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Assume that $X_n$ are independent (but not necessarily of the same distribution) and that $Var[X_n]>0$ for all $n$. We know that $$\frac{X_n-E[X_n]}{n}\to 0 \textrm{ almost surely as $n\to\infty$},$$ and that $E[X_n]>0$ for all $n$. We also know that $$\sum_{n=0}^\infty \frac{Var[X_n]}{n^2}<\infty.$$ How can we prove that $$\sum_{i=1}^n X_i \to \infty\text{ almost surely as $n\to\infty$?}$$ This seems to be intuitively clear, but a formal proof eludes me. But what can we say if $E[X_n] = 0$ for all $n$?

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You cannot: if $X_n = 2^{-n}$ almost surely then the expectations are also $2^{-n}$ and their sum is $1$.

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@50c such condition is usually stated as $\operatorname{Var}(X_n)>0$. Anyway, take $X_n = 2^{-n}$ or $3^{-n}$ with probabilities $\frac12$. –  Ilya May 31 '12 at 14:53
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@50c: have you tried to compute variances in the example I left you in the previous comment and see if it contradicts with your new condition? –  Ilya May 31 '12 at 15:55
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