Intuition of law of iterated logarithm? Let $X_i$ be iid random variables with $EX_i = 0$ and $Var X_i=1$ and $S_n=X_1+\cdots+X_n$. Then the law of the iterated logarithm says almost everywhere we have
$$\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} = \sqrt{2}$$
On the other hand the central limit theorem says
$$\frac{S_n}{\sqrt{n}} \to N(0,1)$$
Can anyone explain why dividing by an extra $\sqrt{\log{\log{n}}}$ should go from giving $N(0,1)$ to something bounded by the constant $\sqrt{2}$?
To try to understand I considered the simple case when each $X_n$ is $N(0,1)$ so that $S_n/\sqrt{n}$ is also normally distributed as $N(0,1)$. Then $S_n/\sqrt{n\log{\log{n}}}$ is distributed as $N(0,1/\log{\log{n}})$. Then it would seem to me that to even have just $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}$ requires either
$$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right) < \infty$$
or if 
$$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right) = \infty$$
then to achieve $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}$ the sets {$ \omega : \frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}$} cannot for example cover the probability space over and over infinitely forever. I don't know the value of $\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right)$ but since it is the sum of the probability of the tail ends of a bunch of normal distributions you would expect there to be no closed form even for partial sums.
In the other direction for $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}}$ to not have a value lower than $\sqrt{2}$ isn't it necessary that something like the following holds
$$\sum_{n=3}^\infty P\left(\sqrt{2}-\epsilon < \frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}\right) = \infty$$
Can anyone explain why this number $\sqrt{2}$ should pop up?
I already asked the above on math.stackexchange (link) but apparently moving it here was impossible hence the duplicate post.
 A: One nice fact that helps intuition is that
the $\log\log n$ factor "disappears on a sparse subsequence".
To be precise, let's first look at the following equivalent statement of the Law of the iterated logarithm (Khintchine 1924):

Let $X=(X_{0},X_{1},\ldots)$ be a random variable on $\{0,1\}^{\mathbb N}$ having the fair-coin distribution.
          Let $S_{n}=\sum_{k=0}^{n-1}X_{k}$. Then with probability one,
          $$
   \limsup_{n\to\infty}\frac{S_{n}-\frac{n}{2}}{\varphi(n)\sqrt{n}}=1,
  $$
          where $\varphi(n)=\sqrt{\frac{1}{2}\log\log n}$.

Now, the question is, since the standard deviation of $S_{n}$ is simply $\sqrt{n}$, why is there that strange $\varphi(n)$?
Michel Weber (Law of the iterated logarithm for subsequences, 1990) gave the following answer: we can replace $\varphi$ by an arbitrarily slow-growing function if we replace $\limsup_{n\rightarrow\infty}$ by $\limsup_{n\in N}$ for a sufficiently sparse set $N\subseteq\mathbb N$. In detail:

Let $N=\{\nu_{1}<\nu_{2}<\cdots\}\subseteq\mathbb N$ and let
              $\{Y_{n}\}$ be an i.i.d. sequence with $\mathbb E(Y_{n})=0$ and $\mathbb E(Y_{n}^{2})=1$. Let $S_{n}=Y_{1}+\cdots+Y_{n}$.
              Let 
              $$
    p_{n}=|\{m\le n: N\cap (2^{m-1},2^{m}]\ne\varnothing\}|,
   $$
              $$
    \mathcal L(k)=\ln p_{n}\quad\text{if}\quad k\in (2^{n-1},2^{n}].
   $$
              Then we have
              $$
    \limsup_{j\to\infty}\frac{S_{\nu_{j}}}{\sqrt{2\nu_{j}\mathcal L(\nu_{j})}}=1
    \quad\text{a.s.}
   $$

For $N=\mathbb N$ we get the usual law of the iterated logarithm. 
For sparse sets $N$, the function $\mathcal L(\nu_{j})$ is an arbitrarily slow-growing function, 
so the dominator is standard deviation ($\sqrt{\nu_{j}}$) times a small factor.
What is happening is that we are looking at the finite sums $S_n$ only for $n\in N$ where $N$ is sparse. But $S_n$ is still $\sum_{k=0}^{n-1}X_k$ with no restriction of $k$ being in $N$. So the computation of $S_n$ is running uninterrupted but we are only opening our eyes to inspect it rather rarely.
A: Since the main part of the density function of Normal distribution is exp(-x^2/2). So we need add the 2 to cancel out the denominator in the exponential part of the density function of Normal distribution.
