The law of iterated logarithm states that for a random walk $$S_n = X_1 + X_2 + ... X_n$$ with $X_i$ independent random variables such that $P(X_i = 1) = P(X_i = -1) = 1/2$, we have
$$\limsup_{n \rightarrow \infty} S_n / \sqrt{2 n \log \log n} = 1, \qquad \rm{a.s.}.$$ Similarly for Brownian motion $B(t)$ the law of iterated logarithm says $$\limsup_{t \rightarrow \infty} B(t)/ \sqrt{2 t \log \log t} = 1, \qquad \rm{a.s.}.$$ However the central limit theorem for brownian motion says $$\frac{S_n}{\sqrt{n}} \to B(t).$$ So how can we have the same scalling in the law of iterated logarithm?