First of all the question in the topic heading is not answerable without further conditions, but I have a specific example in mind.
Let $X_i$ have densities
\begin{equation} f(t):=\frac{\sigma-1}{2}\mathfrak{1}_{\{|t|\geq 1\}}|t|^{-\sigma} \end{equation} with respect to the Lebesgue measure. Let $F_X{a}:=\mathbb{P}(X\leq a)$ be the distribution function for the random variable $X$. Also let $Y_n=S_n/n:=\frac{1}{n}\sum_k X_k$ Then I wish to calculate $\lim_{n\to\infty}F_{Y_n}(t)$ for $\sigma>2$ and $t\neq 0$ as a function of t. Also what can I say about $t=0$?
Now it is easy to see that the finiteness of $\mathbb{E}|X_i^n|$ depends on $\sigma$ and it is an easy calculation that for $\sigma>2$ we have $\mathbb{E}X_i=0$. Now I should be somehow utilizing the weak law of large numbers which in this case states that \begin{equation} \mathbb{P}(|Y_n|>\epsilon)\to0 \end{equation} as $n\to\infty$ for any $\epsilon>0$, but I don't know how to utilize this.
It is easy to see that $E|X_i| <\infty$ and $EX_i=0$. By SLLN, $Y_n \to 0$ almost surely (or, by WLLN, $Y_n$ converges in probability), hence also in distribution. So $P(Y_n \leq t ) \to 1$ for $t >0$ and $0$ for $t <0$.
Since $Y_n$ has a symmetric continuous distribution it follows that $P(Y_n\leq 0)=\frac 1 2$ for all $n$ so the limit is also $\frac 1 2$