# $\sqrt{n}(x_n-x_0) \to N(0,\sigma_0^2)$ in distribution implies $x_n \to x_0$ almost surely

Is this true? I figured that if not, there will some positive probability $$\sigma$$ that $$\sqrt{n}(x_n-x_0)$$ takes $$\sqrt{M} \cdot \epsilon$$ for infinitely many large $$M$$. Even though this "blowing up" is weird, it does not appear to be a contradiction since we don't know how frequent of such event. That is, it could still converge to $$N(0,\sigma_0^2)$$ in distribution with it infinitely often taking some larger and larger values. How to write a formal proof of this(suppose it's true)?