A problem related to basic martingale theory In our probability theory class, we are supposed to solve the following problem:

Let $X_n$, $n \geq 1 $ be a sequence of independent random variables such that 
  $ \mathbb{E}[X_n] = 0, \mathbb{Var}(X_n)  = \sigma_n^2 < + \infty $ and 
  $ | X_n | \leq K, $ for some constant $ 0 \leq K < + \infty,  \ \forall n \geq 1$. 
Use martingale methods to show that 
  $$ \sum \limits_{n = 1}^{\infty} \ X_n \ \mbox{ converges }  \mathbb P-\mbox{a.s.}  \  \Longrightarrow  \ \sum\limits_{n = 1}^{\infty} \ \sigma_n^2 < + \infty .$$

Could anybody give me a hint? 
Thanks a lot for your help!
Regards, Si

Set $S_n:=\sum_{i=1}^n X_i$ and $S_0=0$. $S$ is a martingale (wrt the natural filtration), so $S^2$ is a sub-martingale.
Using the Doob's decomposition we can write $S^2=M+A$ where M is a martingale and A is predictable (increasing) process, both null at 0.
It turns out that $A_n=\sum_{i=1}^n \sigma_i^2$.
Define the stopping time $T_\alpha$ as $T_\alpha=\inf\{n \geq0 \ : \ |S_n|>\alpha\}$.
By OST we have $$E[S_{T_\alpha \wedge n}^2]-E[A_{T_\alpha \wedge n}]=0;\qquad \forall n.$$
Also $|S_{T_\alpha \wedge n}| \leq \alpha+K$, since $|X_i| \leq K $; then
$$E[A_{T_\alpha \wedge n}] \leq (\alpha+K)^2;\qquad \forall n.$$
Since $\sum X_i$ converges a.s., the partial sum $S_n$ are a.s. bounded, then for some $\alpha$ we have that $(T_{\alpha}=\infty)$ has positive probability. So applying this fact to the last inequality, we get $A_{\infty}=\sum\sigma_n^2 < \infty$.
P.S.: I hope it is clear; I am a new user and I want to say hi to everyone.
Also I noticed that I can not comment in some post; is there any restriction?
Thanks.
