Convergence in distribution to Gaussian random variables If $X_n$ is a sequence of random variables such that $X_n\stackrel{d}{\to}N(0,1)$, where $N(0,1)$ is a standard Gaussian random variable, is it true that $\frac{X_n}{n^\alpha}\stackrel{P}{\to}0$ for every $\alpha>0$?
 A: Yes, and this would be also true if the limiting distribution was not Gaussian, say a real-valued random variable $X$.
Indeed, let $R>0$ be a point of continuity of the cumulative distribution function of $\lvert X\rvert$. Fix $\varepsilon>0$. Since for $n$ large enough, $n^\alpha \varepsilon>R$,
$$
\mathbb P\left(\lvert X_n\rvert>\varepsilon n^\alpha\right)\leqslant \mathbb P\left(\lvert X_n\rvert>R\right)
$$
and taking the $\limsup_{n\to+\infty}$ gives
$$
\limsup_{n\to+\infty}\mathbb P\left(\lvert X_n\rvert>\varepsilon n^\alpha\right)\leqslant \mathbb P\left(\lvert X\rvert>R\right)\tag{*}.
$$
Since $R$ is arbitrary and since we can always find a sequence $(R_k)$ of positive numbers which are continuity points of the cumulative distribution function of $\lvert X\rvert$ such that $R_k\to +\infty$, we conclude that $\mathbb P\left(\lvert X_n\rvert>\varepsilon n^\alpha\right)\to 0$.
Note that when the limiting distribution function is continuous, the proof is a lit bit simpler since we can simply choose any $R>0$ and let $R\to+\infty$ in (*).
