This feels like a very basic question, but I wasn't able to find an answer. Apologies if it is already answered and/or trivial...
Assuming that a CLT holds for some sequence $X_n$, i.e. $\sqrt{n}(X_n-\mu)\rightarrow_d N(0,\sigma^2)$, then what can be said about the convergence of $X_n$ towards $\mu$, i.e. if we knock off the $\sqrt{n}$ factor? I know that it converges in probability (e.g. here) , but what about $L_p$ or almost sure convergence? If this does not hold, could someone give a counter example?
We can consider the following example: $$\tag{*}X_n=\mu+\frac 1{\sqrt n}N+\frac 1{\sqrt n}Y_n,$$ where $N$ follows a centered normal distribution with variance $\sigma^2$ and the sequence $\left(Y_n\right)_{n\geqslant 1}$ converges to $0$ in probability. Let $$ Y_n:=c_n\sqrt n\mathbf 1_{A_n} $$ where $c_n\geqslant 1$, $\left(A_n\right)_{n\geqslant 1}$ is independent, $\mathbb P\left(A_n\right)\to 0$ and $\sum_{n=1}^{+\infty}\mathbb P\left(A_n\right)=+\infty$. Then the sequence $\left(Y_n/\sqrt n\right)_{n\geqslant 1}$ does not converge to $0$ almost surely and since $N/\sqrt n\to 0$ almost surely, we infer by (*) that the sequence $\left(X_n-\mu\right)_{n\geqslant 1}$ does not converge to $0$ almost surely. If we choose $c_n$ big enough, we may not have the convergence in $\mathbb L^p$.