\Pr\left( \limsup_n \frac{S_n}{\sqrt{n}} > M \right) \geqslant \limsup_n \Pr\left( \frac{S_n}{\sqrt{n}} > M \right) = \Pr\left( \mathcal{N}(0, 1) > M \right) > 0 .
\limsup_n \frac{S_n}{\sqrt{n}}=\infty \qquad \text{with probability 1.} .