By M. Mursaleen, S.A. Mohiuddine

This book's committed specialize in the 'almost' convergence and statistical convergence of double sequences demystifies the idea that utilizing a number of unveiling examples, highlighting the appliance of double sequences in key parts of natural and utilized arithmetic.

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J,k The existence of the limit BP- lim p,q α(j, k, p, q, s, t)xj k j,k = aj k zj k + BP- lim p,q j,k α(j, k, p, q, s, t) x k − ξ + BP- lim + BP- lim p,q α(j, k, p, q, s, t)(xj − ξ ) j,k p,q j,k α(j, k, p, q, s, t)ξ j,k then would follow if the limits on the right side exist. The third limit α(j, k, p, q, s, t)ξ = ξ v BP- lim p,q j,k exists by (iii). We will show that the first limit equals N ∈ N be such that ε |xj − ξ | ≤ 4M j vj (xj − ξ ). To this end, let ε > 0 and for j ≥ N. By (v) we can choose S ∈ N such that for p, q ≥ S and any s, t ∈ N, N −1 α(j, k, p, q, s, t) − vj ≤ j =1 k ε 4 x ∞ Then α(j, k, p, q, s, t) − vj (xj − ξ ) j k N −1 ≤2 α(j, k, p, q, s, t) − vj x j =1 k ∞ .

16) For the expression in the second bracket, we have m n−1 1 (m + 1)n = = = 1 n 1 n 1 n p=0 q=0 n−1 n−1 m−1 n−1 τpqst p=0 q=0 m m−1 τpqst − (m + 1) p=0 n−1 m−1 τpqst p=0 1 (m + 1)τmqst − m(m + 1) q=0 n−1 τmqst q=0 1 − m(m + 1)n n−1 τmqst − q=0 τpqst p=0 1 mτmqst − m(m + 1) q=0 1 mn 1 mn 1 m m(m + 1) q=0 1 = mn = τpqst − m τpqst p=0 m n−1 τpqst p=0 q=0 1 dm,n−1,s,t . 15), we get dmnst − dm−1,n,s,t − dm,n−1,s,t + dm−1,n−1,s,t = = 1 m(n + 1) n τmqst − q=0 1 n mn(n + 1) 1 mn n−1 τmqst − q=0 n n−1 τmqst − (n + 1) q=0 1 1 dmnst + dm,n−1,s,t m m τmqst − q=0 1 (dmnst − dm,n−1,s,t ) m 32 2 = = = 1 nτmnst − mn(n + 1) n−1 1 (dmnst − dm,n−1,s,t ) m τmqst − q=0 1 (n + 1)τmnst − mn(n + 1) 1 1 τmnst − mn (n + 1) Almost Convergence of Double Sequences n τmqst − q=0 n τmqst − q=0 1 (dmnst − dm,n−1,s,t ) m 1 (dmnst − dm,n−1,s,t ).

17) with the last two inequalities provides yMN − s ≤ 7 + |s| + C if min(M, N ) ≥ max(M1 , M2 , M3 , M4 , M5 ). Since > 0 is arbitrary, this proves the convergence of the A-means ymn to s as M and N tend to ∞, that is, the A-summability of x to the same limit. 2(b) and the boundedness of x. ) Let A = (amnj k ) be a strongly regular matrix. 2(b). 4) is not satisfied. Then there exists > 0 such that, for infinitely many pairs of M and N both tending to ∞, we have ∞ ∞ | 10 aMN mn | ≥ 12 . 18) j =0 k=0 or ∞ ∞ | 10 aM,N,2j +1,k | ≥ 6 .