ASYMPTOTIC EQUIVALENCE OF SEQUENCES AND SUMMABILITY

For a sequence-to-sequence transformation A , let R m A x = ∑ n ≥ m | ( A x ) n | and μ m A x = sup n ≥ m | ( A x ) n | . The purpose of this paper is to study the relationship between the 
asymptotic equivalence of two sequences ( lim n x n / y n = 1 ) and the variations of asymptotic 
equivalence based on the ratios R m A x / R m A y and μ m A x / μ m A y .

(x) and y (y) be infinite sequences, and let A be a sequence-to-sequence transformation.We write x y if lin z,.,/y,., 1.In order to compare rates of convergence of sequences, in [2] Pobyvanets introduced the concept of asymptotically regular matrices, which preserve the asymptotic equivalence of two nonnegative sequences, that is z y implies Ax Ay.Furthermore, in [1] Fridy introduced new ways to compare rates by using the ratios R.x/R.y, ,z/tmy when they tend to zero.In [2] Marouf studied the relationship of these ratios when they have limit one.In the present study we investigate some further properties involved with the ratios such iAx/IAy, RAx/RAy when they have limit one.

NOTATIONS AND BASIC THEOREMS.
For a summability transformation A, we use DA to denote the domain of A: DA {x" , a,kzk converges for such n >_ 0} k=0 and CA to denote the summability field: CA {x x 6 DA, _, ankx converges.}r=O k=O Also Ps={x'x,6>O forall n} and P {x'x,,>O forall n.}For a sequence x in 1 or e , we also define Rmx ,>, Ix,[ and /mx sup.>.[xn[ for We list the following results without proof.THEOREM 1. (Pobyvanets [2]).A nonnegative matrix A is asymptotically regular if and only if for each fixed interger m, lim,_.ooa,,/,__o a,kO.THEOREM 2. A matrix A is a Co-Co matrix (i.e.A preserves zero limits) if and only if (a) lim_oo ak 0 for k 0, 1,2,....
(b) There exists a number M > 0 such that for each n =0 lal < M.
This is a contradiction of tAx Ay.THEOREM 4. Suppose A is a nonnegative matrix; then x py implies Ax #Ay for any bounded sequences x, y E P, for some > 0, if and only if A satisfies the following three conditions: (i) (aki)=o is a bounded sequence dominated by some B; 3---0 (ii) For any j 0,1,2,...  Before we prove this theorem, we shall give some exaznples of A which satisfy the above conditions (i), (ii), and (iii).PROOF OF THEOREM 4. First, assume that for any bounded sequences z,y E P,, for some 6 > 0, /x /y implies iAx #Ay; we wish to prove that A satisfies the conditions (i), (ii) and (iii).Take x V (1,1 ); then z,y are bounded, z,y e Pa, and z y; so IAx IAy.But (tAx),, sups>, .,"=oa,.. Hence, (=oa:.i)=oshould be bounded.This proves (i).To prove (ii) suppose there is a j such that li-" sups,>,, a A for some A > 0. As in the proof of Theorem 3, take > 0 and define y (1,1 and f if l+t if n=j. Then x,y .Pa,z,ya re bounded, and/x kY; so we have pAx pAy.But 8UPk>n i=0 akiYi sup>(tai + Eio a) supk.0 > li"-sups>, aki nsuPk>n i=0 aki tA-1.
Finally, we are going to prove (iii).For any given infinite sequence jl < j2 < we define z and y by and for every n, f 2, if n j, for u 1,2,..., 1, otherwise.
It is easy to see that x, y are bounded, x,y E P and #z #y.This implies #Az #Ay.

SUPk_> =0 akj
On the other hand, it is clear that Hm sup e a < 1.