TRANSLATIVITY OF ROGOSlNSKI SUMMABILITY METHODS OF DIFFERENT ORDERS

The paper deals with the problem of translatlvlty of Rogoslnskl summability methods (R, r),, of orders h, r; 0 < h-r < I. It has been shown by the author (i) that

paper we prove that when h-r e |/2,1], (R h is translatlve, and when ,r h-r e (0, I/2)' (Rh,r) is neither translatlve to the right nor to the left.These results establish both, the open problem and its conjecture which have been given by the author [I].

I. INTRODUCTION.
The series a 0 + a +... of real or complex terms with its partial sums S a 0 + a +...+ a is said to be summable by the Rogoslnskl method n n (R h );0 < h-r < I, if t +t, as n+% where ,r n n t n [ cos(k+r)Snak, k=0 and (1.1) In the special case in which r 0, thls method will reduce to the Rogoslnskl- Bernsteln method (R h) which has been the subject of many papers (see Ai-Madl [I], Agnew [2] and Petersen [3]). A. K. AL-MADI For a summability method A T (translat[ve to the left) if, whenever .an is summable A to s, then so is .an-''1 A Tr if the converse holds.A e T if, and only if A Tiff T r Much work has already been done on translative summability methods (see AI-Madi [1,4,5], Chowdhury [6], and Kuttner [7][8][9][10].
The author [I] proved that if h e I/2 ,i], (Rh, 0) e T, and if 0 < h < I/2(--I), then (Rh, O) T U Tr.The author left the problem unsettled for the rest of the interval (0, I/2 ).He conjectured that when h (0, I/2 ), then (Rh, O) T U Tr. then (R h e T and when h-r (0, I/2 ), then (R h T U T The second result ,r Dr r establishes both the open problem and its conjecture which have been given in [I].
The following result will be used in the proof of Theorems 3.1 and 3.2.We ill show that when h-r E (1_ I], then (Rh,r) and (C,I) are equivalent, and the result follows from the fact that (C,I) it follows form (I.I) that tn =k=O An' kMk' (3.8) n,k Using the same technique as Agnew [2; p. 544-545], we see that if h-r is in (I/3 I], (Rh,r) is equivalent to (C,I).This completes the proof.
PROOF OF THEOREM 3.2.
To prove the result, it is enough to consider some translative summability method, and to show that this method is equivalent to (R h in the case in which h-r =I For this, we consider the sequence-to- sequence method Q given by the transformation n n+l Qn 2n---l Mn-I + Mn' (3.9 where Mn is given by (1.3).Using (1.3) and (3.9) to obtain Qn in terms of Qn' the result is 2n+____1_ Qn+l 2n+3 Qn" (3.10) This implies that Q E T.
Next put h-r =I and write A given in (3 7)  It is easily seen that the transformation given by (3.12) is regular.Hence applying Theorem 3.4 to the matrix given in (3.12), one can easily show that the transformation given by (3.12) is equivalent to convergence, and (Rh,r) is equivalent to Q.This completes the proof.(3.25) We will show that if 0 < h-r <I/2 and 0 < z <I/4, then for sufficiently large n z and n 1/2m j n F is not bounded.This implies that the transformation n,n-j given in (3.21) is not regular, and consequently, when h-r E (0, I ), (Rh, r) Tr. z To prove this, we will show that for sufficiently large n and 2 n the terms of the sum (3.25) alternate in sign, and then we will show tIat the limit of F is n,n-j unbounded.
The inversion formula o[ (3.17 n-u,n-_ wU-J Next, we will show that when  Observe that for any x; (x # -I), we have where B (j+h-r-l)k(x) -xk'(x), (with x instead of w), (3.44) k(x) x + x 2-x3+...+ (-l)J-2x j-2.Hence, when h-r e (0, I/2); that is when w > I, it follows from (3.47)   Using (3.I8) and (3.47), it follows from (3.52) that when 0 < h-r < 1/2, H is not n n-j Z bounded for all sufficiently large n and and n I'z n This implies that the transformation given by (3.48) is not regular; that is when h-r e (0, I/2 (R h T This completes the proof of leorem 3. ,r given by (3.1) is equivalent to convergence.PROOF OF THEOREM 3.1.
it follows from (3.43) that for all sufficiently large n and 2 ] l)J-lwJ-I[1-h+rl[l-2h+2r] ,k and Dn,k are given in the proof of the first part.Using the identity given by (3.23), we have from (3.24) and (3.4q) that