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I am trying to show giving a sequence $(X_n) \subseteq L^2$ of independent random variables with zero mean and $\sum_{n\in N}E[X_n^2]< \infty$ , $\sum_{n\in N}X_n$ converges a.s.

For this I wanna get $\sum_{n\in N} \mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)<\infty$. As if I can show it , then by Borel-Cantelli lemma, $\sum_{n\in N}X_n$ converges to zero almost surely.

But I dont know how to get $\sum_{n\in N} \mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)<\infty$,

I tried this:

By Chebychev's inequality, $\mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)\leq \mathbb{E}[X_n^2]/\epsilon^2 $, follows that $\sum_{n\in N}\mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)\leq\sum_{n\in N}\mathbb{E}[X_n^2]/\epsilon^2$. However, letting $\epsilon\longrightarrow 0$ doesn't make the right side bounded. Seems doesnt work...

I am wondering if anyone can show the correct way to do it. Thanks in advance.

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    I don't understand: if you can show what you said after the "I can get", then almost surely, $\sum_nX_n=0$. But it doesn't need to be the case.2012-08-13
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    I need some help to prove that $\sum_{n\in N} \mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)<\infty$2012-08-13
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    @lindamac This inequality is false. That's why you can't prove it.2012-08-13

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