$\sqrt{n}(X_n-x)\to\mathcal{N}(0,1)$ implies $X_n\to x$ almost surely Let $(X_n)$ be a sequence of real random variables and $x$ a real number such that $\sqrt{n}(X_n-x)$ converges in law to a Gaussian variable $\mathcal{N}(0,1)$ as $n\to\infty$. 
What is the strongest convergence result of $(X_n)$ to $x$ that can be obtained? Can we prove that $(X_n)$ converges almost surely to $x$?
 A: No, in general. The strongest result is that $X_n$ converges to $x$ in probability. Or, equivalently, that there exists a subsequience $n_i$ s.t. $X_{n_i}$ converges to $x$ a.s. 
Let, for instance, $x=0$ and $Y_n\sim \mathcal N(0,1/n)$, so that $\sqrt{n}Y_n\sim\mathcal N(0,1)$. And let $Z_n$ be a sequence of independent Bernoulli r.v.'s such that $\mathbb P(Z_n=1)=1/n=1-\mathbb P(Z_n=0)$. Let also $Y_n$ and $Z_n$ be independent. Then $Z_n$ converges to zero in probability but not a.s., as follows from Borel-Cantelly lemma. 
And $\sqrt{n}Z_n$ also converges to zero in probability and whence weakly, so 
$$\sqrt{X_n}=\sqrt{n}(Y_n+Z_n) \xrightarrow{\mathcal D}\mathcal N(0,1)$$
as a sum of two independent weakly convergent sequences.
But $X_n=Y_n+Z_n$ does not converges a.s. to zero: for $0<\varepsilon<1$,
$$
\mathbb P(|X_n|>\varepsilon)=\mathbb P(|Y_n+Z_n|>\varepsilon)=\frac1n\mathbb P(|Y_n+1|>\varepsilon)+\left(1-\frac1n\right)\mathbb P(|Y_n|>\varepsilon)\geq\frac1n\mathbb P(|Y_n+1|>\varepsilon)=\frac1n\cdot \alpha(n,\varepsilon)
$$
where $\alpha(n,\varepsilon)\to 1$ as $n\to\infty$, so the sum $\sum_{n=1}^\infty \mathbb P(|X_n|>\varepsilon)$ diverges. Then by Borel-Cantelly lemma, the events $\{|X_n|\geq \varepsilon\}$ appear a.s. infinitely often and then $X_n$ does not converge a.s. to zero. 
