TITLE: Series of independent Bernoulli variables
QUESTION [7 upvotes]: Let $X_1, X_2, \ldots$ be independent, identically distributed random variables with distribution $\text{Ber}(\frac{1}{2})$. Define the random varible:
$$Y:=\sum_{n=1}^\infty \frac{X_n}{2^n}$$
Prove that $Y$ is uniformly distributed over the unit interval $[0,1]$.

The only way I currently know how to attack this problem is by using Levy's theorem:
For $Y_n:=\sum_{k=1}^n \frac{X_k}{2^k}$, it's enough to prove that the characteristic function of $Y_n$ converges to the characteristic function of a uniform random variable, i.e.:
$$\lim_{n\to\infty}\mathbb{E}[e^{itY_n}]=\frac{e^{it}-1}{it}$$
Since $X_1, X_2, \ldots$ are independent, we have:
$$\mathbb{E}[e^{itY_n}]=\prod_{k=1}^n \mathbb{E} \left[e^{it\frac{X_k}{2^k}} \right] = \prod_{k=1}^n \frac{1+e^{it/2^k}}{2}.$$
I'm basically stuck because I don't know what to do with the product as $n\to\infty$.
Is there a trick to it? Is there maybe a simpler solution? Thanks in advance!

REPLY [4 votes]: Picking up where you left off:
\begin{align}
\mathbb{E}[e^{itY_n}]=\prod_{k=1}^n \mathbb{E} \left[e^{it\frac{X_k}{2^k}} \right] &= \prod_{k=1}^n \frac{1+e^{it/2^k}}{2} \\
&= \frac{1}{2^n} \prod_{k=1}^n (e^{it/2^k}+1) \\
&= \frac{1}{2^n} (e^{it/2^{n}}+1) \prod_{k=1}^{n-1} (e^{it/2^k}+1) \\
&= \frac{1}{2^n} (e^{it/2^{n}}+1) \frac{(e^{it/2^{n}}-1)}{(e^{it/2^{n}}-1)} \prod_{k=1}^{n-1} (e^{it/2^k}+1) \\
&= \frac{1}{2^n (e^{it/2^n}-1)} (e^{it/2^{n-1}}-1) \prod_{k=1}^{n-1}  (e^{it/2^k}+1)\\
&= \frac{1}{2^n (e^{it/2^n}-1)} (e^{it/2^{n-1}}-1) (e^{it/2^{n-1}}+1) \prod_{k=1}^{n-2}  (e^{it/2^k}+1)\\
&= \frac{1}{2^n (e^{it/2^n}-1)} (e^{it/2^{n-2}}-1) \prod_{k=1}^{n-2}  (e^{it/2^k}+1)\\
& \, \vdots \\
&= \frac{1}{2^n (e^{it/2^n}-1)} (e^{it/2^{n-n}}-1)\\ \\
&= \frac{(e^{it}-1)}{2^n (e^{it/2^n}-1)} .
\end{align}
And, the Taylor expansion of $e^{it/2^n}$ is $1+\frac{1}{2^n}it+O(t^2)$, where $O(t^2)$ is such that $2n \cdot O(t^2) \to 0$, as $n \to \infty$. Therefore, $\mathbb{E}[e^{itY_n}]$ becomes:
\begin{align}
\mathbb{E}[e^{itY_n}] &= \frac{(e^{it}-1)}{2^n (1+2^{-n}it+O(t^2)-1)} \\
&= \frac{(e^{it}-1)}{(it+2^n O(t^2))} \\
& \to \frac{(e^{it}-1)}{it}, \quad \mbox{as } n \to \infty. \quad \blacksquare
\end{align}