Suppose $W_t$ is a standar brownian motion. Call
$$ X_t = W_{t+\Delta}-W_t $$
the variation of $W$ in an interval of length $\Delta$. Call $$ \sigma = \sqrt{2}\,\Delta $$
the standard deviation of $X_t^2$. I want to approximate
$$ X_t^2 =\mathbb{E}\left[X_t^2\right]+O_p\left(\sigma^\alpha\right) =\Delta+O_p\left(\sigma^\alpha\right) $$
for a suitable choice of $\alpha$. Chebyshev's inequality immediately delivers $\alpha=1$, however Chebyshev's inequality it is generic and does not exploit the fact that $X_t^2-\Delta$ has a normal distribution. In fact
$$ \mathbb{P}\left[\left|X_t^2-\Delta\right|\geq M\,\sigma^{\alpha}\right] =\text{erfc}\left(\frac{M\,\sigma^{\alpha-1}}{\sqrt{2}}\right) $$
and, since for large $M$ the function $\text{erfc}\left(\frac{M\,\sigma^{\alpha-1}}{\sqrt{2}}\right)$ goes to zero I would conclude that
$$ (W_{t+\Delta}-W_t)^2 = \Delta+O_p\left(\Delta^\alpha\right) $$
for any $\alpha>1$. Is there some flaw in my logic?