Concentration inequality for the law of iterated logarithm The following question arose in one of my research projects. Before stating it, let me give a short background. We all know the law of iterated logarithm. It states that if $X_1,\ldots,X_n$ are i.i.d. mean $0$, variance $1$ random variables and if $S_n := X_1 + \ldots + X_n$, then almost surely,
$$\limsup_{n\rightarrow \infty} \frac{\pm S_n}{\sqrt{2n\log \log n}}=1~.$$
My question is as follows:
Is there a finite sample exponential concentration inequality for the quantity $\left|S_n/\sqrt{2n\log\log n}\right|?$ That is, suppose that $t > 1$ is fixed. Then can we bound the probability something like:$$\mathbb{P}\left(\left|\frac{S_n}{\sqrt{2n\log \log n}}\right|> t\right) \leq e^{-n^\alpha}$$ for some $\alpha > 0$? 
Any help will be greatly appreciated.
 A: As was noted in the comments by Yuval and Kevin, even if $X_1$ is bounded, the best upper bound on the probability in question is a negative power of $\ln n$. To get such a bound (and even an asymptotics), it is actually enough to assume that $E|X_1|^k<\infty$ for some $k>2$. Indeed, a theorem due to S. Nagaev states this: 

Suppose that $X_1,X_2,\dots$ are zero-mean unit-variance iid random variables, with $S_n:=\sum_1^n X_i$. Let $Z\sim N(0,1)$. Take any real $k>2$. Then the condition $E|X_1|^k<\infty$ is sufficient for the asymptotic relation $P(S_n\ge z\sqrt n)\sim P(Z\ge z)$ (as $n\to\infty$) to hold in the zone $0\le z\le\sqrt{(\frac k2-1)\ln n}$ and necessary for this relation to hold in the zone $0\le z\le\sqrt{(k+1)\ln n}$. 

So, assuming that indeed $E|X_1|^k<\infty$ for some $k>2$, and letting $z=t\sqrt{2\ln\ln n}$, we see that 
$$P\Big(\Big|\frac{S_n}{\sqrt{2n\ln\ln n}}\Big|> t\Big)
\sim P(Z\ge z)\sim\frac1{z\sqrt{2\pi}}e^{-z^2/2}
=\frac1{2t\sqrt{\pi\ln\ln n}}(\ln n)^{-t^2}  
$$
for each $t>0$ as $n\to\infty$. 
