Let $X_{n}$ for $n\ge 1$ be i.i.d. Rademacher ($\pm 1$) random variables. For what $p$ does
$$\frac{1}{n^{p}}\sum_{k=1}^{n}X_{k}$$
converge almost surely and for what $p$ does it diverge almost surely? It follows from the $3$-series theorem and Kronecker's lemma that for $p > \frac{1}{2}$, it converges almost surely to $0$.
Best Answer
The law of the iterated logarithm states that \begin{align} \limsup_n\frac{S_n}{\sqrt{n}\log\log n}\to1 \text{ a.s.},\\ \liminf_n\frac{S_n}{\sqrt{n}\log\log n}\to-1 \text{ a.s.}, \end{align} where $S_n=\sum_1^nX_k$. In words, for each $\omega$ in a set of probability 1, the sample path $S_n(\omega)/(\sqrt{n}\log\log n)$ is arbitrary close to $1$ infinitely often and to $-1$ i.o. as $n\to\infty$. Therefore, $n^\epsilon\log\log n\to\infty$ for $\epsilon\ge 0$ implies that \begin{align} \limsup_n (n^\epsilon\log\log n) S_n(\omega)/(\sqrt{n}\log\log n)=\limsup_nS_n(\omega)/n^{1/2-\epsilon}=\infty\\ \liminf_n (n^\epsilon\log\log n) S_n(\omega)/(\sqrt{n}\log\log n)=\liminf_n S_n(\omega)/n^{1/2-\epsilon}=-\infty. \end{align} So for $p\le 1/2$ the process fails to converge almost surely.