By the central limit theorem we have that for a an iid sequence $X_i$ with mean $\mu$ and variance $\sigma^2$ that,
$$\sqrt{n}\frac{\overline{X}_n-\mu}{\sigma}\rightarrow_d N(0,1)$$ as $n\rightarrow\infty$
But this also implies that for sufficiently large $n$ we have approximately,
$$ \overline{X}_n \sim N(\mu,\sigma^2/n) $$
And as $n \rightarrow\infty$ we have $\overline{X}_n \sim N(\mu,0)=\mu$
So we have $P(\lim_{n\rightarrow{\infty}}\overline{X}_n=\mu)=1$. Which would mean that the CLT implies the LLN. This reasoning is probably false since the convergence for the CLT is weak. But still it seems plausible since the limiting distribution will just be a constant. Can someone point out what is wrong with this reasoning?
If $\sqrt{n}(\bar{X}_n-\mu)\xrightarrow{d}N(0, \sigma^2)$, then for any $\epsilon>0$, $$ \mathsf{P}(|\bar{X}_n-\mu|> \epsilon)=\mathsf{P}(|\sqrt{n}(\bar{X}_n-\mu)|> \sqrt{n}\epsilon)\to 0. $$ as $n\to\infty$. That is, $\bar{X}_n\xrightarrow{p}\mu$.
As for the a.s. convergence assume, in addition, that $v_3:=\mathsf{E}|X_1|^3<\infty$. Then, using a non-uniform Berry-Esseen bound for i.i.d. sequences, \begin{align} \sum_{n\ge 1}\mathsf{P}(|\sqrt{n}(\bar{X}_n-\mu)|> \sqrt{n}\epsilon)&\le 2\sum_{n\ge 1}\bar{\Phi}\!\left(\sqrt{n}\epsilon/\sigma\right)+\frac{Cv_3}{(1+(\sqrt{n}\epsilon/\sigma)^3)\sqrt{n}} \\ &\le 2\int_0^{\infty}\bar{\Phi}\!\left(\sqrt{x}\epsilon/\sigma\right)+\frac{Cv_3}{(1+(\sqrt{x}\epsilon/\sigma)^3)\sqrt{x}}dx \\ &=\frac{\sqrt{2\pi}\sigma^2}{\epsilon^2}+\frac{C8\sqrt{3}\pi \sigma v_3}{\epsilon}<\infty, \end{align} where $C>0$ is a constant. That is, $\mathsf{P}(|\bar{X}_n-\mu|> \epsilon\text{ i.o.})=0$.