Distribution of infinite sum of Bernoulli Let $(X_n)$ be a sequence of independent Bernoulli random variables of parameter $1/2.$
Let $Y_n=\sum_{k=1}^n\frac{X_k}{2^k}.$
We have $Y_n\overset{n\to\infty}\to\sum_{k=1}^{+\infty}\frac{X_k}{2^k},$

I would like to compute the distribution and to prove that $\vert F_n(x)-x\vert\le 1/2^n$ for all $x\in [0,1].$

We have $P_{Y_n}=\sum_{x\in Y_n(\Omega)}P(Y_n=x)\delta_x.$
If I am not mistaken we have $Y_n(\Omega):=\{\frac{k}{2^n}: k=0,1,\ldots,2^{n-1}\}.$
Now I just have to compute $P(Y_n=\frac{k}{2^n});$ if I write $Y_n=0.X_1X_2\ldots X_n$ I imagine we have $P(Y_n=\frac{k}{2^n})=P(X_1=a_1)P(X_2=a_2)\cdots P(X_n=a_n)=\frac{1}{2^n}.$ 
Therefore, $$P_{Y_n}=\frac{1}{2^n}\sum_{k=1}^{2^n-1}\delta_{k/2^n}.$$
I have no idea to prove that $\vert F_n(x)-x\vert\le 1/2^n,$ I tried wrting $x=\sum_k a_k/2^k$ but...
Question: How can I prove formally that $$P(Y_n=\frac{k}{2^n})=\frac{1}{2^n}\; \mbox{and}\; \vert F_n(x)-x\vert\le 1/2^n ?$$
 A: It holds: $$Y_n=\sum_{k=1}^n\frac{X_k}{2^k} = Y_n= \frac{1}{2^n}\sum_{k=1}^n2^{n-k}X_k$$
So it holds $$0 \le Y_n \le \frac{1}{2^n}\sum_{k=1}^n2^{n-k} = \frac{2^n - 1}{2^n}$$
And we have $Y_n = k$ with $k\in \{0,\ldots, 2^n - 1\}$ (what's slightly different than what you already had)
And because each representation of $k$ is unique as $\sum_{k=1}^n2^{n-k}X_k$ we have $$P(Y_n = k) = \frac{1}{2^n}$$
Now $F_n(x)$ is piecewise constant with jumps of height $\frac{1}{2^n}$ at $\frac{k}{2^n}$ and the max-difference of $F_n(x)$ and $x$ is reached at the jumppoints. Hence $|F_n(x) - x|$ is equal or less the jump-height of $\frac{1}{2^n}$
Or if you want to have it written as formula: We have
$$F_n(x) = \frac{k+1}{2^n} \text{ for } x\in \left[\frac{k}{2^n},\frac{k+1}{2^n}\right)$$
And we have for $x\in \left[\frac{k}{2^n},\frac{k+1}{2^n}\right)$ that $$|F_n(x) - x| \le \frac{k+1}{2^n} - \frac{k}{2^n} = \frac{1}{2^n}$$ 
Hence: $$|F_n(x) - x| \le \sup_{k} \left\{\sup_{x\in \left[\frac{k}{2^n},\frac{k+1}{2^n}\right)}|F_n(x) - x|\right\} = \frac{1}{2^n}$$
A: I'm not sure if this formal enough for you, but if you think of $\frac{X_k}{2^k}$ as the $k-$ bit in a binary expansion, you can see that $Y_n=\sum_{k=1}^n\frac{X_k}{2^k}$ can be regarded as as real number in $[0,1)$ where each one of the fractional bits (up to bit $n$) is uniformly distributed (the rest are zero).
Now, the event $Y_n=\frac{k}{2^n}$ corresponds to one (and only one) tuple $(X_1, X_2 \cdots X_n)$ (the one that gives the binary expansion of integer $k$). Then, because all $2^n$ tuples are equiprobable, $P(Y_n=\frac{k}{2^n}) = 2^{-n}$.
And, again, thinking of $x\in[0,1)$ in terms of its binary expansion, and calling $\tilde x$ its binary truncation up to $n$ fractional binary digits, we get $$F_n(x) = P(Y_n \le x) =P(Y_n \le \tilde x) = \frac{1+\tilde x 2^n}{2^n}=\frac{1}{2^n} +\tilde x$$
Hence 
$$F_n(x) -x = \frac{1}{2^n} - (x-\tilde x)$$
But $0\le (x-\tilde x)<\frac{1}{2^n}$, then ... (you should be able to go on from here).
