Sum of $X_k$ with $\mathbb{P}(X_k=\pm 1) = 1/2\pm 1/(2\sqrt{k})$

Question

Let $\{X_k\}$ be a sequence of mutually independent random variables with \begin{align} \mathbb{P}(X_k = 1) & = \frac{1}{2} + \frac{1}{2\sqrt{k}}, \\ \mathbb{P}(X_k = -1) & = \frac{1}{2} - \frac{1}{2\sqrt{k}} \end{align} for each $k \ge 1$.

Question. What are the distributions of $\liminf_k\sum_{\ell=1}^k X_\ell$ and $\limsup_k\sum_{\ell=1}^k X_\ell$?

Intuition. It is not difficult to construct a sequence of i.i.d. random variables $\{U_k\}$ uniformly distributed on $\{-1,1\}$ and satisfy $U_k \le X_k$ for each $k\ge 1$. Note that $\sum_{\ell=1}^k U_\ell$ oscillates between $-\sqrt{2k\log\log k}$ and $\sqrt{2k\log\log k}$ following the law of the iterated logarithm, and $\mathbb{E}\left(\sum_{\ell=1}^k(X_\ell - U_\ell)\right) = \mathcal{O}(\sqrt{k})$. Would this mean that $\sum_{\ell}^k U_\ell$ is "strong" enough to dominate $\sum_{\ell=1}^k(X_\ell - U_\ell)$ and reveal the limiting behavior of $\sum_{\ell=1}^k X_\ell$?

Motivation. This question is motivated by the discussions under a MathSE question. For a sequence of independent random variables $\{X_k\}$ taking values from $\{-1, 1\}$ with $\mathbb{E}(X_k) = Ck^{-\alpha}$ ($C>0$, $\alpha >0$), the behavior of $\liminf_k\sum_{\ell=1}^k X_\ell$ and $\limsup_k\sum_{\ell=1}^k X_\ell$ are fairly clear if $\alpha < 1/2$ (where $\sum_{\ell=1}(X_\ell-U_\ell)$ "dominates") or $\alpha > 1/2$ (where $\sum_{\ell=1}U_\ell$ "dominates"), but the critical case with $\alpha = 1/2$ seems more challenging. I hope it will not turn out too trivial for MathOverflow.

Application. I am a computational/applied mathematician with limited knowledge of probability theory. This problem arises when I analyze a class of randomized algorithms, where $X_k$ indicates whether the $k$th iteration succeeds (+1) or fails (-1). The convergence of the algorithms needs $\sum_{\ell=1}^\infty X_\ell = +\infty$.

Attempts. Under the MathSE question, several attempts have been made, applying Hoeffding's inequality, the law of the iterated logarithm, and, most intriguingly, Kakutani's dichotomy theorem, yet none of them covers $\alpha = 1/2$. See MathSE for more details.

Answer 1

The answer is immediate if we apply Kolmogrov's law of the iterated logarithm, which renders $$ \limsup_{k\to \infty} \frac{\sum_{\ell=1}^k{[X_\ell-\mathbb{E}(X_\ell)]}}{\sqrt{2B_k\log\log B_k}} = 1 \quad \text{a.s.}, $$ where $$ B_k =\sum_{\ell=1}^k \mathrm{Var}(X_\ell). $$ The theorem is applicable because $\{X_k\}$ is uniformly bounded. Indeed, Kolmogrov's law needs only $B_k\to \infty$ and $\|X_k - \mathbb{E}(X_k)\|_{\infty} = o(\sqrt{B_k/\log\log B_k})$ (see also Theorems 7.1--7.3, Petrov 1995).

Since $B_k = \sum_{\ell}^k(1 - 1/\ell)$, it is clear that $(B_k\log\log B_k)/(k\log\log k) \to 1$. Hence we also have $$ \limsup_{k\to \infty} \frac{\sum_{\ell=1}^k{[X_\ell-\mathbb{E}(X_\ell)]}}{\sqrt{k\log\log k}} = 1 \quad \text{a.s.} $$ When $\sum_{\ell=1}^k \mathbb{E}(X_k) = o(\sqrt{k\log\log k})$, which is the case here, it holds that $$ \limsup_{k\to \infty} \frac{\sum_{\ell=1}^k{X_\ell}}{\sqrt{k\log\log k}} = 1 \quad \text{a.s.} $$ Thus $$ \limsup_{k\to \infty} \sum_{\ell=1}^k X_\ell = \infty \quad \text{a.s.} $$ Applying the same argument to $\{-X_k\}$, we obtain $$ \liminf_{k\to \infty} \sum_{\ell=1}^k X_\ell = -\infty \quad \text{a.s.} $$