l1 (e) m,n k1,l1 a m,n,k,l K, for all m, n, where K is some constant. Proof of necessity. (a) Define a sequence (x m,n ) as follows: 1, if m p, n q; x m,n 0, otherwise. Then lim x m,n 0, y m,n a m,n,p,q. Hence in order that lim y m,n 0, it is necessary m,n m,n that lim a m,n,p,q 0 for each p and q. Thus condition (a) is necessary. m,n (b) Consider the sequence (x m,n ) defined as follows : x m,n 1 m, n. Then lim x m,n 1, y m,n m,n m,n k1, l1 a m,n,k,l. Since by regularity of T lim y m,n 1, condition (b) is necessary. m,n (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 fixed integer) the sequence m a m,n,k,l0 0, k1 l l 0 (some 36

42 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 m 1 and n 1 such that m 1 k1 Choose m 2 > m 1 ; n 2 > n 1 and such that m 1 k1 a m1,n 1,k,l 0 > h. a m2,n 2,k,l 0 h m 2 2, a m2,n 2,k,l 0 > h, k1 and in general choose m p > m p 1 ; n p > n p 1 and such that From (3.2) we have m p km p 1 +1 m p 1 k1 a mp,n p,k,l 0 < h 2 p 1, m p k1 a mp,n p,k,l 0 > h. (3.2) a mp,np,k,l 0 > h h ( 2 h 1 1 ). (3.3) p 1 2 p 1 Define a sequence (x m,n ) as follows : 0, if n l 0 ; sgn a m1,n 1,k,l 0, m m 1 ; (3.4) sgn a m2,n 2,k,l 0, m 1 < m m 2 ; x m,n sgn a mp,n p,k,l 0, m p 1 < m m p ; Here lim x m,n 0 since lim m,n x m,n lim m (lim n x m,n ) lim m (0) 0. For m,n this sequence (x m,n ) we have y mp,n p m p a mp,np,k,l 0 x k,l0 k1 m p 1 a mp,np,k,l 0 x k,l0 + k1 m p km p 1 +1 a mp,n p,k,l 0 x k,l0. Since m p 1 k1 it follows that a mp,n p,k,l 0 sgn a mp,n p,k,l 0 m p 1 k1 a mp,n p,k,l 0, so we have from (3.2), (3.3), and (3.4) 37

43 m p 1 a mp,np,k,l 0 x k,l0 k1 m p km p 1 +1 a mp,n p,k,l 0 x k,l0 m p 1 k1 m p 1 a mp,n p,k,l 0 amp,n p,k,l 0 k1 m p km p 1 +1 sgn a mp,n p,k,l 0 h 2 p 1, ( amp,n p,k,l 0 h p 1 ). Hence ( y mp,np h 1 1 ) h ( 2 p 1 2 h 1 1 ). p 1 2 p 2 Thus y m,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. m 1 and n 1 such that m 1,n 1 k1,l1 a m 1,n 1,k,l 1. Choose m 2 > m 1, n 2 > n 1, and such that m 1,n 1 a m 2,n 2,k,l 2, k1,l1 m 2,n 2 k1,l1 a m 2,n 2,k,l 24, and, in general, choose m p > m p 1, n p > n p 1, such that Choose m p 1,n p 1 k1,l1 From (3.5) we have a m p,n p,k,l 2p 1, m p,n p k1, l1 a m p,n p,k,l 22p. (3.5) m p 1,n p k1, ln p 1 +1 amp,n p,k,l + m p,n p 1 km p 1 +1,l1 amp,n p,k,l + m p,n p km p 1 +1, ln p 1 +1 amp,n p,k,l 2 2p 2 p 1 2 2p 2 2p 1 2 2p 1. 38

