TITLE: Can the Law of the Iterated Logarithm be strengthened?
QUESTION [12 upvotes]: http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm

.$\quad$1.  Can the independence assumption be weakened, similar to this?

.$\quad$2.  Can the identically distributed assumption be dropped/weakened, in the latter case similar to this?

.$\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 \;  \;  $ ?

REPLY [6 votes]: For your third question, see a paper of Erdös, 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:
$$\limsup_{n\to \infty} \frac{\frac{S_n}{\sqrt{2n\log\log n}}-1}{\frac{\log\log\log n}{\log\log n}}=\frac 34$$