# Counterexample for Almost Surely Convergence

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 $$\frac{\sum_{i=1}^n X_i}{\sqrt{n} \log n} \stackrel{a.s.}{\nrightarrow}0?$$

• What have you tried? Do you think there is such a sequence? Oct 5 '20 at 0:25

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