44 We now have two sequences of integers m 1 < m 2 < m 3 < m 4, n 1 < n 2 < n 3 < n 4, such that m p 1, n p k1,ln p 1 +1 amp,n p,k,l + m p 1,n p 1 amp,n p,k,l k1,l1 m p, n p 1 km p 1 +1,l1 amp,n p,k,l 2 p 1, p > 1, (3.6) + m p,n p km p 1 +1, ln p 1 +1 amp,n p,k,l 2 2p 1. (3.7) Define a sequence (x m,n ) as follows: x m,n sgn a m1,n 1,k,l, k m 1, l n 1, m 1 < k m 2, 1 l n 1 x m,n 1 2 sgn a m 2,n 2,k,l 1 k m 1, m 1 < l n 2 m 1 < k m 2, n 1 < l n 2 (3.8) x m,n 1 2 p 1 sgn a m p,n p,k,l m p 1 < k m p, 1 l n p 1 1 k m p 1, n p 1 < l n p m p 1 < k m p, n p 1 < l n p Here lim x m,n 0. Consider m,n y mp,n p m p,n p k1, l1 + a mp,n p,k,l x k,l m p,n p 1 km p 1 +1, l1 m p 1,n p 1 k1, l1 a mp,n p,k,l x k,l + a mp,n p,k,l x k,l + m p,n p km p 1 +1, ln p 1 +1 m p 1,n p k1, l n p 1 +1 a mp,n p,k,l x k,l. a m,n,k,l x k,l From (3.7) and (3.8) we have m p 1,n p 1 m p 1,n p 1 a mp,n p,k,l x k,l k1, l1 k1, l1 amp,n p,k,l 2p 1, 39

45 m p 1,n p k1, ln p 1 +1 a mp,n p,k,l x k,l + [ 1 mp 1,n p 2 p 1 k1, ln p p 1 22p 1 2 p. Hence Thus m p,n p 1 km p 1 +1, l1 amp,n p,k,l + a mp,n p,k,l x k,l + m p,n p 1 km p 1 +1, l1 amp,n p,k,l y mp,n P 2 p 2 p 1 2 p 1. lim y m p,n P. p m p,n p km p 1 +1, ln p m p,n p km p 1 +1, ln p 1 +1 a mp,n p,k,l x k,l amp,n p,k,l Since the subsequence of the (y m,n ) sequence does not converge, the sequence (y m,n ) has no limit and thus condition (e) is necessary. Proof of sufficiency. Let the limit of the convergent sequence (x m,n ) be x; then From condition (b) we may write m,n k1, l1 y m,n x where r m,n is any sequence such that m,n k1, l1 a m,n,k,l x k,l x. a m,n,k,l + r m,n 1, (3.9) ] lim r m,n 0. m,n (3.10) Therefore, from (3.9), we have y m,n x + p,q k1, l1 m,q y m,n x m,n k1, l1 a m,n,k,l (x k,l x) + a m,n,k,l (x k,l x) + kp+1, l1 a m,n,k,l (x k,l x) r m,n x; p,n k1, lq+1 m,n a m,n,k,l (x k,l x) a m,n,k,l (x k,l x) + r m,nx. (3.11) kp+1, lq+1 40

46 Since x m,n x, we can choose p and q so large that for any preassigned small constant ɛ, x k,l x < ɛ, whenever k p, l q. 5A Let L max{ x k,l x : k, l}. Now choose M and N such that whenever m M, n N, the following inequalities are satisfied: (i) p,q k1, l1 (ii) r m,n < (iii) (iv) m k1 n l1 a m,n,k,l < ɛ 5pqL ɛ 5 x (from condition (a)), (from equation (3.10)), a m,n,k,l < ɛ, l 1, 2,, q (from condition (c)), 5qL a m,n,k,l < ɛ, k 1, 2,, p (from condition (d)). 5pL Hence whenever m M, n N we have from ( 3.11) that y m,n x ɛ 5pqL pql + ɛ 5pL pl + ɛ 5qL ql + ɛ 5A A + ɛ x ɛ. Thus 5 x lim y m,n x 0, m,n or y m,n x, which proves that the transformation is regular, and hence the theorem. Example [7] We define a transformation of type T as follows: We have a m,n,k,l 1, m, n, k, l N. 2 k n (a) (b) (c) lim a m,n,k,l 0, m,n lim m,n m,n k1, l1 lim m,n k1 a m,n,k,l lim m,n n k1 m 1 a m,n,k,l lim m,n n 0, 1 2 k 1, 41

