A converse to the law of the iterated logarithm for random walk

Let $\{S_n\}$ be a random walk with independent increments. Then $\mathbf{E}S_1=0$, $\mathbf{E}S_1^2=1$ iff
$$ \limsup_{n\to\infty}\frac{S_n}{\sqrt{2n\log\log n}}=1\qquad\text{almost surely.} $$