Probability of $ \sum_{n=1}^\infty \frac{x_n}{2^n} \leq p$ for Bernoulli sequence Probability of $ \sum_{n=1}^\infty \frac{x_n}{2^n} \leq p$  for a  Bernoulli(1/2) sequence $(x_n)$ and $p \in [0,1]$. I know that the answer should be just $p$ but how can you prove it?
Say for $p= 1/2$ I can look at it as 1-( Probability of being greater than 1/2) which is 1/2 since if our first term in the sequence is 1 then we're done. But how one shows for general $p$?

Your knowledge about the answer suggest that the the random variable $Y=\sum_{n=1}^\infty X_n/2^n$ has a uniform distribution on the interval $(0,1)$ and by the means of moment generating functions is possible to prove so:
 for every $t\in\mathbb{R}$, we have
$$
e^{tY}\leq e^{t\wedge0}
$$
Since, clearly, $0\leq Y\leq 1$.
By independence, it follows from the dominated convergence theorem that
$$
\mathbb{E}[e^{tY}]=\prod_{n=1}^\infty\mathbb{E}[\exp({\frac{t}{2^n}X_n})]
$$
Note that the factors in the right hand side are the MGF of Bernoulli(1/2) random variables evaluated at $t/2^n$, so they are equal to $2^{-1}(\exp(2^{-n}t)+1)$. Therefore, we need only to find the product
$$
\prod_{n=1}^\infty 2^{-1}(\exp(2^{-n}t)+1)
$$
After expanding the first partial products, it is not hard to see that the expression for the partial products is given by
$$
\prod_{n=1}^m2^{-1}(\exp(2^{-n}t)+1)=2^{-m}\sum_{n=0}^{2^m-1} \exp(n2^{-m}t)
$$
(This can be proven by induction). The latter is a sum of powers which equals to $(e^t-1)/(e^{2^{-m}t}-1)$. 
We have then
$$
\prod_{n=1}^\infty 2^{-1}(\exp(2^{-n}t)+1)=(e^t-1)\lim\limits_{m\to\infty}\frac{2^{-m}}{e^{2^{-m}t}-1}
$$
And using L'Hôpital's rule it's easy to see that the limit in the right hand side is $1/t$. We therefore get
$$
\mathbb{E}[e^{tY}]=\frac{e^t-1}{t}
$$
Which is the moment generating function of a Uniform(0,1) distribution. By uniqueness of the MGF, $Y$ has such distribution and it follows that, $\mathbb{P}[Y\leq p]=p$ for every $p\in [0,1]$.
