Convergence of Sum of Random variable to another - Cantor function Let $(X_{n})_{n\geq1}$ be i.i.d. Ber$\left(\frac{1}{2}\right)$. I want to show that $$\sum_{{n\geq1}}\frac{2X_{n}}{3^{n}}$$ converges almost surely to a random variable $X$, without saying that this random variable distribution is Cantor-Distribution. I guess we should use Borel–Cantelli somehow - but I am not sure about it. Can you please help?
Thanks.
 A: Note that the $X_n$ are bounded by $1$, so the absolute convergence follows from the convergence of the sum $\sum \limits_{n = 1}^\infty \frac{2}{3^n}$.
A: Formally, there is a probability space $Ω$ (with a $σ$-algebra $\mathcal F$ and a probability measure $P$) and the random variables $X_n$ map $Ω$ to $\mathbb R$ (with $\mathcal B(\mathbb R)$) as follows
\begin{align}X_n:Ω&\mapsto \mathbb R\\[0.2cm]ω &\to \{0,1\} \end{align} with $P(\{ω:X_n(ω)=1\})=P(\{ω:X_n(ω)=1\})=1/2$. Now, take an $ω \in Ω$ and see what happens to the series $$\sum_{n\ge 1}\frac{2}{3^n}X_n(ω)$$ By definition $X_n(ω)=0$ or $=1$ and therefore certainly $X_n(ω)\le 1$. Hence, $$\sum_{n\ge 1}\frac{2}{3^n}X_n(ω)\le \sum_{n\ge 1}\frac{2}{3^n}=2\cdot\frac{1}{1-\frac13}=3$$ and therefore by comparison test, the series converges. This is true for any $ω\in Ω$, therefore in particular almost surely. (Note: we did not calculate the limit of the series, we just showed that is converges for almost all (actually for all) $ω\in Ω$).
