<http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm>
  
  
  
  

.$\quad$1. Can the independence assumption be weakened, similar to [this](http://en.wikipedia.org/wiki/Central_limit_theorem#CLT_under_weak_dependence)?
  
  
  

.$\quad$2. Can the identically distributed assumption be dropped/weakened, in the latter case similar to [this](http://en.wikipedia.org/wiki/Central_limit_theorem#Lyapunov_CLT)?
  
  
  

.$\quad$3. Can the result be fine-tuned, presumably to something of the form
  
  

$\displaystyle\limsup\_{n\to \infty} \frac{\frac{S\_n}{\sqrt{2\cdot n\cdot \operatorname{log}(\operatorname{log}(n))}}-1}{f(n)} \; \; $ some\_relation\_symbol $\;$ some\_constant $\qquad$ almost surely $\qquad \; \; $ ?



Can the Law of the Iterated Logarithm be strengthened?
For your third question, see a [paper of Erdös](http://www.ams.org/mathscinet-getitem?mr=6630), where he proves an even more precise result (at least, in the special case $S\_n=\sum\_{i=1}^nY\_i$ where $Y\_i$ (independent) are $\pm 1$ with probability $\frac 12$), namely that for $\delta>0$, the following holds with probability one:
$$S\_n>\left(\frac{2n}{\log\log n}\right)^{1/2}(\log\log n+\frac 34\log\log\log n+\frac 12\log\log\log\log n+\cdots+(\frac 12-\delta)\log^{(k)}n)\qquad\text{for infinitely many $n$}$$
and
$$S\_n>\left(\frac{2n}{\log\log n}\right)^{1/2}(\log\log n+\frac 34\log\log\log n+\frac 12\log\log\log\log n+\cdots+(\frac 12+\delta)\log^{(k)}n)\qquad\text{for only finitely many $n$}$$
In particular, this implies that:
$$\limsup\_{n\to \infty} \frac{\frac{S\_n}{\sqrt{2n\log\log n}}-1}{\frac{\log\log\log n}{\log\log n}}=\frac 34$$


