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Suppose $X_i$ are iid with mean $0$ and variance $1$. Is there such a sequence of random variables that satisfying $$ \frac{\sum_{i=1}^n X_i}{n} \stackrel{a.s.}{\rightarrow}0, $$ but $$ \frac{\sum_{i=1}^n X_i}{\sqrt{n} \log n} \stackrel{a.s.}{\nrightarrow}0? $$

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  • $\begingroup$ What have you tried? Do you think there is such a sequence? $\endgroup$ Oct 5 '20 at 0:25
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It is well known (kolmogorov theorem) that if $X_k$ are independent random variables with variances $Var(X_k)$ (we don't assume equal distribution) and $b_k$ is a sequence divergent to infinity (of positive integers), then if

$$ \sum_{k=1}^\infty \frac{Var(X_k)}{b_k^2} < \infty$$ it holds that $$ \frac{1}{b_n} \sum_{k=1}^n X_k \to 0$$ almost surely.

In your example $b_n = \sqrt{n}\ln(n)$ and $Var(X_k)=1$, hence $$ \sum_{k=2}^\infty \frac{Var(X_k)}{b_k^2} = \sum_{k=2}^\infty \frac{1}{k\ln^2(k)} < \infty $$ so it must holds that $$\frac{1}{\ln(n)\sqrt{n}}\sum_{k=1}^n X_k \to 0$$ almost surely

(Obviously first condition is satisfied (even without assumption about variances) by Strong law of large numbers)

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