Let $\{X_n\}_{n\geq 1}$ be a sequence of independent random variables such that $E(X_n) = 0$, $E(X_n^2) = \sigma_n^2$, and let $s_n^2 = \sum_{k = 1}^n \sigma_k^2 \to \infty$. Show that for any $a > 1/2$, $Y_n = s_n^{-2}(\log{s_n^2})^{-a} \sum_{k = 1}^n X_k$ converges to zero almost surely.
I know that this is equivalent to showing that $\sum_{n = 1}^\infty P(|Y_n| > \epsilon) < \infty$, and this sum can be estimated via Markov's inequality, but I am unable to get nice bounds.
I have also tried using results by Kronecker and Levy to see that it is enough to show that $\sum_{n = 1}^\infty \operatorname{var}(Z_n) < \infty$, where $Z_n = X_n s_n^{-2}(\log{s_n}^2)^{-a}$, but this doesn't work either.
Use your notations. For $ a>\frac12$, denote \begin{equation*} Z_n=\frac{X_n}{s_n(\log s_n^2)^{a}} \end{equation*} then $\mathsf{E}Z_n=0 $ and \begin{align*} \mathsf{Var}Z_n&=\frac{\mathsf{Var}X_n}{s_n^2(\log s_n^2)^{2a}} =\frac{s_n^2-s_{n-1}^2}{s_n^2(\log s_n^2)^{2a}} \le \int_{s_{n-1}^2}^{s_n^2}\frac{\mathrm{d} x}{x(\log x)^{2a}}\\ &=\frac{1}{(2a-1)(\log s_{n-1}^2)^{2a-1}}-\frac1{(2a-1)(\log s_{n}^2)^{2a-1}}.\quad (s_0=0) \end{align*} Hence \begin{equation*} \sum_n\mathsf{Var}Z_n < \infty,\quad\text{and}\quad \sum_n \frac{X_n}{s_n(\log s_n^2)^{a}} \quad \text{converge a.s.}. \end{equation*} Furthermore, by Kronecker Lemma, \begin{equation} \frac{\sum_{k=1}^n X_k}{s_n(\log s_n^2)^{a}}\stackrel{\text{a.s.}}\longrightarrow 0. \end{equation}