The Berry-Esseen theorem provides a second-order approximation to the central limit theorem (itself a first order approximation.) Higher order approximations are available, see here. If the random variables are Bernouilli($p$), better approximations are available, this is a topic of active research. Before stating my question, Let me mention a recent result, which is the topic of a PhD thesis (the full thesis available here):
It seems that my question (below) is just a particular case of this result, not sure. Probably it is, but maybe I got it wrong.
Question
The proportion $q_n$ of binary digits of $\sqrt{2}$ (or any other normal number for that matter, say $\pi$ or $\ln 2$), among the first $n$ digits, satisfies
$$\sqrt{n}\cdot\Big|q_n-\frac{1}{2}\Big|<\frac{1}{\sqrt{2\pi}}\approx 0.398942 \mbox{ as } n\rightarrow\infty.$$
This assumes $p=\frac{1}{2}$. Below is the chart showing $\sqrt{n}\cdot|q_n-\frac{1}{2}|$ up to $n = 1,000,000$.
I don't expect an answer for $\sqrt{2}$ nor for any other normal number. All I want to known is, if the digits were i.i.d Bernouilli($\frac{1}{2}$), would this be correct? See related questions on Math.StackExchange, here and here.
The goal is to try to prove a much weaker result for $\sqrt{2}$ hoping it would be easier than proving the result stated in my question, maybe something like $\log \log n \cdot |q_n - \frac{1}{2}| = o(1)$. That would be enough to prove that the binary digits of $\sqrt{2}$ are zero 50% of the time.
