Let$\{X_n\}_{n\ge1}$ be an arbitrary sequence of random variables. Show that the series $\sum\limits_{n\ge1}a_nX_n$ converges absolutely a.s. for some constants $a_n\ne0$.
My attempt at a solution:
For any $n$ there exists a $b_n\in\mathbb{R}$ such that $P\big(\big|\frac{X_n}{b_n}\big|>\frac{1}{2^n}\big)<\frac{1}{2^n}$ \begin{align*} &\implies \sum_{n\ge1}P\bigg(\bigg|\frac{X_n}{b_n}\bigg|>\frac{1}{2^n}\bigg)\le\sum_{n\ge1}\frac{1}{2^n}<\infty\\ &\implies P\bigg(\bigg|\frac{X_n}{b_n}\bigg|>\frac{1}{2^n}\quad\text{i.o.}\bigg)=0\\ &\implies P\bigg(\bigg|\frac{X_n}{b_n}\bigg|\le\frac{1}{2^n}\quad\text{eventually}\bigg)=1\\ &\implies\exists\,\,N\in\mathbb{N}\,\,\text{such that}\,\,\bigg|\frac{X_n}{b_n}\bigg|\le\frac{1}{2^n}\,\,\text{a.s. for all}\,\,n\ge N\\ \end{align*} Then, \begin{align*} \sum_{n\ge1}\bigg|\frac{X_n}{b_n}\bigg|<\infty&\iff\sum_{n\ge N}\bigg|\frac{X_n}{b_n}\bigg|<\infty\\ &\iff\sum_{n\ge N}\frac{1}{2^n}<\infty,\,\,\text{which is true.} \end{align*} And so the result is shown with $a_n=\frac{1}{b_n}$
Does this solution work? Also, is there a simpler or more elegant way to show this?