Let $\{X_n\}$ be a sequence of independent random variables with distribution functions $\{F_n\}$. Define $S_n=\sum_j X_j$. Let $\{b_n\}$ be a given sequence of real numbers increasing to $+\infty$. Suppose that we have
(1) $\displaystyle\sum_{j=1}^n\int_{|x|>b_n}\,dF_j(x)=o(1)$,
(2) $\displaystyle\frac{1}{b_n^2}\sum_{j=1}^n\int_{|x|\leq b_n}x^2\,dF_j(x)=o(1)$;
then if we put $$a_n=\sum_{j=1}^n\int_{|x|\leq b_n}x\,dF_j(x),$$ we have $$\frac{1}{b_n}(S_n-a_n)\rightarrow0\mbox{ in probability.}$$
$\bullet$ Proof.
Define for each $n\geq1$ and $1\leq j\leq n$, $$Y_{n,j}=\begin{cases}X_j,&\mbox{if }|X_j|\leq b_n;\\ 0,&\mbox{if }|X_j|>b_n,\end{cases}$$and write $T_n=\sum_{j=1}^nY_{n,j}$. We need to show our goal through the definition of $Y_{n,j}$ and the follows:
(a) $S_n$ and $T_n$ are equivalent;
(b) $\sigma^2\left(\frac{T_n}{b_n}\right)=o(1)$ such that $\frac{T_n-\mathscr{E}(T_n)}{b_n}\rightarrow0$ in probability;
(c) $\mathscr{E}(T_n)=a_n$.
For (a), by Boole's inequality, $$\begin{array}{rl}\mathscr{P}\{T_n\neq S_n\}&=\mathscr{P}\left\{\bigcup_{j=1}^n(Y_{n,j}\neq X_j)\right\}\\ & \leq\sum_{j=1}^n\mathscr{P}\{Y_{n,j}\neq X_j\}\\ & =\sum_{j=1}^n\mathscr{P}\{|X_j|>b_n\}=\sum_{j=1}^n\int_{|x|>b_n}\,dF_j(x)=o(1)\end{array}$$due to assumption (1). Thus, we have that $S_n$ and $T_n$ are equivalent.
For (b), $$\begin{array}{rl}\sigma^2\left(\frac{T_n}{b_n}\right)
&=\sum_{j=1}^n\sigma^2\left(\frac{Y_{n,j}}{b_n}\right)\\
&\leq\sum_{j=1}^n\mathscr{E}\left[\left(\frac{Y_{n,j}}{b_n}\right)^2\right]\\
&=\frac{1}{b_n^2}\sum_{j=1}^n\int_{|x|\leq b_n}x^2\,dF_j(x)=o(1)\end{array}$$due to assumption (2). Thus we have that $\frac{T_n-\mathscr{E}(T_n)}{b_n}\rightarrow0$ in probability.
Put (a) and (b) together, then $$\frac{S_n-\mathscr{E}(T_n)}{b_n}\rightarrow0\mbox{ in probability.}$$Finally, since $$\mathscr{E}(T_n)=\sum_{j=1}^n\mathscr{E}(Y_{n,j})=\sum_{j=1}^n\int_{|x|\leq b_n}x\,dF_j(x)=a_n,$$we have $$\frac{S_n-a_n}{b_n}\rightarrow0\mbox{ in probability.}$$
$\Box$
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