Does $\pi$ satisfy the law of the iterated logarithm? It is widely conjectured that $\pi$ is normal in base $2$.


  But what about the law of the iterated logarithm?


Namely, if $x_n$ is the $n$th binary digit of $\pi$, does it seem likely (from computer experiments for example) that the following holds? $$\limsup_{n\rightarrow\infty} \frac{S_n }{\sqrt{n\log\log n}}=\sqrt{2}\quad\text{where}\quad S_n=2(x_1 + \ldots + x_n) - n$$


  What about other (conjectured) normal numbers like $e$ and $\sqrt{2}$?




I am sorry if this is too easy, but I tried to search for it and I could not find in on the Internet.  I suppose I could run an experiment myself, but I assumed this is well known, and I would need to brush up on my programming skills to do so...

Update 8/9/2013:

I found a website with the first 32,000 binary digits of $\pi$ and (using a spreadsheet program) graphed out the average of the bits $S_n/n$, comparing it to $\sqrt{\frac{2 \log \log n}{n}}$.  The results were inconclusive.  The average never got close to $\sqrt{\frac{2 \log \log n}{n}}$ (except at the very beginning when it was way past it). However, I had the same result with a source of randomness (the one built into the spreadsheet program).  My conclusion is that 32,000 bits is not enough to see if the law of the iterated logarithm (experimentally) holds for $\pi$. (The picture in the Wikipedia article uses at least $10^{50}$ bits, and the pattern is clear at about $10^{12}$ bits.  However, I don't know where to get even 1,000,000 binary digits of $\pi$ on the Internet.

[End Update]



Also, I am sorry that I really don't know how to properly tag this.
I favourite'd this two years ago and recently stumbled upon it again. Here's some code I wrote to answer this numerically.




The dotted line is $\sqrt{2 \log\log n/n}$
The solid line is $S_n/n$
$S_n = 2 \sum_{k=1}^n x_k - n$ where $x_k$ is the k-th decimal digit of Pi in binary.


You should be able to zoom in by running the code. You'll need matplotlib and numpy. I used y-cruncher to generate the digits.
I am not sure if this is the answer you are looking for (or if you want something more computational specifically for $\pi$), but if a number is normal in base 2, then the sum $S_n$ that you defined above will satisfy the law of the iterated logarithm.

First, let $X_i$ be the $i$th binary digit.  Assuming normality, the digits $X_i$ are iid and $X_i\sim\text{Bernoulli}(1/2)$.  Then, let $Y_i = 2X_i-1$.  Consequently, $\mathrm{EY_i=0}$, $\text{Var}(Y_i)=1$, and $S_n = Y_1+\ldots+Y_n$.  Consequently,
$$
  \mathrm{P}\left(
    \limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt{2n\log\log n}}=1
  \right)=1.
$$
This result and an accompanying proof, which relies on the result for Brownian motion and Skorokhod embedding, can be found in Section I.16 of Rogers and Williams' Diffusions, Markov Processes and Martingales: Volume 1.

In general, if a number is normal in base $b$, then the digits $X_i$ have a discrete uniform distribution on $0,1,2,\ldots,b-1$.  Hence, $\mathrm{E}X_i=\frac{1}{2}(b-1)$ and $\text{Var}(X_i) = \frac{1}{12}(b^2-1)$.  Therefore, setting 
$$
  Y_i = \frac{2\sqrt{3}\left(X_i-\frac{1}{2}(b-1)\right)}{\sqrt{b^2-1}},
$$
gives $Y_i$ zero mean and unit variance.  The sum of the $Y_i$
$$
  S_n = \sum_{i=1}^n Y_i = \frac{2\sqrt{3}}{\sqrt{b^2-1}}\sum_{i=1}^n X_i -n\sqrt{3}\sqrt{\frac{b-1}{b+1}},
$$
once again satisfies the law of the iterated logarithm.