47 (d) (e) lim m,n m,n k1, l1 n a m,n,k,l l1 a m,n,k,l lim m,n m,n k1, l1 1 2 k 1 2 k 0, a m,n,k,l 1. Before considering the regularity of a transformation of type lemmas. Lemma [7] If a m,n 0 m, n, sequence ɛ m,n such that (i) ɛ m,n 0,, m1, n1 S, we will prove certain a m,n diverges, it is possible to find a bounded (ii) (iii) lim ɛ m,n 0, m,n, k1, l1 a k,l ɛ k,l diverges. Proof. Let S m,n S m,n, hence Then m,n k1, l1 a k,l. Then S m+1,n+1 S m,n+1 S m,n, S m+1,n+1 S m+1,n lim S m,n +. We define m,n ɛ m,n (i) ɛ m,n is bounded and 0 m, n, (ii) lim ɛ m,n 0. m,n From some value of m, n onward, 1 S m,n, if S m,n 0, 0, if S m,n 0. ɛ m,n ɛ m+1,n ɛ m+1,n+1, ɛ m,n ɛ m,n+1 ɛ m+1,n+1. For a fixed m, n, where ɛ m,n 0, we have 42

48 p,q km+1, ln+1 a k,l ɛ k,l + ɛ m+p, n+q [ m,q k1, ln+1 p,q km+1, ln+1 a k,l ɛ k,l + a k,l + p,m km+1, l1 m,q k1, ln+1 a k,l + a k,l ɛ k,l p,n km+1, l1 a k,l ] ɛ m+p, n+q (S m+q,n+q S m,n ). (3.12) In the above summation ɛ 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 Then inequality (3.12) becomes p,q m,q a k,l ɛ k,l + a k,l ɛ k,l + km+1, ln+1 k1, ln+1 S m,n 1 2 S m+p,n+q. p,n km+1, l1 a k,l ɛ k,l 1 2 ɛ m+p,n+ps m+p,n+q 1 2. Setting m 1 p, n 1 q, this process can be repeated. Since this can be carried out an infinite number of times, we have, Lemma [7] If, m1, n1 m1, n1 a m,n a m,n ɛ m,n. bounded sequence x m,n such that x m,n 0 and Proof. Under the hypothesis, lemma with regard to the series a m,n x m,n a m,n ɛ m,n sgn a m,n as wished to prove., m1, n1, is not absolutely convergent, it is possible to choose a m1, n1, m1, n1 a m,n x m,n diverges to +. a m,n diverges. Now choose ɛ m,n as in the preceding a m,n. We define x m,n ɛ m,n sgn a m,n ; then ɛ m,n a m,n. By the preceding lemma We now proceed to consider transformations of type S., m1, n1 a m,n x m,n diverges, 43

49 Theorem [7] In order that whenever a bounded sequence (x m,n ) possesses a limit,, x, a m,n,k,l x k,l shall converge and lim a m,n,k,l x k,l x, it is necessary and k1, l1 sufficient that (a) lim a m,n,k,l 0 for each k and l, m,n m,n k1, l1 (b), k1, l1 a m,n,k,l converges for each m and n, (c) (d) (e) lim m,n lim m,n lim m,n, k1, l1 a m,n,k,l 1, a m,n,k,l 0 for each l, k1 a m,n,k,l 0 for each k, l1 (f), k1, l1 a m,n,k,l A for all m and n, and for some A > 0. Proof of necessity. (a) Define a sequence (x m,n ) as follows: 1, if m p, n q, x m,n 0, otherwise. Here y m,n a m,n,p,q. Hence condition (a) is necessary., (b) Choose any fixed m and n and assume k1, l1 a m,n,k,l diverges; then there exists by the preceding Lemma a bounded sequence x m,n having the limit zero, and such, that a m,n,k,l x k,l diverges. This contradicts our hypothesis; hence condition (b) is k1, l1 necessary. (c) Consider the sequence (x m,n ) defined as follows:, x m,n 1, m, n N. Then y m,n a m,n,k,l ; and thus condition (c) is necessary. k1, l1 (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