Brownian motion nowhere differentiable

Question

I'm reconstructing a proof that the Brownian Motion is nowhere differentiable. They chose $3$ points and because the increments are independent and $\sim \mathcal{N}(0,1/n)$ they take instead a random variable $Z\sim \mathcal{N}(0,1)$. The equality is $$P\left(\bigcap_{j=1}^3\{|W_{\frac{k+j}{n}}-W_{\frac{k+j-1}{n}}|\leq \frac{C}{n}\}\right)=\left[P\left(|Z|\leq \frac{C}{\sqrt{n}}\right)\right]^3\leq \frac{C}{n^{3/2}}$$ Why is there a $\sqrt{n}$ in the middle and not an $n$?