Probability that sum of independent random (non iid) variables exceeds the variance infinitely often Let $X_1,X_2,\ldots$ be independent random variables, with $\Pr[X_i=\sigma_i]=\Pr[X_i=-\sigma_i]=1/2$ ($\sigma_i\geq 0$).  Suppose that:
(a) There is an upper bound $B$ such that $\sigma_i\leq B$ for all $i$. 
(b) There is a lower bound $A>0$ such that $\sigma_i>A$ for infinitely many $i$'s.
Set $S_n=\sum_{i=1}^n X_i$ and $(\Sigma_n)^2=\sum_{i=1}^n \sigma_i^2$. Then, $(\Sigma_n)^2=Var(S_n)$ (the variance). By assumption (b) $\Sigma_n$ goes to $\infty$.
Consider $0<\alpha < 1$.  I want to understand the probability of the event $S_n \geq \alpha (\Sigma_n)^2$. 
Since $\Sigma_n \rightarrow \infty$ the central limit theorem holds and we have that 
$\Pr[S_n \geq \alpha (\Sigma_n)^2]=\Pr[\frac{S_n}{\Sigma_n} \geq \alpha \Sigma_n] \approx
\frac{1}{\sqrt{2\pi}}\int_{\alpha\Sigma_n}^{\infty} e^{-x^2/2}dx $,
which goes to $0$ since $\Sigma_n$ goes to $\infty$.  So, the probability of the event with which I am concerned goes to $0$. 
Now the question is about the probability that the event happens infinitely often.  Specifically, I would like to have that
$\Pr[S_n \geq \alpha (\Sigma_n)^2\; \mbox{infinitely often}]=0 .
$
Is this true? For applying the Borel Cantelli lemma I would need to estimate the sum of the integrals, which I do not know how to do.  Any ideas? counter example?
Thanks.
 A: I think I have a proof:
Denote $n(i)$ such that $\Sigma_{n(i)}=i$. 
Denote by $A_k$ the event that there exists $i\in \{n(k^2),\ldots,n((k+1)^2\}$ for which $S_i\geq \alpha(\Sigma_i)^2$.   We shall prove that there exists $c$ for which $\Pr[A_k]\leq c/k^2$ for all $k$. 
Set $\beta=\alpha/2$. Then, for any $k$, considering $n(k^2)$,
$\Pr[ S_{n(k^2)} > \beta\cdot(\Sigma_{n(k^2)})^2 ] \approx \frac{1}{\sqrt{2\pi}}\int_{\beta\cdot k^2}^{\infty} e^{-x^2/2}dx < \frac{1}{\sqrt{2\pi}} \cdot \frac{1}{\beta\cdot k^2} =\frac{c_1}{k^2},
$
for the appropriate $c_1$.
Now consider the random variables $X_i$ for $i=n(k^2)+1,\ldots,n((k+1)^2)$.  Set $D_{j}=\sum_{i=n(k^2)+1}^{j} X_i$.  Then
$Var(D_{n((k+1)^2)})=(\Sigma_{n((k+1)^2)})^2-(\Sigma_{n(k^2)})^2=(k+1)^4-k^4 = O(k^3).
$
So, by the Kolmogorov inequality 
$
\Pr[\sup_{n(k^2)<j\leq n((k+1)^2)} D_j \geq \beta(k+1)^4 ] \leq \frac{Var(D_{n((k_1)^2)})}{(\beta(k+1)^4)^{2}}= \frac{O(k^3)}{\beta^2(k+1)^8} \ll \frac{c_2}{k^2}
$
for some constant $c_2$.
So, for any $k$, 
$
\Pr[A_k]\leq \Pr[\sup_{n(k^2)<j\leq n((k+1)^2)} S_{j} \geq \alpha (k+1)^4 ]\leq 
$
$
\Pr[S_{n(k^2)}>\beta (k+1)^4]+Pr[\sup_{n(k^2)<j\leq n((k+1)^2)} D_{j} \geq \beta(k+1)^4 ]\leq 
$
$
 \leq \frac{c_1+c_2}{k^2} .
$
So, 
$
\sum_{k=1}^{\infty}\Pr[A_k] < \infty ,
$
and the result follows by the Borel Cantelli lemma.
Is this right?
