Iterated logarithm and gaussian concentration : a paradox Let $G_1, \dots, G_n$ be iid random variables, such that $G_1 \sim \mathcal N(0,1)$
Let $$S_n = \sum_{i=1}^n G_i\quad \text{and} \quad\tilde{S}_n = \frac{1}{\sqrt{2n\log\log n}}S_n$$
It is easy to see that $\tilde{S}_n \sim N(0,\sigma_n)$, where $\sigma_n = \frac{1}{\sqrt{2\log\log n}}$.
By the iterated logarithm law, $$\limsup_{n \rightarrow \infty} \ \tilde{S}_n = 1 \quad \text{a.s} $$
So what is troubling for me, is that I cannot prove by direct means that if $(Y_n)_{n \in \mathbb N}$ are indepedent random variables such that $Y_n \sim \mathcal N(0,\sigma_n)$ then it holds that
$$\limsup_{n \rightarrow \infty} \ Y_n = 1 \quad \text{a.s} $$
For instance, only trying to prove that :  a.s, $Y_n  > 1+\epsilon$ occurs only a finite number of times :
Using Borell-Cantelli we want to prove that $$\sum_{n} \mathbb P(Y_n>1+\epsilon) <\infty $$
However, Gaussian concentration only gives : $$\mathbb P(Y_n>x) = P(\sigma_nG>x) \leq e^{-\frac{x^2}{2\sigma_n^2}} $$
One can check it is far from summable. So where do we loose information ? Is the inequality too loose ? 
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In fact, $\limsup_{n \to \infty} Y_n = \infty\ne1$ a.s. More specifically, 
\begin{equation}
 \limsup_{n \to \infty} \frac{Y_n}{\sqrt{\ln n/\ln\ln n}} = 1
\end{equation}
a.s. or, equivalently,
\begin{equation}
 \limsup_{n \to \infty} \frac{Z_n}{\sqrt{2\ln n}} = 1 \tag{*}
\end{equation}
a.s., where $Z_n:=Y_n/\sigma_n\sim N(0,1)$. 
Indeed, $\P(Z_1>z)=\exp\{-z^2/(2+o(1))\}$ as $z\to\infty$. So, for any $\ep\in(0,1)$,
\begin{equation}
 \sum_n\P(Z_n>\sqrt{(2+\ep)\ln n})= \sum_n\exp\{-(2+\ep)\ln n/(2+o(1))\}<\infty
\end{equation}
and 
\begin{equation}
 \sum_n\P(Z_n>\sqrt{(2-\ep)\ln n})= \sum_n\exp\{-(2-\ep)\ln n/(2+o(1))\}=\infty. 
\end{equation}
So, (*) follows by the Borel--Cantelli lemma. 
(There is no paradox here, since your $Y_n$'s are independent, and the $\tilde{S}_n$'s are rather strongly dependent.) 
