Convergence in Distribution to Normal Limit Implies Convergence in If $\sqrt{n} (X_n - \beta) \overset{d}{\to} N(0, \Sigma)$ is it the case that $X_n \overset{p}{\to} \beta$? 
 A: 
It is.

Fix some $x\gt0$ and, for every $n$, let $A_n=[\|X_n-\beta\|\geqslant x]$. Then the task is to show that 
$$
\lim\limits_{n\to\infty}\mathbb P(A_n)=0.
$$
For every $t\geqslant0$, there exists some finite $N_t$ such that $A_n\subseteq [\sqrt{n}\|X_n-\beta\|\geqslant t]$ for every $n\geqslant N_t$ (choose any $N_t\geqslant (t/x)^2$). Hence $\mathbb P(A_n)\leqslant\mathbb P(\sqrt{n}\|X_n-\beta\|\geqslant t)$ for every $n\geqslant N_t$. 
By hypothesis, $\mathbb P(\sqrt{n}\|X_n-\beta\|\geqslant t)\to\mathbb P(\|Y\|\geqslant t)$ when $n\to\infty$, where $Y$ is normal, hence $\limsup\limits_{n\to\infty}\mathbb P(A_n)\leqslant\mathbb P(\|Y\|\geqslant t)$. Since this inequality holds for every $t\geqslant0$ and $\inf\limits_{t\geqslant0}\mathbb P(\|Y\|\geqslant t)=0$, one sees that $\limsup\limits_{n\to\infty}\mathbb P(A_n)\leqslant0$, that is, $\lim\limits_{n\to\infty}\mathbb P(A_n)=0$.
This proves that $\lim\limits_{n\to\infty}\mathbb P(\|X_n-\beta\|\geqslant x)=0$, for every $x\gt0$. Hence $X_n\to\beta$ in probability.
