Let $\{a_n\}$ be a sequence of complex numbers and let $\{A_n\}$ be a non-descreasing sequence of positive numbers, tending to infinity. We assume that $\sup\limits_{n\ge 1} {\dfrac{\sum\nolimits_{k=1}^{n} {|a_k|}}{A_n}}<\infty$.
Let $\{X_n\}$ be a sequence of integrable, centered, independent and identically distributed random variables. I want to prove that $\frac{1}{A_n} \sum\nolimits_{k=1}^{n} {\mathbb{E}[a_k X_k \mathbf{1}_{\{|X_k|\le A_k/|a_k|\}}]}\xrightarrow{n\to \infty}0.$
In the proof, author wrote: Since $A_n/|a_n|\to \infty$ we have $\mathbb{E}[X_n \mathbf{1}_{\{|X_n|\le A_n/|a_n|\}}]\to 0$. (I understood)
For every $n\ge 1$, put $\alpha_{n,k}=a_k/A_n$ for $1\le k \le n$ and $\alpha_{n,k}=0$ for $k>n$. By assumption we have $\sup\limits_{n\ge 1} {\dfrac{\sum\nolimits_{k=1}^{n} {|a_k|}}{A_n}}<\infty$, so we obtain that $\sup\limits_{n\ge 1} {{\sum\nolimits_{k=1}^{\infty} {|\alpha_{n,k}|}}}<\infty$. (I don't understand).
Now by usual summability arguments we obtain that $\frac{1}{A_n} \sum\nolimits_{k=1}^{n} {\mathbb{E}[a_k X_k \mathbf{1}_{\{|X_k|\le A_k/|a_k|\}}]}\xrightarrow{n\to \infty}0.$ (don't understand).
Can you help me explain the parts I don't understand??