A Kind of Complete Moment Convergence for Sums of Independent and Nonidentically Distributed Random Variables

and Applied Analysis 3 Obviously, it follows from Lemma 4 that 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 P( 󵄨󵄨󵄨Sn 󵄨󵄨󵄨 B n ≥ ε√n) − P (|N| ≥ ε√n) 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 = 2Δ n (ε√n) ≤ 2AB −2−δ n (1 + 󵄨󵄨󵄨ε√n 󵄨󵄨󵄨 2+δ ) −1 n

Recently, Liu and Lin [15] have introduced a new kind of complete moment convergence and obtained the following result.
Theorem A (see [15]).Suppose that {,   ,  ≥ 1} is a sequence of independent and identically distributed (i.i.d.) random variables.Then holds if and only if However, the condition of identical distribution is very strong and rather difficult to verify in some real cases.The following theorem gives a sufficient condition of complete moment convergence for independent nonidentically distributed random variables.Theorem 1.Let {,   ,  ≥ 1} be independent random variables such that   = 0 and [ which verifies that {,   ,  ≥ 1} satisfies the Lyapunov condition (3).
Remark 3. Suppose that {,   ,  ≥ 1} is a sequence of independent and identically distributed (i.i.d.) random variables such that where  > 0 is a constant.Then, from Remark 2, we know that it satisfies Lyapunov's condition.Therefore, by Theorem 1, we have lim Obviously, this case is the result of Liu and Lin [15].Therefore, our condition of Theorem 1 is different from the conditions of Theorem A, and our result partly extends and improves those given in Liu and Lin [15].

Proof of Theorem 1
In this section, we will prove Theorem 1.We first present the following two lemmas, which play a key role in the proof of Theorem 1.
Lemma 4 (see [17]).Suppose that {,   ,  ≥ 1} are independent random variables with   = 0 and [ To prove Theorem 1, we only need to study  1 and  2 .We will divide the proof into two steps.
Step 1.We first prove the equality as follows: In fact, it follows from Proposition 2.1.
By (12), we obtain lim To establish the equality (11), from (13) it follows from Toeplitz's lemma (page 120 of [20]) that lim On the other hand, it follows from Lemma 5 that Noting that the inequality (20) yields By ( 22), we obtain lim Combining ( 19) and ( 23), we see that the equality ( 11) is satisfied.
Step 2. Next, we need to prove the following equality: Obviously, it follows from Proposition 3.1 of [15] that To establish (24), from (25) we only need to prove Hence, by ( 27) and ( 28 Noting the fact that the weighted average of a sequence that converge to 0 also converges to 0, we have which implies that (24) is satisfied.
Therefore, from (11) and (34), we see that (4) is true.This completes the proof of Theorem 1.