COMPLETE CONVERGENCE FOR SUMS OF ARRAYS OF RANDOM ELEMENTS

Let {Xni} be an array of rowwise independent B-valued random elements and {an} constants such that 0 < an ↑ ∞. Under some moment conditions for the array, it is shown that ∑n i=1Xni/an converges to 0 completely if and only if ∑n i=1Xni/an converges to 0 in probability.


Introduction.
Let (B, ) be a real separable Banach space.A separable Banach space B is said to be of type r , 1 ≤ r ≤ 2, if there exists a constants C r such that for all independent B-valued random elements X 1 ,...,X n with mean zero and finite r th moments.
A sequence {X n , n ≥ 1} of B-valued random elements is said to converge completely to zero if for each > 0, (1.2) Note that complete convergence implies almost surely by the Borel-Cantelli lemma.Now let {X n , n ≥ 1} be a sequence of independent random variables.Let ψ(t) be a positive, even and continuous function such that Chung [3] strong law of large numbers (SLLN) states that if Recently, Hu and Taylor [6] proved Chung type SLLN for arrays of rowwise independent random variables.More specifically, let {X ni , 1 ≤ i ≤ n, n ≥ 1} be an array of rowwise independent random variables and let {a n , n ≥ 1} be a sequence of real numbers with 0 < a n ↑ ∞.Let ψ(t) be a positive, even and continuous function such that for some integer p ≥ 2. Furthermore, assume that where k is a positive integer.Then the conditions (1.6), (1.7), (1.8), and (1.9) imply Many classical theorems hold for B-valued random elements under the assumption that the weak law of large numbers (WLLN) holds (see, Kuelbs and Zinn [8], de Acosta [4], Choi and Sung [1,2], Wang, Rao and Yang [10], Kuczmaszewska and Szynal [7], and Sung [9]).
In this paper, we apply de Acosta [4] inequality to obtain Hu and Taylor's [6] result in a general Banach space under the assumption that WLLN holds.

Main Result.
To prove our main theorem, we need the following lemma which is due to de Acosta [4].
Lemma 2.1.For each p ≥ 1, there exists a positive constant C p such that for separable Banach space B and any finite sequence for some s > 0. Then the following statements are equivalent. ( It follows that The proof will be completed by showing that From (i) and (2.6), we have (2.9) Thus, to prove (2.8), it is enough to show that First consider the case of 1 ≤ p ≤ 2. From Markov's inequality and Lemma 2.1(i), we have ↓ for each q ≥ p.Let q = max{p, 2s}.Then we have by Markov's inequality and Lemma 2.1(ii) that (2.12) Since q ≥ p, ψ(|t|)/|t| p ↓ implies ψ(|t|)/|t| q ↓, and so since q ≥ 2s and (2.5).Combining (2.12), (2.13), and (2.14) yields (2.10).Thus (i) ⇒(ii) is proved.Since the implications (ii) ⇒(iii) and (iii) ⇒(iv) are obvious, it remains to show that (iv) ⇒(i).
Assume that (iv) holds.From Lemma 2.1(i) and (2.5) which entails It follows by (iv) that E n i=1 X ni /a n → 0, and so (i) holds.Thus the proof of Theorem 2.2 is completed.
The following theorem states that Theorem 2.2 holds even if the condition (2.5) is replaced by (2.17) for some 1 ≤ r ≤ 2 and s > 0.
Proof.Let {Y ni } and {Z ni } be as in the proof of Theorem 2.2.From the proof of (i) ⇒(ii) in Theorem 2.2, we have [5]) Thus {Y ni } satisfies the conditions of Theorem 2.2, and so (2.19) holds by Theorem 2.2.Let {X ni , 1 ≤ i ≤ n, n ≥ 1} be an array of rowwise independent B-valued random elements and {a n , n ≥ 1} constants such that 0 < a n ↑ ∞.Assume that EX ni = 0 and B is of type r (1 ≤ r ≤ 2).Then(2.4)and(2.17)implythatSinceB is of type r and EX ni = 0, it follows by (2.17) that The condition (2.3) is weaker than (1.6).Hu and Chung[5]proved Corollary 2.4 under the stronger condition (1.6).