Let $W_n = \sum_{i = 1}^{n}X_i$ be a random walk on $\mathbb{R}$, where the increments $X_i$ are i.i.d., symmetric around the origin ($X\sim -X$), such that $-1\leq |X(\omega)| \leq 1$ $\forall\omega\in\Omega$, $\mathbb{E}(X) = 0$ and $Var(X) = \mathbb{E}(X^2) = \sigma^2$. I would like to bound from above the following probability as a function of $k$: $$ P\Bigl(\bigcap_{n = 1}^{k}\Bigl\{W_n \leq\sqrt{n}\Bigr\}\Bigr)\leq h(k) \mbox{ ?} $$ where $\lim_{k\to\infty} h(k) =0$, because I manage to prove it using zero-one Kolmogorov's law and the fact that $\frac{W_n}{\sqrt{n}}$ converges in distribution to a normal random variable.

I know that it may be a really simple question, although I just can not find the right bound.. I have tried lots of different techniques but none of them seem to work properly. I think that a useful fact is the simmetry ( since $\forall n\in\mathbb{N},W_n\sim-W_n$ ) and the fact that $W_n$ has zero mean and (in general) bounded moments.